Exercises I
1. sin4θ-sin2θ=sin6θ
⇒ 2sinθ.cos3θ =sin6θ
⇒ 2sinθ.cos3θ = 2sin3θ.cos3θ
⇒ 2cos3θ(sinθ-sin3θ) = 0
⇒ cos3θ = 0
⇒ 3θ = 2nπ±π½
⇒ θ = 1/3[2nπ±π½]
sinθ - sin3θ = 0
⇒sinθ - [3sinθ - 4sin³θ] = 0
⇒sinθ - 3sinθ + 4sin³θ = 0
⇒4sin³θ - 2sinθ = 0
⇒2sinθ[2sin²θ - 1] = 0
⇒sinθ = 0 or cos2θ = 0
⇒θ = nπ or cos2θ = 0
⇒θ = nπ or 2θ = 2nπ ± π½
⇒θ = nπ or θ = nπ ± π¼
Therefore θ = 1/3[2nπ ± π/2] or nπ or nπ ± π¼
2. cos3θ + cos2θ + cosθ = 0
⇒ cos3θ + cosθ + cos2θ = 0
⇒ 2cos2θ.cosθ + cos2θ = 0
⇒ cos2θ[2cosθ + 1] = 0
⇒ cos2θ = 0 or 2cosθ + 1 = 0
Now cos2θ = 0 ⇒ 2θ = 2nπ ± π½ ⇒θ = nπ ± π¼
2cosθ + 1 = 0 ⇒ 2cosθ = -1 ⇒ cosθ = -½
⇒ cosθ = cos(π-π/3)
⇒ cosθ = cos(2π/3)
⇒ θ = 2nπ ± 2π/3
∴ θ = nπ ± π/4 or 2nπ ± 2π/3
3. cos2θ + 5cosθ = 2
⇒ 2cos²θ - 1 + 5cosθ -2 = 0
⇒ 2cos²θ + 5cosθ - 3 = 0
⇒ cosθ = (-5 ± √49)/4 = (-5±7)/4
⇒ cosθ = 1/2 or -3
But -3 is numerically > 1.
Hence there is no real value of θ.
∴ cosθ = 1/2 ⇒ θ = 2nπ ± π/3
4. cos3x - cos4x = cos5x - cos6x
⇒ 2sin(7x/2).sin(x/2) = 2 sin(11x/2).sin(x/2)
⇒ sin(x/2)[sin(7x/2) - sin(11x/2)] = 0
⇒ sin(x/2) = 0 ⇒x/2 = nπ ⇒ x = 2nπ
or sin(7x/2) - sin(11x/2) = 0
⇒ 2cos(9x/2).sin(-x) = 0
⇒ -2cos(9x/2).sinx = 0
⇒ cos(9x/2) = 0 ⇒9x/2 = 2nπ ± π/2 ⇒ x = (4nπ ± π)/9
sinx=0 ⇒ x = nπ
∴ x = 2nπ or nπ or (4nπ ± π)/9
5. sin2θ - sinθ = cos2θ - cosθ
⇒ 2cos(3θ/2).sin(θ/2) = 2sin(3θ/2).sin(-θ/2)
⇒ cos(3θ/2).sin(θ/2) = -sin(3θ/2).sin(θ/2)
⇒ sin(θ/2).[cos(3θ/2) + sin(3θ/2)] = 0
⇒ sin(θ/2) = 0 or cos(3θ/2) + sin(3θ/2) = 0
sin(θ/2) = 0 ⇒ θ/2 = nπ ⇒ θ = 2nπ
cos(3θ/2) + sin(3θ/2) = 0
⇒ cos(3θ/2) + cos(π/2-3θ/2) = 0
⇒ 2 cos (π/4). cos(3θ/2-π/4) = 0
⇒ (1/√2).cos(3θ/2 - π/4) = 0
⇒ cos(3θ/2 - π/4) = 0
⇒ sin(π/2 - (3θ/2 - π/4)) = 0
⇒ π/2 - 3θ/2 + π/4 = nπ
⇒ 3π/4 - 3θ/2 = nπ
⇒ θ = π/2 - 2nπ/3
∴ θ = 2nπ or π/2 - 2nπ/3
6. 2sin²x + 3cosx - 3 = 0
⇒ 2(1-cos²x) + 3 cosx - 3 = 0
⇒ 2 -2cos²x + 3 cosx - 3 = 0
⇒ -2cos²x + 3cosx - 1 = 0
⇒ 2cos²x - 3cosx + 1 = 0
⇒ cosx = (3±1)/4 = 1 or ½
cosx = 1 ⇒ x = 2nπ
cosx = ½ ⇒ x = 2nπ ± π/3
∴ x = 2nπ or 2nπ ± π/3
7. sinx + cos2x = 0.2
⇒ sinx + 1 - 2sin²x = 0.2
⇒ -2sin²x + sinx + 1 - 0.2 = 0
⇒ -2sin²x + sinx + 0.8 = 0
⇒ 2sin²x - sinx - 0.8 = 0
⇒ 20sin²x - 10sinx - 8 = 0 (multiplying by 10)
⇒ 10sin²x - 5sinx - 4 = 0 (dividing by 2)
⇒ sinx = (5 ± √185)/20 = 0.93 or -0.43
⇒ sinx = 0.93 = 68°26′
⇒ x = (-1)ⁿ.68°26′ + nπ
sinx = -0.43 = 25°28′ ⇒ x = (-1)^(n+1).25°28′ + nπ
∴ x = (-1)^n.68°26′ + n180° or x = (-1)^(n+1).25°28′ + n180°
8. 2cos3θ + 6cosθ + 1 = 0
⇒ 2cos3θ + 6cosθ = -1
⇒ cos3θ + 3cosθ = -½
⇒ 4cos³θ - 3cosθ + 3cosθ = -½
⇒ 4cos³θ = -½
⇒ cos³θ = -1/8
⇒ cosθ = -½ = cos(2π/3)
⇒ θ = 2nπ ± 2π/3
9. 2sinx + cosecx = 3
⇒ 2sin²x + 1 = 3sinx [multiplying both sides by sinx]
⇒ 2sin²x - 3sinx + 1 = 0
⇒ sinx = (3±1)/4 = 1 or ½
⇒ x = (-1)ⁿπ/2 + nπ or (-1)ⁿπ/6 + nπ
10. 5sin²x + 3sinx - 2 = 0
⇒ sinx = (-3 ±√49)/10 = -1 or 0.4
sinx = -1 ⇒ x = (-1)^(n+1).π½ + nπ or sinx = (-1)^n.23° 34′ + n180°
11. sinθ + sin2θ = sin3θ
⇒ sinθ + 2sinθ.cosθ = 3sinθ - 4sin³θ
⇒ 3sinθ - 4sin³θ - sinθ - 2sinθ.cosθ = 0
⇒ -4sin³θ + 2sinθ -2sinθ.cosθ = 0
⇒ 4sin³θ - 2sinθ + 2sinθ.cosθ = 0
⇒ 2sin³θ - sinθ + sinθ.cosθ = 0
⇒ sinθ[2sin²θ - 1 + cosθ] = 0
⇒ sinθ = 0 or 2sin²θ - 1 + cosθ = 0
sinθ = 0 ⇒ θ = nπ
2sin²θ - 1 + cosθ = 0
⇒ 2(1 - cos²θ) - 1 + cosθ = 0
⇒ 2 -2cos²θ - 1 + cosθ = 0
⇒ -2cos²θ + cosθ + 1 = 0
⇒ 2cos²θ - cosθ - 1 = 0
⇒ cosθ = (1 ± 3)/4 = 1 or -½
⇒ cosθ = 1 ⇒ θ = 2nπ
cosθ = -½ ⇒ θ = 2nπ ± 2π/3
Therefore θ = nπ or 2nπ or 2nπ ± 2π/3
12. cos2θ + 5cosθ = 2
⇒ 2cos²θ - 1 + 5cosθ - 2 = 0
⇒ 2cos²θ + 5cosθ - 3 = 0
⇒ cosθ = (-5±7)/4 = ½ or -3
But -3 is numerically > 1. Hence there is no real value of θ.
cosθ = ½ ⇒ θ = 2nπ ± π/3
13. 3sinx + 4cosx = 2
Let rcosα = 3 and rsinα = 4 ⇒ r² = 25 ⇒ r = 5
cosα = 3/5 and sinα = 4/5 ⇒ α = 53°7′
The equation takes the form rcosα.sinx + rsinα.cosx = 2
⇒ r[sinx.cosα + cosx.sinα] = 2
⇒ 5.sin(x + α) = 2 ⇒ sin(x + α) = 2/5 = 0.4
⇒ sin(x + 53°7′) = 0.4
Let θ = x + 53°7′ ⇒ sinθ = 0.4 ⇒ θ = (-1)^n.23°34′ + n180°
⇒ x + 53°7′ = (-1)^n.23°34′ + n.180°
14. 2sinx + 5cosx = 3
Let rcosα = 2 and rsinα = 5 ⇒ r² = 4 + 25 = 29 ⇒ r = 5.385164807
therefore α = 68°11′
∴ The equation takes the form rcosα.sinx + rsinα.cosx = 3
⇒ r[sinx.cosα + cosx.sinα] = 3
⇒ r.sin(x + α) = 3
⇒ sin(x + α) = 3/r =3/5.385164807
⇒ x + α = (-1)^n.33°51′ n.180°
⇒ x + 68°11′ = (-1)^n.33°.51′ + n.180°
15. sinx + √3.cosx = 1
½.sinx + ½√3.cosx = ½ (multiplying both sides by ½)
⇒ sin(π/6).sinx + cos(π/6).cosx = ½
⇒ cos(x - π/6) = ½
⇒ x - π/6 = 2nπ ± π/3
⇒ x = 2nπ + π/3 + π/6 = 2nπ + π/2 or
x = 2nπ - π/3 + π/6 = 2nπ - π/6
16. 2cosx - 3sinx = 2
⇒ 2cosx - 3√1-cos²x = 2
⇒ -3√1-cos²x = 2 - 2cosx
Squaring, 9(1-cos²x) = 4 + 4cos²x - 8cosx
⇒ 9 - 9cos²x - 4 - 4cos²x + 8cosx = 0
⇒ -13cos²x + 8cosx + 5 = 0
⇒ 13cos²x - 8cosx - 5 = 0
⇒ cosx = (8±18)/26 = (8+18)/26 =1 ⇒x = 2nπ
or cosx = -10/26 = -5/13 = -0.39 ⇒ x = ±67°2′ + n360°
∴ x = nπ or n360° ± 67°2′
17. 5cosx - 4sinx = 6
⇒ Let rcosα = 5 and rsinα = 4
∴ r² = 41 ⇒ r = 6.4
∴ α = 38°39′
The given equation takes the form rcosα.cosx - rsinα.sinx = 6
⇒ r[cosα.cosx - sinα.sinx] =6
⇒ 6.4cos(α + x) = 6 ⇒ cos(α+ x) = 6/6.4 = 0.9375
⇒ cos(38°39′ + x) = 0.9375
Let θ = x + 38°39′ ⇒ cosθ = 0.9375 ⇒ θ = 2nπ ± 20°21′
∴ x + 38°39′ =n.360° ± 20°21′
18. cosx + sinx = 2cos2x
⇒ cosx + sinx = 2(cos²x - sin²x) = 2(cosx - sinx)(cosx + sinx)
⇒ cosx + sinx - 2(cosx - sinx)(cosx + sinx) = 0
⇒ (cosx + sinx)[1 -2(cosx - sinx)] = 0
⇒ cosx + sinx = 0 ⇒1/√2.cosx + 1/√2.sinx =0 ⇒ sin(π/4).cosx + cos(π/4).sinx = 0
⇒ sin(π/4 + x) = 0 ⇒ π/4 + x = nπ ⇒ x = nπ - π/4
(or) 1 - 2cosx + 2sinx = 0
⇒ -2cosx + 2sinx = -1
⇒ 2cosx - 2sinx = 1
⇒ cosx - sinx = ½
Let rcosα = 1 and rsinα = -1
∴ tanα = -1 and α = -π/4, r = √2 = 1.414
∴ rcosα.cosx - rsinα.sinx = ½
⇒ rcos(α + x) = ½
⇒ √2.cos(x - π/4) = ½
⇒ cos(x - π/4) = 1/2√2 = 0.35
⇒ x - π/4 = 2nπ ± 69°19′
⇒ x = 2nπ ± 69°19′ + π/4
∴ x = nπ - π/4 or 2nπ ± 69°19′ + π/4
19. tanx + cotx = 3
⇒sinx/cosx + cosx/sinx = 3
⇒ (sin²x + cos²x)/sinx.cosx = 3
⇒ 1/sinx.cosx = 3
⇒ 2/2sinx.cosx = 3
⇒ 2/sin2x = 3
⇒ sin2x = ⅔ = 0.67
∴ x = 20°54′ + nπ
20. 1 + sin²θ = 3sinθ.cosθ
(1 + sin²θ)² = 9sin²θ.cos²θ (squaring both sides)
⇒ 1 + sin⁴θ + 2sin²θ = 9sin²θ.cos²θ
⇒ 1 + sin⁴θ + 2sin²θ = 9sin²θ.(1 - sin²θ)
⇒ 1 + sin⁴θ + 2sin²θ = 9sin²θ - 9sin⁴θ
⇒ 1 + sin⁴θ + 2sin²θ - 9sin²θ + 9sin⁴θ = 0
⇒ 1 + 10sin⁴θ _ 7sin²θ = 0
Let t= sin²θ . ∴ t² = sin⁴θ
∴ 1 + 10t² - 7t = 0
⇒ 10t² -5t -2t + 1 = 0
⇒ 5t(2t - 1) -1(2t - 1) = 0
⇒ (2t - 1)(5t - 1) = 0
⇒ t = ½ or t = ⅕
∴ sin²θ = ½ ⇒ sinθ = 1/√2 = 0.707213578
∴ θ = n.180° + 45°
sin²θ = ⅕ ⇒ sinθ = 1/√5 = 0.447227191 ⇒ θ = 26.56 ⇒ θ = n.180° + 26°34′
21. tanx + tan2x + tan3x = 0
Let t = tanx
∴ t + 2t/(1 - t²) + (3t - t³)/(1 - 3t²) = 0
⇒ [t(1 - t²)(1 - 3t²) + 2t(1 - 3t²) + (3t - t³)(1 - t²)]/(1 - t²)(1 - 3t²) = 0
⇒ t(1 - t²)((1 - 3t²) + 2t - 6t³ + 3t - 3t³ - t³ + t⁵ = 0
⇒ t - 3t³ - t³ + 3t⁵ + t⁵ - 10t³ + 5t = 0
⇒ 4t⁵ - 14t³ + 6t = 0
⇒ 2t⁵ - 7t³ + 3t = 0
⇒ t(2t⁴ - 7t² + 3) = 0
⇒ t = 0 (or) 2t⁴ - 7t² + 3 = 0
put t² = k ∴ 2k² - 7k + 3 = 0
⇒ k = (7±5)/4 = 3 or ½
∴ t² = 3 or t² = ½ = 0.5
⇒ t = √3 or t = 0.707106781
⇒ tanx = √3 or tanx = 0.707106781
⇒ x = 60° ± n.180° or x = 35°15′ ± n.180°
22. tanθ = 1 - sec2θ
= 1 - (1 + tan²θ)/(1 - tan²θ) ∵ cos2θ = (1 - tan²θ)/(1 + tan²θ)
= (-2tan²θ)/(1 - tan²θ)
tanθ - tan³θ = -2tan²θ
-tan³θ + 2tan²θ + tanθ = 0
tan³θ -2tan²θ - tanθ = 0
tanθ[tan²θ - 2tanθ - 1] = 0
tanθ = 0 ⇒ θ = n.180°
tan²θ - 2tanθ - 1 = 0
⇒ tanθ = 1 ± √2 ⇒ tanθ = 2.4144 or 0.4144
⇒ θ = 67°30′+ n180° or θ = 22°30′ + n180°
23. cosec⁴θ + 9.cosec²θ -6cosec²θ.cotθ = 10
cosec⁴θ = (cosec²θ)² = (1 + cot²θ)² = 1 + cot⁴θ + 2cot²θ
9 cosec²θ = 9 + 9cot²θ
6.cosec²θ.cotθ = 6cotθ + 6cot³θ
Gn eqn.becomes
1 + cot⁴θ + 2cot²θ + 9 + 9cot²θ - 6cotθ - 6cot³θ = 10
⇒ cot⁴θ - 6cot³θ + 11cot²θ - 6cotθ = 0
⇒ cotθ[cot³θ - 6cot²θ + 11cotθ -6] = 0
⇒ cotθ = 0 ⇒ θ =n180° + 90
Let Z = cotθ
The above eqn becomes z³ - 6z²+ 11z - 6 = 0
⇒(z-1)(z² -5z + 6) = 0 ⇒z = 1,3,2
z = 1 ⇒ cotθ = 1 ⇒ tanθ = 1 ⇒θ = n180° + 45
z = 3 ⇒ cotθ = 3 ⇒ tanθ = ⅓ ⇒ θ = 18.43494865 ⇒θ = n180° + 18°26′
z = 2 ⇒ cotθ = 2 ⇒ tanθ = ½ ⇒ θ = 26.56505118 ⇒ θ = 26°33′ + n180°
24. tanA = -cot2A
⇒ sinA/cosA = -cos2A/sin2A
⇒ sinA.sin2A = -cosA.cos2A
⇒ cos2A.cosA + sin2A.sinA = 0
⇒ cos(2A - A) = 0 ⇒ cosA = 0 ⇒ A =2nπ ± π½
25 tanθ + 4cos2θ + 1 = 0
⇒ tanθ + 4(1 - tan²θ)/(1 + tan²θ) + 1 = 0
⇒ tanθ( 1 + tan²θ) + 4(1-tan²θ) + 1 + tan²θ = 0
⇒ tanθ + tan³θ + 4 - 4tan²θ + 1 + tan²θ = 0
⇒ tan³θ - 3.tan²θ + tanθ + 5 = 0
⇒ (tanθ + 1)(tan²θ - 4tanθ + 5) = 0
⇒ tanθ = -1 ⇒ θ = n.180° - 45
tan²θ - 4tanθ + 5 = 0
⇒ tanθ = (4±2i)/2 = 2 ± i
considering the real value
tanθ = 2 ⇒ θ = tan⁻2 = 63.43494882 = 63°26′
∴ θ = n.180° + 63°26′
26. √3.(tanθ + secθ) = 4
⇒ secθ + tanθ = 4/√3 —(1)
Multiplying Nr and Dr by secθ - tanθ
(sec²θ - tan²θ)/(secθ - tanθ) = 4/√3
⇒ 1/(secθ - tanθ) = 4/√3
⇒ secθ - tanθ = √3/4 — (2)
(1) - (2) ⇒ 2tanθ = 4/√3 - √3/4 = 13/4√3
⇒ tanθ = 13/8√3 = 0938194187
⇒ θ = 43.1735511 + nπ
⇒ θ = 43° 10′ + n180°
27 Sec²θ - (√3 + 1)tanθ + √3 - 1 = 0
⇒ 1 + tan²θ - √3tanθ - tanθ + √3 - 1 = 0
⇒ tan²θ - tanθ[√3 + 1] + √3 = 0
⇒ tanθ = (√3 + 1 + sqrt(4-2√3))/2 = 1.732050807
⇒ θ = nπ + π/3
Also tanθ = (√3 + 1 - sqrt(4-2√3))/2 = 1
⇒ θ = nπ + π/4
28. tanpx = cotqx
⇒ tanpx = tan(π/2 - qx)
⇒ px = π/2 - qx + nπ
⇒ px + qx = π/2 + nπ
⇒ x(p + q) = π(2n + 1)/2
⇒ x = π(2n + 1)/2(p + q)
29. If cos(A - B) = 1/2 and sin(A + B) = 1/2, find the smallest positive values of A and B and their most general values.
Solution : cos(A - B) = 1/2 ⇒ A - B = 60°
sin(A + B) = 1/2 ⇒ A + B = 30° or 150°
A - B = 60°
A + B = 30°
⇒ A = 45°
When A = 45, B =-15 not +ve.
∴ A - B = 60
A + B = 150
⇒ A = 105
∴ B = 45
General value of A - B = 2nπ ± π/3––(1)
General value of A + B = (-1)^m.π/6 + mπ - - (2)
(1) + (2) ⇒ A = nπ ± π/6 + (-1)^m.π/12 + mπ/2
⇒ A = π[n ± 1/6 + (-1)^m/12 + m/2]
(1) - (2) ⇒ B = π[-n ± 1/6 + (-1)^m/12 + m/2]
30. Explain why the same series of two angles are given by the equations.
θ + π/4 = nπ + (-1)^n.π/6 and θ - π/4 = 2nπ ± π/3
Solution : Given θ + π/4 = nπ + (-1)^n.π/6 — (a)
sin(θ + π/4) = sin(nπ + (-1)^n.π/6) = ½
⇒sinθ.cosπ/4 + cosθ.sinπ/4 = ½
⇒ 1/√2 . cosπ/4 + 1/√2.cosθ = ½
⇒ sinθ + cosθ = 1/√2
Also given that θ - π/4 = 2nπ ± π/3 —(b)
cos(θ - π/4) = cos(2nπ ± π/3)
⇒ cos(θ - π/4) = ½
⇒ cosθ.cosπ/4 + sinθ.sinπ/4 = ½
⇒ cosθ.1/√2 + sinθ.1/√2 = ½
⇒ cosθ + sinθ = 1/√2
Equations (a) and (b) are general solution to the equation cosθ + sinθ = 1/√2.
The general solution can frequently be obtained in many ways. The various
forms which the result takes are merely different modes of expressing the same series of angles.
Exercises IX
(1)
(i) sin6θ
cos6θ + isin6θ = (cosθ+isinθ)6
= cos6θ + 6c1 cos5θ.(isinθ) + 6c2.cos4θ.(isinθ)2 + 6c3.cos3θ(isinθ)3
+ 6c4.cos2θ.(isinθ)4 + 6c5.cosθ.(isinθ)5 + 6c6.(isinθ)6
= cos6θ + 6i.cos5θ.sinθ - 15.cos4θ.sin2θ-20i.cos3θ.sin3θ + 15.cos2θ.sin4θ + 6i.cosθ.sin5θ - sin6θ
Equating the imaginary parts, we have
sin6θ = 6.cos5θ.sinθ - 20.cos3θ.sin3θ + 6.cosθ.sin5θ
= sinθ.(6.cos5θ - 20.cos3θ.sin2θ + 6.cosθ.sin4θ)
= sinθ.[6.cos5θ - 20.cos3θ.(1 - cos2θ) + 6.cosθ.(sin2θ)2]
= sinθ.[6.cos5θ - 20.cos3θ +20.cos5 + 6.cosθ.(1 - cos2θ)2]
= sinθ.[26.cos5θ - 20.cos3θ + 6.cosθ.(1 + cos4θ - 2.cos2θ)
= sinθ.[26.cos5θ - 20.cos3θ + 6.cosθ + 6.cos5θ - 12.cos3θ]
= sinθ.[32.cos5θ - 32.cos3θ + 6.cosθ]
(ii) cos5θ
cos5θ + isin5θ = (cosθ + isinθ)5 = cos5θ + 5c1.cos4θ.(isinθ) + 5c2.cos3θ.(isinθ)2 + 5c3.cos2θ.(isinθ)3 + 5c4.cosθ.(isinθ)4 + 5c5.(isinθ)5
= cos5θ + 5.cos4
Equating the real parts, we have
cos5θ = cos
= cos
= cos
= 11.cos
= 11.cos
= 16.cos
(iii) cos9θ
cos9θ + isin9θ = (cosθ + isinθ)9
= cos9θ + 9c1.cos8θ.(isinθ) + 9c2.cos7θ.(isinθ)2 + 9c3.cos6θ(isinθ)3
+ 9c4.cos5θ.(isinθ)4 + 9c5.cos4θ.(isinθ)5 + 9c6.cos3θ.(isinθ)6
+ 9c7.cos2θ.(isinθ)7 + 9c8.cosθ.(isinθ)8 + 9c9.(isinθ)9
= cos9θ + 9i.cos8θ.sinθ - 36.cos7θ.sin2θ - 84i.cos6θ.sin3θ
+ 126.cos5θ.sin4θ + 126i.cos4θ.sin5θ - 84.cos3θ.sin6θ
- 36i.cos2θ sin7θ + 9.cosθ.sin8θ + i.sin9θ
Equating the real parts, we have
cos9θ = cos9θ - 36.cos7θ.sin2θ + 126.cos5θ.sin4θ - 84.cos3θ.sin6θ + 9.cosθ.sin8θ
= cos9θ - 36.cos7θ.(1 - cos2θ) + 126.cos5θ.(sin2θ)2 - 84.cos3θ.(sin2θ)3 + 9.cosθ.(sin2θ)4
= cos
= 37.cos
= 37.cos
= 247.cos
= 247.cos9θ - 540.cos7θ + 378.cos5θ - 84.cos3θ + 9.cosθ + 9.cos9θ + 36.cos5θ + 18.cos5θ - 36.cos7θ - 36.cos3θ
= 256.cos9θ - 576.cos7θ + 432.cos5θ - 120.cos3θ + 9.cosθ
(iv) cos4θ
cos4θ + i.sin4θ = (cosθ + isinθ)4
= cos4θ + 4c1.cos3θ.(isinθ) + 4c2.cos2θ.(isinθ)2 + 4c3.cosθ.(isinθ)3 + 4c4.(isinθ)4
= cos4θ + 4.cos3θ.(isinθ) - 6.cos2θ.sin2θ - 4i.cosθ.sin3θ + sin4θ
Equating the real parts, we have
cos4θ = cos4θ - 6.cos2θ.sin2θ + sin4θ
= cos4θ - 6.cos2θ.(1 - cos2θ) + (sin2θ)2
= cos4θ - 6.cos2θ + 6.cos4θ + (1 - cos2θ)2
= 7.cos4θ - 6.cos2θ + 1+ cos4θ - 2.cos2θ
= 8.cos4θ - 8.cos2θ + 1
(v) tan4θ
we know that tannθ = (nc1.tanθ - nc3.tan3θ + ...)/(1 - nc2.tan2θ + nc4.tan4θ - ...)
Put n=4 in the above
tan4θ = (4.tanθ - 4.tan3θ)/(1 - 6.tan2θ + tan4θ)
= 4.tanθ(1 - tan2θ)/(1 - 6.tan2θ + tan4θ)
(vi) tan5θ
= (5c1.tanθ - 5c3.tan3θ + 5c5.tan5θ)/(1 - 5c2.tan2θ + 5c4.tan4θ)
= (5.tanθ - 10.tan3θ + tan5θ)/(1 - 10.tan2θ + 5.tan4θ)
(vii) tan9θ
Nr = 9c1.tanθ - 9c3.tan3θ + 9c5.tan5θ - 9c7.tan7θ + tan9θ
= 9.tanθ - 84.tan3θ + 126.tan5θ - 36.tan7θ + tan9θ
Dr = 1 - 9c2.tan2θ + 9c4.tan4θ - 9c6.tan6θ + 9c8.tan8θ
= 1 - 36.tan2θ + 126.tan4θ - 84.tan6θ + 9.tan8θ
tan9θ
=(9.tanθ - 84.tan3θ + 126.tan5θ - 36.tan7θ + tan9θ)/(1 - 36.tan2θ + 126.tan4θ - 84.tan6θ + 9.tan8θ)
(2) (cosθ + isinθ)11 = cos11θ + 11c1.cos10θ.(isinθ) + 11c2.cos9θ.(isinθ)2 + 11c3.cos8θ.(isinθ)3 + ...+ 11c10.cosθ.(isinθ)10 + (isinθ)11
= cos
Equating the real parts, last term in cos11θ is -11.cosθ.sin10θ.
(cosθ + isinθ)9 = cos9θ + 9c1.cos8θ.(isinθ) + 9c2.cos7θ.(isinθ)2 + 9c3.cos6θ.(isinθ)3 + .... + 9c8.cosθ.(isinθ)8 + (isinθ)9
= cos9θ + i.9.cos8θ.sinθ - 36.cos7θ.sin2θ - i.84.cos6θ.sin3θ + .... 9.cosθ.sin8θ + i.sin9θ
Equating the imaginary parts, last term in the expansion of sin9θ is sin9θ.
(3) (a) cos7θ + isin7θ = (cosθ + isinθ)7
= cos
Equating the imaginary parts
sin7θ = 7.cos
sin7θ/sinθ = 7.cos
cos
cos
sin7θ/sinθ = 7(1 - 3.sin
= 7 - 21.sin
= 7 - 56.sin
3 (b) From (1) (iii), Equating the imaginary parts
sin9θ = 9.cos8θ.sinθ - 84.cos6θ.sin3θ + 126.cos4θ.sin5θ - 36.cos2θ.sin7θ + sin9θ
sin9θ/sinθ = 9.cos8θ - 84.cos6θ.sin2θ + 126.cos4θ.sin4θ - 36.cos2θ.sin6θ + sin8θ – (1)
sin4θ = (sin2θ)2 = (1 - cos2θ)2 = 1 + cos4θ - 2.cos2θ
sin6θ = sin4θ.sin2θ = (1 + cos4θ - 2.cos2θ).(1 - cos2θ) = 1 - cos2θ + cos4θ - cos6θ - 2.cos2θ + 2.cos4θ = 1 - 3.cos2θ + 3.cos4θ - cos6θ
sin8θ = sin6θ.sin2θ = (1 - 3.cos2θ + 3.cos4θ - cos6θ).(1 - cos2θ)
= 1 - cos2θ - 3.cos2θ + 3.cos4θ + 3.cos4θ - 3.cos6θ - cos6θ + cos8θ = 1 - 4.cos2θ + 6.cos4θ - 4.cos6θ + cos8θ
84.cos6θ.sin2θ = 84.cos6θ.(1 - cos2θ) = 84.cos6θ - 84.cos8θ
126.cos4θ.sin4θ = 126.cos4θ.(1 + cos4θ - 2.cos2θ)
= 126.cos4θ + 126.cos8θ - 252.cos6θ
36.cos2θ.sin6θ = 36.cos2θ.(1 - 3.cos2θ + 3.cos4θ - cos6θ)
= 36.cos2θ - 108.cos4θ + 108.cos6θ - 36.cos8θ
From (1),
sin9θ/sinθ = 9.cos8θ - 84.cos6θ + 84.cos8θ + 126.cos4θ - 252.cos6θ + 126.cos8θ - 36.cos2θ + 108.cos4θ - 108.cos6θ + 36.cos8θ + 1 - 4.cos2θ + 6.cos4θ - 4.cos6θ + cos8θ
= 256.cos8θ - 448.cos6θ + 240.cos4θ + 1 - 40.cos2θ
(4) (cosθ + isinθ)
case 1 : When n is even
Let n = 2m, where m is a positive integer
Then the last term in the binomial expression (1) is
(isinθ)n = in.sinnθ = (i)2m.sinnθ = (i2)m.sinnθ = (-1)m.sinnθ
= (-1)n/2.sinnθ.which is real and the last but one term is
ncn-1.cosθ.(isinθ)n-1 = n.cosθ.in-1.(sinθ)n-1
= n.cosθ.i2m-1.(sinθ)n-1
= n.cosθ.i2m-2.i.(sinθ)n-1(multiplying nr and dr by i)
= i.i2(m-1).n.cosθ.(sinθ)n-1
= i.(i2)m-1.n.cosθ.(sinθ)n-1
= i.(-1)m-1.n.cosθ.(sinθ)n-1
= i.(-1)n/2-1.n.cosθ.(sinθ)n-1 which is imaginary
Hence the last term in the expansion of cosnθ is (-1)n/2.sinnθ and the last term in the expansion of sinnθ is (-1)n/2-1.n.cosθ.(sinθ)n-1
Case (2) : When n is odd
Let n = 2m + 1, where m is a positive integer
Then (i)n = (i)2m+1 = i2m.i = (i2)m.i = (-1)m.i = (-1)(n-1)/2
(i)
(isinθ)
n
= n.cosθ.(sinθ)
Hence the last term in the expansion of cosnθ is n.cosθ.(sinθ)
and sinnθ is (-1)
(5) Given equation is tan2θ = λ.tan(θ+α)
2.tanθ/(1 - tan2θ) = λ.(tanθ + tanα)/(1 - tanθ.tanα)
⇒ 2.tanθ(1 - tanθ.tanα) = λ.(tanθ + tanα).(1 - tan²θ)
⇒ 2.tanθ - 2.tan²θ.tanα = λ.(tanθ - tan³θ + tanα - tanα.tan²θ)
⇒ 2.tanθ - 2.tan²θ.tanα = λ.tanθ - λ.tan³θ + λ.tanα - λ.tanα.tan²θ
⇒ 2.tanθ - 2.tan²θ.tanα - λ.tanθ + λ.tan³θ - λ.tanα + λ.tanα.tan²θ = 0
⇒ λ.tan³θ + λ.tanα.tan²θ - 2.tanα.tan²θ + 2.tanθ - λ.tanθ - λ.tanα = 0
⇒ λ.tan³θ + tan²θ.(λ.tanα - 2.tanα) + tanθ.(2 - λ) - λ.tanα = 0
⇒ λ.tan³θ - tanα.tan²θ.(2 - λ) + tanθ.(2 - λ) - λ.tanα = 0
This is a cubic equation is tanθ.
Its roots are given as tanθ1,tanθ2 and tanθ3.
S1 =∑tanθ1 = (2 - λ).tanα/λ
S2 = ∑tanθ1.tanθ2 = (2 - λ)/λ and S3 = tanθ1.tanθ2.tanθ3 = tanα
tan(θ1 + θ2 + θ3) = (S1 - S3)/( 1 - S2)
S1 - S3 = (2 - λ).tanα/λ - tanα
= [(2 - λ).tanα - λ.yanα]/λ
= (2.tanα - λ.tanα - λ.tanα)/λ
= (2.tanα - 2λ.tanα)/λ
= 2.tanα.(1 - λ)/λ
1 - S2 = 1 - (2 - λ)/λ
= 2.(λ-1)/λ
tan(θ1 + θ2 + θ3) = (S1 - S3)/(1-S2)
= 2.tanα.(1-λ).λ/λ.2.(λ-1)
= -tanα except when λ=1
when λ=1, fraction takes indeterminate form.
Thus tan(θ1+θ2+θ3) = -tanα = tan(-α)
⇒ θ1+θ2+θ3 = nπ-α, where n is any integer
or θ1+θ2+θ3
Thus θ1+θ2+θ3 + α is a multiple of π
(6) Given tan(θ+π∕4) = 3.tan3θ
⇒ [tanθ+tan(π/4)]/[1-tanθ.tan(π/4)] = 3.[3.tanθ-tan³θ]/[1-3.tan²θ]
⇒ (1+tanθ)/(1-tanθ) = 3.[3.tanθ-tan³θ]/[1-3.tan²θ]
Let tanθ=x
∴ (1+x)/(1-x) = 3.(3x-x³)/(1-3x²)
⇒ (1+x)(1-3x²)=3(1-x)(3x-x³)
⇒ 1-3x²+x-3x³ = 3(3x-x³-3x²+x4)
⇒ 1-3x³-3x²+x = 9x-3x³-9x²+3x4
⇒ 3x4-3x³-9x²+9x-1+3x³+3x²-x=0
⇒ 3x4+0.x³-6x²+8x-1=0
It is a 4th degree equation. Hence 4 roots
Coefficient of x³ is zero. α,β,γ and δ are the roots of the equation.
∑tanα = -(coeff. of x³)/coeff.of x4 =0
⇒ tanα+tanβ+tanγ+tanδ=0
(7) Given sin(θ+α)=a.sin2θ+b
⇒sinθ.cosα+cosθ.sinα = a.2sinθ.cosθ+b
⇒2t/(1+t²).cosα +(1-t²)/(1+t²).sinα = 2a.2t/(1+t²).(1-t²)/(1+t²) + b
where t = tan(θ/2)
⇒2t.cosα+(1-t²).sinα=4at.(1-t²)/(1+t²)+b(1+t²)
⇒2t.(1+t²).cosα+(1-t4).sinα=4at.(1-t²)+b.(1+t²)²
⇒2t(1+t²)cosα+(1-t4)sinα=4at(1-t²)+b(1+t4+2t²)
⇒2tcosα+2t³cosα+sinα-t4sinα=4at-4at³+b+bt4+2bt²
⇒t4(b+sinα)-2t³(cosα+2a)+2bt²+2t(2a-cosα)+b-sinα=0
Let the four roots of the above equation be t1,t2,t3 and t4.
S1=∑t1=2(2a+cosα)/(b+sinα)
S2=∑t1t2=2b/(b+sinα)
S3=∑t1t2t3=-2(2a-cosα)/(b+sinα)
S4=t1t2t3t4=(b-sinα)/(b+sinα)
S1-S3=(4a+2cosα+4a-2cosα)/(b+sinα)=8a/(b+sinα)
1-S2+S4=1-2b/(b+sinα)+(b-sinα)/(b+sinα)
=(b+sinα-2b++b-sinα)/(b+sinα)
=0/(b+sinα)=0
tan[(θ1+θ2+θ3+θ4)/2]=(S1-S3)/(1-S2+S4)
= 8a/0=∞
⇒(θ1+θ2+θ3+θ4)/2=kπ+π/2
⇒θ1+θ2+θ3+θ4=2kπ+π=(2k+1)π
(8) cos2θ+a.cosθ+b.sinθ+c=0
Let tan(θ/2)=t
sinθ=[2tan(θ/2)]/[1+tan²(θ/2)]
cosθ=[1-tan²(θ/2)]/[1+tan²(θ/2)]
Let tan(θ/2)=t
Then sinθ=2t/(1+t²) and cosθ=(1-t²)/(1+t²)
sin²θ= 4t²/(1+t²)²
cos2θ=1-2.sin²θ = 1-2.4t²/(1+t²)²
= [(1+t²)²-8.t²]/(1+t²)²
= (1+t4+2.t²-8.t²)/(1+t²)²
= (1+t4 -6.t²)/(1+t²)²
Using the above values in the given equation we have
(1+t4-6.t²)/(1+t²)² + a.(1-t²)/(1+t²)+b.2t/(1+t²) + c=0
⇒1+t4-6.t²+a(1-t²)(1+t²)+b.2t(1+t²)+c.(1+t²)²=0
⇒1+t4-6.t²+a(1-t4)+2bt+2bt³+c(1+t4+2t²)=0
⇒1+t4-6t²+a-at4+2bt+2bt³+c+ct4
⇒t4(1-a+c)+t³.2b+t².(2c-6)+t.2b+1+a+c=0-–––––(1)
This is a fourth degree equation in t
ie., tan(θ/2) giving four roots α,β,γ and δ.
ie.,tan(α/2),tan(β/2),tan(γ/2) and tan(δ/2).
By theory of equations (1) gives
S1=∑α=∑tan(α/2) = -2b/(1-a+c)
S2 = ∑αβ = ∑tan(α/2).tan(β/2) = (2c-6)/(1-a+c)
S3 = ∑αβγ = ∑tan(α/2).tan(β/2).tan(γ/2)
= -2b/((1-a+c) = S1
S4 = αβγδ = tan(α/2)tan(β/2)tan(γ/2)tan(δ/2)
= (1+a+c)/(1-a+c)
tan(α+β+γ+δ/2) = (S1-S3)/(1-S2+S4)
= (S1-S1)/(1-S2+S4)––––––(2)
1-S2+S4 = 1 - (2c-6)/(1-a+c) + (1+a+c)/(1-a+c)
= (1-a+c-2c+6+1+a+c)/(1-a+c)
= 8/(1-a+c) ≠ zero
(2)⇒ tan(α+β+γ+δ/2) = 0 = tan0
⇒(α+β+γ+δ)/2=0=nπ,n is an integer
⇒α+β+γ+δ = 2nπ
(9) Given asin2θ+bsinθ+c=0
⇒a.2sinθ.cosθ+b.sinθ+c=0
⇒2a.2t/(1+t²).(1-t²)/(1+t²)+b.2t/(1+t²)+c=0 where t =tan(θ/2)
⇒4at(1-t²)+2b(1+t²)+c(1+t²)²=0
⇒4at-4at³+2b+2bt²+c(1+t4 +2t²)=0
⇒ct4-4at³+t²(2b+2c)+4at+2b+c=0–––––(1)
This is a fourth degree equation in t.
ie., tan(θ/2) giving four roots
Let t1=tan(θ1/2),t2=tan(θ2/2),t3=tan(θ3/3) and t4=tan(θ4/4)
By theory of equations (1) gives
S1=∑t1=∑tan(θ1/2)=4a/c
S2=∑t1t2=(2b+2c)/c
S3=∑t1t2t3=-4a/c
S
1-S
S
tan[(θS1-S3)/(1-S2+S4)
=8a/c÷0=∞=tan(π/2)
⇒(θ1+θ2+θ3+θ4)/2=π/2 + nπ
⇒θ1+θ2+θ3+θ4 = 2nπ+π=π(2n+1) is an odd multiple of π
(10) a.cos2θ+b.sin2θ+c.cosθ+d.sinθ=0
⇒a.(1-2sin²θ)+b.2sinθ.cosθ+c.cosθ+d.sinθ=0––––(1)
Let t=tan(θ/2)
1-2sin²θ=1-2(2t/1+t²)²
=1-2.4t²(1+t²)²
=[(1+t²)²-8t²]/(1+t²)²
=(1+t4+2t²-8t²)/(1+t²)²
=(1+t4-6t²)/(1+t²)²
2sinθcosθ=2.2t/(1+t²).(1-t²)/(1+t²)=4t(1-t²)/(1+t²)²
c.cosθ=c.(1-t²)/(1+t²)
d.sinθ=d.2t/(1+t²)
Substituting the above values in (1) we get
a.(1+t4-6t²)+4bt.(1-t²)+c.(1-t4)+2dt.(1+t²)=0
⇒t4(a-c)+t³(2d-4b)-6at²+t(2d+4b)+a+c=0
This is a fourth degree equation in t
ie.,tan(θ/2) giving four roots t1,t2,t3 and t4.
ie., tan(α/2),tan(β/2),tan(γ/2) and tan(δ/2)
S1=∑t1=∑tan(α/2)=(4b-2d)/(a-c)
S2=∑t1t2=∑tan(α/2).tan(β/2)=-6a/(a-c)
S3=∑t1t2t3=∑tan(α/2).tan(β/2).tan(γ/2)
=-(2d+4b)/(a-c)
S4=tan(α/2).tan(β/2).tan(γ/2).tan(δ/2)
=(a+c)/(a-c)
S1-S3=(4b-2d)/(a-c)+(2d+4b)/(a-c)
=8b/(a-c)
1-S2+S4=1+6a/(a-c)+(a+c)/(a-c)=8a/(a-c)
tan(α+β+γ+δ/2)=(S1-S3)/(1-S2+S4)
=8b/(a-c)÷8a/(a-c)
= b/a
(11) Given a²cos²θ+b²sin²θ+2gacosθ+2fb.sinθ+c=0
Let t=tan(θ/2)
sinθ=2.tan(θ/2)/(1+tan²(θ/2))=2t/(1+t²)
cosθ=(1-tan²(θ/2))/(1+tan²θ/2)=(1-t²)/(1+t²)
Using the above values in the given equation we get
a²[(1-t²)/(1+t²)]²+b².[2t/(1+t²)]²+2ga.(1-t²)/(1+t²)+2fb.2t/(1+t²)+c=0
⇒t4(a²-2ga+c)+t³.4fb+t²(4b²-2a²+2c)+4fb.t+a²+2ga+c=0
This is a fourth degree equation in t
ie., tan(θ/2) giving four roots t1,t2,t3 and t4.
ie., tan(θ1/2),tan(θ2/2),tan(θ3/2) and tan(θ4/2)
By theory of equations we have
S1=∑t1=∑tan(θ1/2)=-4bf/(a²-2ga+c)
S2=∑t1t2=∑tan(θ1/2)tan(θ2/2)=(4b²-2a²+2c)/(a²-2ga+c)
S3=∑t1t2t3=∑tan(θ1/2).tan(θ2/2).tan(θ3/2)
=-4fb(a²-2ga+c)=S1
S4=∑t1t2t3t4=tan(θ1/2).tan(θ2/2).tan(θ3/2).tan(θ4/2)
=(a²+2ga+c)/(a²-2ga+c)
tan(θ1+θ2+θ3+θ4/2)=(S1-S3)/(1-S2+S4)
=(S1-S1)/(1-S2+S4)
1-S2+S4=1-(4b²-2a²+2c)/(a²-2ga+c)+(a²+2ga+c)/(a²-2ga+c)
=(4a²-4b²)/(a²-2ga+c)≠0
Hence tan(θ1+θ2+θ3+θ4/2)=0=tan0
⇒θ1+θ2+θ3+θ4/2=0=nπ,n being an integer
⇒θ1+θ2+θ3+θ4=2nπ,an even multiplier of π
(12) tanx/tan3x=tanx÷(3tanx-tan³x)/(1-3.tan²x)
=tanx.(1-3.tan²x)/(3tanx-tan³x)
=(1-3tan²x)/(3-tan²x)=n say
⇒1-3tan²x=n(3-tan²x)
⇒1-3tan²x=3n-ntan²x
⇒n.tan²x-3.tan²x=3n-1
⇒tan²x.(n-3)=3n-1
⇒tan²x=(3n-1)/(n-3)=(1-3n)/(3-n)
These two values of tan²x must be positive, and therefore
n must be greater than 3 or less than 1/3.
(13) Let x=cosθ+isinθ
1/x=cosθ-i.sinθ
x-1/x=2i.sinθ; x+1/x=2.cosθ
xn=cosnθ+i.sinnθ
1/xn=cosnθ-i.sinnθ
xn+1/xn=2.cosnθ––––––(1)
(x-1/x)8=(2i.sinθ)8
(x-1/x)8=x8+8c1.x7.(-1/x)+8c2.x6.(-1/x)²+8c3.x5.(-1/x)³
+8c4.x4.(-1/x)4+8c5.x³.(-1/x)5+8c6.x².(-1/x)6
+8c7.x.(-1/x)7+8c8.(-1/x)8
= x8-8.x6+28.x4-56.x²+70-56.(1/x²)+28.(1/x4)-8.(1/x6)+1/x8
= x8+1/x8 - 8.(x6+1/x6)+28.(x4+1/x4)-56.(x²+1/x²)+70
28.i8.sin8θ=2.cos8θ-8.2.cos6θ+28.2.cos4θ-56.2.cos2θ+70 using (1)
⇒27.sin8θ=cos8θ-8.cos6θ+28.cos4θ-56.cos2θ+35
⇒sin8θ=(1/128)[cos8θ-8.cos6θ+28.cos4θ-56.cos2θ+35]–––––(2)
(x+1/x)8=28.cos8θ
(x+1/x)8=x8+8c1.x7(1/x)+8c2.x6.(1/x)²+8c3.x5.(1/x)³+8c4.x4.(1/x)4
+8c5.x³(1/x)5+8c6.x².(1/x)6+8c7.x.(1/x)7+8c8(1/x)8
=x8+8.x6+28.x4+56.x²+70+56.(1/x²)+28.(1/x4)+8.(1/x6)+1/x8
= x8+1/x8 + 8.(x6+1/x6)+28.(x4+1/x4)+56.(x²+1/x²)+70
= 2.cos8θ+8.2.cos6θ+28.2.cos4θ+56.2.cos2θ+70
28.cos8θ= 2.cos8θ+8.2.cos6θ+28.2.cos4θ+56.2.cos2θ+70
⇒27.cos8θ=cos8θ+8.cos6θ+28.cos4θ+56.cos2θ+35
⇒cos8θ=(1/128).[cos8θ+8.cos6θ+28.cos4θ+56.cos2θ+35]––––(3)
(2) +(3)⇒128.cos8θ+128.sin8θ=2.cos8θ+56.cos4θ+70
⇒64.(cos8θ+sin8θ)=cos8θ+28.cos4θ+35
(14) cos3θ+sin3θ=0
⇒4.cos³θ-3.cosθ+3.sinθ-4.sin³θ=0
⇒4(cos³θ-sin³θ)-3(cosθ-sinθ)=0
⇒4(cosθ-sinθ)(cos²θ+cosθ.sinθ+sin²θ)-3(cosθ-sinθ)=0
⇒4(cosθ-sinθ)(1+cosθ.sinθ)-3(cosθ-sinθ)=0
⇒4(1+cosθ.sinθ)-3=0
⇒4+4cosθ.sinθ-3=0
⇒1+2.2sinθ.cosθ=0
⇒1+2.sin2θ=0
⇒2.sin2θ=-1
⇒sin2θ=-1/2
⇒2.tanθ/(1+tan²θ)=-1/2
⇒-4.tanθ=1+tan²θ
⇒tan²θ+4.tanθ+1=0–––––(1)
⇒tanθ=(-4±√12)/2=-2+√3 or -2-√3
Let x=tanθ
(1)⇒x²+4x+1=0
⇒x=-2±√3
tan(π/6)=1/√3
⇒tan(2π/12)=1/√3
⇒2.tan(π/12)/(1-tan²π/12)=1/√3
⇒2t/(1-t²)=1/√3 where t = tan(π/12)
⇒t²+2t.√3-1=0
⇒t=-√3±2
since tan(π/12)>0,tan(π/12)=-√3+2
-tan(π/12)=-2+√3
tan(5π/12)=tan(π/4+π/6)
=[tan(π/4)+tan(π/6)]/[1-tan(π/4).tan(π/6)]
=[1+1/√3]/[1-1.1/√3]
=(√3+1)/(√3-1)
=2+√3
-tan(5π/12)=-2-√3
Hence -tan(π/12) and -tan(5π/12) are the roots of the equation x²+4x+1=0
(15) (i) 2(1+cos8θ)=2(1+cos2(4θ))
=2.2cos²4θ
=(2.cos4θ)²––––(1)
From Q.no.(1)(iv)
cos4θ=8.cos4θ-8.cos²θ+1
Using the above value in (1)
2(1+cos8θ)=[2.(8cos4θ-8cos²θ+1)]²
=(16.cos4θ-16.cos²θ+2)²
= [(2cosθ)4-4.4cos²θ+2]²
=[(2cosθ)4-4.(2cosθ)²+2]²
=(x4-4x²+2)² where x = 2cosθ
(15)(ii) (1+cos10θ)/(1+cos2θ)
1+cos10θ=1+cos2(5θ)=2cos²5θ
From (1)(ii) ,cos5θ=16cos5θ-20cos³θ+5cosθ
cos²5θ=(16cos5θ-20cos³θ+5cosθ)²
=[cosθ(16cos4θ-20cos²θ+5)]²
=cos²θ(16cos4θ-20.cos²θ+5)²
2.cos²5θ=2.cos²θ(16.cos4θ-20.cos²θ+5)²
(1+cos10θ)/(1+cos2θ)=(16.cos4θ-20.cos²θ+5)²
=[(2cosθ)4-5.(2cosθ)²+5]²
=(x4-5x²+5)² where x = 2cosθ
(15)(iii) (1+cos9θ)/(1+cosθ)=2.cos²(9θ/2)/2cos²(θ/2)
=2cos²(9θ/2).2sin²(θ/2)/2cos²(θ/2).2sin²(θ/2) [multiplying Nr and Dr by sin²(θ/2)]
=[2cos(9θ/2).sin(θ/2)]²/[2cos(θ/2).sin(θ/2)]²
=[(sin5θ-sin4θ)/sinθ]²
sin5θ=5.cos4θ.sinθ-10.cos²θ.sin³θ+sin5θ
sin4θ=4.cos³θ.sinθ-4.cosθ.sin³θ
sin5θ-sin4θ=5.cos4θ.sinθ-10.cos²θ.sin³θ+sin5θ-4.cos³θ.sinθ+4.cosθ.sin³θ
[sin5θ-sin4θ]/sinθ=5.cos4θ-10.cos²θ.sin²θ+sin4θ-4.cos³θ+4.cosθ.sin²θ
=5.cos4θ-10.cos²θ.(1-cos²θ)+(sin²θ)²-4.cos³θ+4.cosθ(1-cos²θ)
=5.cos4θ-10.cos²θ+10.cos4θ+(1-cos²θ)²-4.cos³θ+4.cosθ-4.cos³θ
=15.cos4θ-10.cos²θ+1+cos4θ-2.cos²θ-8.cos³θ+4.cosθ
=16.cos4θ-8.cos³θ-12.cos²θ+4.cosθ+1
=(2.cosθ)4-(2.cosθ)³-3.(2.cosθ)²+2.2cos+1
(1+cos9θ)/(1+cosθ)=([sin5θ-sin4θ]/sinθ)²
=(x4-x³-3.x²+2x+1)², where x=2cosθ
(15)(iv) (1+cos7θ)/(1+cosθ)
=2.cos²(7θ/2)/2.cos²(θ/2)
=2.cos²(7θ/2).2.sin²(θ/2)/2.cos²(θ/2).2.sin²(θ/2)
[Multiplying Nr. and Dr by 2.sin²(θ/2)]
=[2.cos(7θ/2).sin(θ/2)]²/[2.cos(θ/2).sin(θ/2)]²
=[(sin4θ-sin3θ)/sinθ]²–––––(1)
sin4θ=4.cos³θ.sinθ-4.cosθ.sin³θ
sin3θ=3.sinθ-4.sin³θ
(sin4θ-sin3θ)/sinθ=4.cos³θ-4.cosθ.sin²θ-3+4.sin²θ
=4.cos³θ-4.cosθ.(1-cos²θ)-3+4.(1-cos²θ)
=4.cos³θ-4.cosθ+4.cos³θ-3+4-4.cos²θ
=8.cos³θ-4.cos²θ-4.cosθ+1
=(2.cosθ)³-(2.cosθ)²-2.(2cosθ)+1
=x³-x²-2x+1 , where x=2cosθ
Using the above value in (1) we get
(1+cos7θ)/(1+cosθ)=(x³-x²-2x+1)²
Exercise X
-–––––––––
(1)We know that sin18°=(√5-1)/4 cos36°=(√5+1)/4 LHS=sin(π/5).sin(2π/5).sin(3π/5).sin(4π/5)
=sin36°.sin72°.sin108°.sin144°
=sin36°.sin72°.sin(180-72).sin(180-36) [sin(180-θ)=sinθ]
=sin36°.sin72°.sin72°.sin36°
=sin²36°.sin²72°
=(sin36°.sin72°)²
=(2.sin36°.sin72°/2)²
=¼(2.sin36°.sin72°)²
=¼[cos(36-72)-cos(36+72)]² [2.sinA.sinB=cos(A-B)-cos(A+B)]
=¼[cos(-36)-cos108]²
=¼[cos36-cos(90+18)]² [cos(-θ)=cosθ]
=¼[cos36-(-sin18)]² [cos(90+θ)=-sinθ]
=¼[cos36+sin18]²
=¼[(√5+1)/4+(√5-1)/4]²
=¼[(2√5/4)]²=5∕16=RHS
(2) Let θ=nπ/7
⇒7θ=nπ
⇒sin7θ=sin(nπ)=0
From Page no.68 Worked example 1
sin7θ=7.sinθ-56.sin³θ+112.sin5θ-64.sin7θ
sin7θ=0
⇒7.sinθ-56.sin³θ+112.sin5θ-64.sin7θ=0
⇒sinθ(7-56.sin²θ+112.sin4θ-64.sin6θ)=0
⇒sinθ=0 or 7-56.sin²θ+112.sin4θ-64.sin6θ=0
⇒64.sin6θ-112.sin4θ+56.sin²θ-7=0
⇒64.x6-112.x4+56.x²-7=0 where x=sinθ––––(1)
sinθ=0 corresponds to n=0
Therefore the roots of the equation (1) are
sin(π/7),sin(2π/7),sin(3π/7),sin(4π/7),sin(5π/7)
and sin(6π/7).
Since sin(π-θ)=sinθ
sin(6π/7)=sin(π-π/7)=sin(π/7)
sin(5π/7)=sin(π-2π/7)=sin(2π/7)
sin(4π/7)=sin(π-3π/7)=sin(3π/7)
Product of the roots
=sin(π/7).sin(2π/7).sin(3π/7).sin(4π/7).sin(5π/7).sin(6π/7)
=sin²(π/7).sin²(2π/7).sin²(3π/7)=7/64
[Negattive sign is discarded since all the terms of the
expression on the left side are positive]
(3) Let θ denote any of the angles
π/10,5π/10,9π/10,13π/10 and 17π/10.
5θ=π/2,5π/2,9π/2,13π/2 and 17π/2.
=(2nπ+π/2),n=0,1,2,3,4
sin5θ=sin(2nπ+π/2)=1
⇒5.sinθ-20.sin³θ+16.sin5θ=1
⇒16.sin5θ-20.sin³θ+5.sinθ=1––––(1)
Let x=sinθ
(1)⇒16.x5-20.x³+5x=1 has five roots viz.,
sin(π/10),sin(5π/10),sin(9π/10),sin(13π/10) and sin(17π/10)
(i) Sum of the roots
=sin(π/10)+sin(5π/10)+sin(9π/10)+sin(13π/10)+sin(17π/10)
=-0/16=0––––(2)
(ii) From (2)
sin18+sin90++sin162+sin234+sin306=0 [π/10=18]
⇒sin18+1+sin(180-18)+sin(180+54)+sin(360-54)=0
⇒sin18+1+sin18-sin54-sin54=0[sin(180-θ)=sinθ,sin(180+θ)=-sinθ, sin(360-θ)=-sinθ]
⇒2.sin18+1-2.sin54=0
⇒sin18+½-sin54=0
⇒sin18+½=sin54
(4) Given cos4θ=½
⇒8.cos4θ+1-8.cos²θ=½
⇒16.cos4θ-16.cos²θ+1=0
⇒16c4-16c²+1=0 where c=cosθ
cos4θ=½
⇒4θ=cos-(½)
⇒θ=¼.cos-(½)
=¼[(π/3)+2nπ,(5π/3)+2nπ]
when n=0,θ=(π/12),(5π/12)
when n=1,θ=(7π/12),(11π/12)
After this cosine value repeats.
Hence roots of the given equation are cos(π/12),cos(5π/12),cos(7π/12) and cos(11π/12).
Let x=c²
Then the original eqn becomes 16x²-16x+1=0
Then cos²(π/12) and cos²(5π/12) are the roots of the equation
it is not necessary to mention cos2(7π/12) and cos2(11π/12)
they differ only in sign, so their squares are the same roots of the quadratic.
⇒θ=2nπ/7,n=1,2,3,4,5,6,7
⇒7θ=2nπ
⇒4θ+3θ=2nπ
⇒4θ=2nπ-3θ
⇒cos4θ=cos(2nπ-3θ)
⇒cos4θ=cos3θ
⇒2.cosθ²2θ-1=4.cos³θ-3.cosθ [cos2θ=2.cos²θ-1]
⇒2.[cos2θ]²-1=4.cos³θ-3.cosθ
⇒2.[2.cos²θ-1]²-1=4.cos³θ-3.cosθ
⇒2[4cos4θ+1-4cos²θ]-1=4cos³θ-3cosθ
⇒8cos4θ+2-8cos²θ-1-4cos³θ+3cosθ=0
⇒8cos4θ-4cos³θ-8cos²θ+3cosθ+1=0
Let x=cosθ
8x4-4x³-8x²+3x+1=0
⇒(x-1)(8x³+4x²-4x-1)=0––––––(1)
⇒x-1=0
⇒x=1
⇒cosθ=1
⇒θ=14π/7=2π
Hence the roots of the equation 8x³+4x²-4x-1=0 are the
remaining six roots namely
cos(2π/7),cos(4π/7),cos(6π/7),cos(8π/7),cos(10π/7) and cos(12π/7)
But cos(12π/7)=cos(2π-2π/7)=cos(2π/7)
cos(10π/7)=cos(2π-4π/7)=cos(4π/7)
cos(8π/7)=cos(2π-6π/7)=cos(6π/7)
Hence cos(2π/7),cos(4π/7) and cos(6π/7) are the roots
of the equation 8x³+4x²-4x-1=0
Sum of the roots=cos(2π/7)+cos(4π/7)+cos(6π/7)=-4/8
⇒1-2.sin²(π/7)+1-2.sin²(2π/7)+1-2.sin²(3π/7)=-1/2
⇒3-2[sin²(π/7)+sin²(2π/7)+sin²(3π/7)]=-1/2
⇒-2[sin²(π-π/7)+sin²(2π/7)+sin²(π-3π/7)]=-½-3 [sin(π-θ)=sinθ]
⇒-2[sin²(6π/7)+sin²(2π/7)+sin²(4π/7)]=-7∕2
⇒sin²(2π/7)+sin²(4π/7)+sin²(6π/7)=7∕4
(6) Let θ=2π/7 or 4π/7 or 6π/7 or 8π/7 or 10π/7 or 12π/7 or 14π/7
⇒θ=2nπ/7,n=1,2,3,4,5,6,7
⇒7θ=2nπ
⇒4θ+3θ=2nπ
⇒4θ=2nπ-3θ
⇒cos4θ=cos(2nπ-3θ)
⇒cos4θ=cos3θ
⇒2.cosθ²2θ-1=4.cos³θ-3.cosθ [cos2θ=2.cos²θ-1]
⇒2.[cos2θ]²-1=4.cos³θ-3.cosθ
⇒2.[2.cos²θ-1]²-1=4.cos³θ-3.cosθ
⇒2[4cos4θ+1-4cos²θ]-1=4cos³θ-3cosθ
⇒8cos4θ+2-8cos²θ-1-4cos³θ+3cosθ=0
⇒8cos4θ-4cos³θ-8cos²θ+3cosθ+1=0
Let x=cosθ
8x4-4x³-8x²+3x+1=0
⇒(x-1)(8x³+4x²-4x-1)=0––––––(1)
⇒x-1=0
⇒x=1
⇒cosθ=1
⇒θ=14π/7=2π
Hence the roots of the equation 8x³+4x²-4x-1=0 are the
remaining six roots namely
cos(2π/7),cos(4π/7),cos(6π/7),cos(8π/7),cos(10π/7) and cos(12π/7)
But cos(12π/7)=cos(2π-2π/7)=cos(2π/7)
cos(10π/7)=cos(2π-4π/7)=cos(4π/7)
cos(8π/7)=cos(2π-6π/7)=cos(6π/7)
Hence cos(2π/7),cos(4π/7) and cos(6π/7) are the roots
of the equation 8x³+4x²-4x-1=0
Put 2x=y in the above equation
y³+y²-2y-1=0 has roots 2cos(2π/7),2cos(4π/7),2cos(6π/7)
(7) Let θ=π/11,3π/11,5π/11,7π∕11,....19π/11,21π/11
(11 values)
Let z=cosθ+isinθ
z11=cos11θ+isin11θ=-1
⇒z11+1=0
⇒(z+1)(z10-z9+z8-z7+z6-z5+z4
⇒z+1=0⇒z=-1⇒θ=11π/11=π
z+1/z=2.cosθ
Let x=cosθ
Therefore z+1/z=2x
z10-z9+z8-z7+z6-z5+z4
Dividing the above equation by z
(z
z²+1/z²=(z+1/z)²-2=4x²-2
z³+1/z³=(z+1/z)³-3(z+1/z)=8x³-3.2x=8x³-6x
now
(z²+1/z²)(z³+1/z³)=z
(4x²-2)(8x³-6x)=z
32x
z
Substituting these values in (1)
32x
⇒32x
where x=cosθ has 10 values π/11,3π/11,...19π/11,21π/11
cos(21π/11)=cos(2π-π/11)=cos(π/11)
cos(19π/11)=cos(2π-3π/11)=cos(3π/11)
cos(17π/11)=cos(2π-5π/11)=cos(5π/11)
cos(15π/11)=cos(2π-7π/11)=cos(7π/11)
cos(13π/11)=cos(2π-9π/11)=cos(9π/11)
Hence the equation has five roots
cos(π/11),cos(3π/11),cos(5π/11),cos(7π/11) and cos(9π/11)
cos(π/11)+cos(3π/11)+cos(5π/11)+cos(7π/11)+cos(9π/11)=-(-16/32)=½
(sum of the roots)
(8) (i) When n is odd
nc1.tanθ-nc3.tan³θ+nc5.tan5θ...+(-1)(n-1)/2.tannθ
tan nθ= ––––––––––––––––––––––––––––––––––––––––––––––––
1-nc2.tan²θ+nc4.tan4θ...+n(-1)(n-1)/2.tann-1θ
When n is even
nc1.tanθ-nc3.tan³θ+nc5.tan5θ...+n(-1)(n-2)/2.tann-1θ
tan nθ= –––––––––––––––––––––––––––––––––––––––––––––––––––
1-nc2.tan²θ+nc4.tan4θ...+(-1)(n/2).tannθ
Let θ=nπ/7,n=0,1,2,3,4,5,6
⇒7θ=nπ
⇒tan7θ=tannπ=0
7c1.tanθ-7c3.tan³θ+7c5.tan5θ-7c7.tan7θ
⇒ –––––––––––––––––––––––––––––––––––––––=0
1-7c2.tan²θ+7c4.tan4θ-7c6.tan6θ
⇒7c1.tanθ-7c3.tan³θ+7c5.tan5θ-7c7.tan7θ=0––––(1)
7c1=7,7c3=35,7c5=21,7c7=1
(1)⇒7.tanθ-35.tan³θ+21.tan5θ-tan7θ=0
⇒tanθ(tan6θ-21.tan4θ+35.tan²θ-7)=0
Let tanθ=x
x(x6-21x4+35x²-7)=0
omitting x=tanθ=0 we have,
x6-21x4+35x²-7=0 whose roots are
tan(π/7),tan(2π/7),tan(3π/7),tan(4π/7),tan(5π/7) and tan(6π/7).
But tan(4π/7)=tan(π-3π/7)=-tan(3π/7)
tan(5π/7)=tan(π-2π/7)=-tan(2π/7) and
tan(6π/7)=tan(π-π/7)=-tan(π/7).
Product of the roots
=tan(π/7).tan(2π/7).tan(3π/7).tan(4π/7).tan(5π/7).tan(6π/7)=-7
⇒[-tan(6π/7)].tan(2π/7).[-tan(4π/7)].tan(4π/7).[-tan(2π/7)].tan(6π/7)=-7
⇒tan²(2π/7).tan²(4π/7).tan²(6π/7)=7
⇒tan(2π/7).tan(4π/7).tan(6π/7)=√7
In Q.No.5 we have proved
cos(2π/7),cos(4π/7) and cos(6π/7) are the roots of
8x³+4x²-4x-1=0
Product of the roots:
cos(2π/7).cos(4π/7).cos(6π/7)=-(-1/8)=1/8–––––(2)
To prove:
sin(2π/7).sin(4π/7).sin(6π/7)=√7/8
tan(2π/7).tan(4π/7).tan(6π/7)=√7(proved already)
⇒sin(2π/7).sin(4π/7).sin(6π/7)/cos(2π/7).cos(4π/7).cos(6π/7)=√7
⇒sin(2π/7).sin(4π/7).sin(6π/7)÷1/8=√7(using (2))
⇒sin(2π/7).sin(4π/7).sin(6π/7)=√7/8
(ii) When n is odd,
tannθ=[nc1.t-...+(-1)(n-1)/2.ncn.tn]/[1-nc2.t²+...],where t=tanθ
and is zero for θ=rπ/n
Hence the values of tan(rπ/n) for r=0 to (n-1) are the roots of
nc1.t-...-(-1)(n-1)/2.nc(n-2).t(n-2)+(-1)(n-1)/2.ncn.tn=0
⇒tn-nc2.t(n-2)+...=0
Sum of the products two together = coeff. of t(n-2)=-½n(n-1)=n(1-n)/2
(9) tan13θ=0 is satisfied by θ=nπ/13 where n is
any integer or zero.
considering the nr for tan13θ
13t-286.t³+1287.t
where t = tanθ
t(t
The factor t corresponds to θ=0
It follows that ±tan(π/13),±tan(2π/13),±tan(3π/13),±tan(4π/13),
±tan(5π/13) and ±tan(6π/13) are the roots of
t12-78.t10+715.t8-1716.t6+1287.t4-286.t²+13=0
Product of the roots
tan²(π/13).tan²(2π/13).tan²(3π/13).tan²(4π/13).tan²(5π/13).tan²(6π/13)=13
⇒tan(π/13).tan(2π/13).tan(3π/13).tan(4π/13).tan(5π/13).tan(6π/13)=√13
(10) Let θ=π/7,2π/7,3π/7,4π/7,5π/7,6π/7
7θ=any multiple of π and hence tan7θ=0
⇒[7.tanθ-35.tan³θ+21.tan5θ-tan7θ]/[1-21.tan²θ+35.tan4θ-7.tan6θ]=0
⇒7.tanθ-35.tan³θ+21.tan5θ-tan7θ=0
⇒tanθ[7-35.tan²θ+21.tan4θ-tan6θ]=0
If tanθ=0 then θ=π=7π/7 which gives none of the
above angles.
Therefore tan6θ-21.tan4θ+35.tan²θ-7=0–––––(1)
Roots of the above equation are tan(π/7),tan(2π/7),
tan(3π/7),tan(4π/7),tan(5π/7) and tan(6π/7).
tan(π-4π/7)=tan(4π/7)=tan(3π/7) [tan(π-θ)=tanθ]
tan(5π/7)=tan(π-5π/7)=tan(2π/7)
tan(6π/7)=tan(π-6π/7)=tan(π/7)
Hence roots of (1) are tan²(π/7),tan²(2π/7) & tan²(3π/7)
Put tan²θ=x in (1)
Then x³-21.x²+35x-7=0 whose roots are
tan²(π/7),tan²(2π/7) & tan²(3π/7).
(11) Let θ denote any of the angles 2π/7,4π/7,...14π/7
so that 7θ=an even multiple of π=2nπ
⇒4θ+3θ=2nπ
⇒4θ=2nπ-3θ
⇒sin4θ=sin(2nπ-3θ)=-sin3θ
⇒2.sin2θ.cos2θ=-(3.sinθ-4.sin³θ)
⇒2.2sinθ.cosθ.((1-2sin²θ)=4sin³θ-3sinθ
⇒4cosθ(1-2sin²θ)=4sin²θ-3 if sinθ≠0 ie., if θ≠14π/7
⇒4√1-sin²θ(1-2sin²θ)=4sin²θ-3
Lets sinθ=x
4.√1-x²(1-2x²)=4x²-3
⇒16(1-x²)(1-2x²)²=(4x²-3)² [squaring both sides]
⇒16(1-x²)(1+4x4-4x²)=16x4+9-24x²
⇒16(1+4x4-4x²-x²-4x6+4x4)=16x4+9-24x²
⇒16(1-4x6+8x4-5x²)=16x4+9-24x²
⇒16-64x6+128x4-80x²=16x4+9-24x²
⇒16x4+9-24x²-16+64x6-128x4+80x²=0
⇒64x6-112x4+56x²-7=0
⇒(8x³)²-(4√7x²-√7)²=0
⇒(8x³+4√7x²-√7)(8x³-4√7x²+√7)=0
⇒(x³+√7x²/2-√7/8)(x³-√7x²/2+√7/8)=0––––(1)
The roots of the equation (1) are
sin(2π/7),sin(4π/7),...sin(12π/7)
sin(8π/7)=sin(2π-6π/7)=-sin(6π/7)
sin(10π/7)=sin(2π-4π/7)=-sin(4π/7)
sin(12π/7)=sin(2π-2π/7)=-sin(2π/7)
Equation (1) resolved into two equations
x³-(√7/2)x²+√7/8=0–––––(2) and
x³+(√7/2)x²-√7/8=0–––––(3)
If we replace x by -x in equation (2), we get equation (3).
So the roots of the equation (3) are the roots of the equation (2) with their signs changed.
Thus if α,β,γ are the roots of (2), then -α,-β,-γ are the roots of (3).
So to write the roots of the equation (2) we have to select one out of
sin(2π/7),sin(12π/7),one out of sin(4π/7),sin(10π/7) and one out of
sin(6π/7),sin(8π/7).
For equation (2) sum of the roots is is positive,sum of the product
of the roots taken two at a time is zero and the product of the roots
is negative.
To satisfy these requirements one root of the equation (2) should be negative and two positive.
The values of sin(2π/7),sin(4π/7),sin(6π/7) are positive
while those of sin(8π/7),sin(10π/7),sin(12π/7) are negative.
The following are the three possibilities to select the roots of (2):
sin(2π/7),sin(4π/7),sin(8π/7)––––(4)
(or) sin(2π/7),sin(6π/7),sin(10π/7)–––––(5)
(or) sin(4π/7),sin(6π/7),sin(12π/7)–––––(6)
If we select the values (4) as the roots,the sum of
the products of the roots taken two at a time
= sin(2π/7).sin(4π/7)+sin(2π/7).sin(8π/7)+sin(4π/7).sin(8π/7)
=½[cos(2π/7)-cos(6π/7)+cos(6π/7)-cos(10π/7)+cos(4π/7)-cos(12π/7)]
=½[cos(2π/7)-cos(10π/7)+cos(4π/7)-cos(12π/7)]
=½[cos(2π-2π/7)-cos(10π/7)+cos(2π-4π/7)-cos(12π/7)]
=½[cos(12π/7)-cos(10π/7)+cos(10π/7)-cos(12π/7)]
=½*0=0
But when we consider the sum of the products of the roots taken two
at a time by taking the values (5) and (6), none of these comes out to
be zero.
Hence set (4) represents the roots of (2).
Hence sin(2π/7),sin(4π/7),sin(8π/7) are the roots of
x³-(√7/2)x²+√/8=0
∴ sin(2π/7)+sin(4π/7)+sin(8π/7)=√7/2 and
sin(2π/7).sin(4π/7).sin(8π/7)=√7/8
(12) (i) sin9θ/sinθ
=256cos8θ-448cos6θ+240cos4θ-40cos²θ+1
[Ref Exercise IX Q.No.3(b)]
Let θ=π/9,2π/9,3π/9 and 4π/9
in each of these cases sin9θ=0
256x8 - 448x6 + 240x4 - 40x2 + 1 = 0 where x= cosθ
Let x²=1/z
then the above equation reduces to
z
hence sec²(π/9)+sec²(2π/9)+sec²(3π/9)+sec²(4π/9)=40
⇒sec²(π/9)+sec²(2π/9)+4+sec²(4π/9)=40
⇒sec²(π/9)+sec²(2π/9)+sec²(4π/9)=36
Prod of the roots:
sec²(π/9).sec²(2π/9).sec²(3π/9).sec²(4π/9)=256
sec²(π/9).sec²(2π/9).4.sec²(4π/9)=256
⇒sec(π/9).sec(2π/9).sec(4π/9)=8
(ii) cos6θ=cos6θ-6c2.cos4θ.sin²θ+6c4.cos²θ.sin4θ-6c6.sin6θ
=cos6θ-15.cos4θ.sin²θ+15.cos²θ.sin4θ-sin6θ
=cos6θ-15.cos4θ.(1-cos²θ)+15.cos²θ.(sin²θ)²-(sin²θ)³
=cos6θ-15.cos4θ+15.cos6θ+15.cos²θ.(1-cos²θ)²-(1-cos²θ)³
=16.cos6θ-15.cos4θ+15.cos²θ.(1+cos4θ-2.cos²θ)-(1-3.cos²θ+3.cos4θ-cos6θ)
=16.cos6θ-15.cos4θ+15.cos²θ+15.cos6θ-30.cos4θ-1+3.cos²θ-3.cos4θ+cos6θ
=32.cos6θ-48.cos4θ+18.cos²θ-1
Let cos6θ=½
then θ=π/18,5π/18,7π/18,11π/18,13π/18 and 17π/18.
cos6θ=½
⇒2.cos6θ=1
⇒2.[32.cos6θ-48.cos4θ+18.cos²θ-1]=1
⇒64.cos
⇒64.cos6θ-96.cos4θ+36.cos²θ-3=0-–––(1)
-cos(11π/18)=cos(π-11π/18)=cos(7π/18)
-cos(13π/18)=cos(π-13π/18)=cos(5π/18)
-cos(17π/18)=cos(π-17π/18)=cos(π/18)
Put cos²θ=x in (1)
Then (1) becomes 64x³-96x²+36x-3=0
Hence cos²(π/18),cos²(5π/18) and cos²(7π/18)
are the roots of the above equation.
Put x=1/z in the above equation
Then 64.(1/z³)-96.(1/z²)+36.(1/z)-3=0
⇒64-96.z+36.z²-3z³=0
⇒3z³-36z²+96z-64=0
Hence sec²(π/18)+sec²(5π/18)+sec²(7π/18)=-(-36)/3=12
[x=cos²θ=1/z ⇒z=sec²θ & sum of the roots=12]
Product of the roots
=sec²(π/18).sec²(5π/18).sec²(7π/18)=-(-64)/3=64/3
⇒sec(π/18).sec(5π/18).sec(7π/18)=8/√3
Exercise XI
––––––––––
(1) Let x=cosθ+isinθ; xn=cosnθ+isinnθ
1/x=cosθ-i.sinθ;1/xn=cosnθ-i.sinnθ
x+1/x=2.cosθ––––––(1)
xn+1/xn=2.cosnθ–––––(2)
Raising both sides of (1) to the power of 7, we have
(2.cosθ)7=(x+1/x)7
=x7+7c1.x6.(1/x)+7c2.x5.(1/x)²+7c3.x4.(1/x)³
+7c4.x3.(1/x)4+7c5.x².(1/x)5+7c6.x.(1/x)6+7c7.(1/x)7
=x7+7.x5+21.x³+35.x+35.1/x+21.(1/x³)+7.(1/x5)+1/x7
⇒27.cos7θ=(x7+1/x7)+7.(x5+1/x5)+21.(x³+1/x³)+35.(x+1/x)
=2.cos7θ+7.2.cos5θ+21.2.cos3θ+35.2.cosθ
⇒26.cos7θ=cos7θ+7.cos5θ+21.cos3θ+35.cosθ
(2) Let x=cosθ+isinθ; xn=cosnθ+isinnθ
1/x=cosθ-i.sinθ;1/xn=cosnθ-i.sinnθ
x+1/x=2.cosθ––––––(1)
xn+1/xn=2.cosnθ–––––(2)
Raising both sides of (1) to the power of 8, we have
(2.cosθ)
=x8+8c1.x7.(1/x)+8c2.x6.(1/x)²+8c3.x5.(1/x)³+8c4.x4.(1/x)4
+8c5.x³.(1/x)5+8c6.x².(1/x)6+8c7.x(1/x)7+8c8.(1/x)8
=x8+8.x6+28.x4+56.x²+70+56.(1/x²)+28.(1/x4)+8.(1/x6)+1/x8
=x8+1/x8+8.(x6+1/x6)+28(x4+1/x4)+56(x²+1/x²)+70
28.cos8θ=2.cos8θ+8.2.cos6θ+28.2.cos4θ+56.2.cos2θ+70
⇒27.cos8θ=cos8θ+8.cos6θ+28.cos4θ+56.cos2θ+35
(3) Let x=cosθ+isinθ; xn=cosnθ+isinnθ
1/x=cosθ-i.sinθ;1/xn=cosnθ-i.sinnθ
x+1/x=2.cosθ––––––(1)
xn+1/xn=2.cosnθ–––––(2)
(2i.sinθ)8=(x-1/x)8
=x8+8c1.x7.(-1/x)+8c2.x6.(-1/x)²+8c3.x5.(-1/x)³
+8c4.x4.(-1/x)4+8c5.x³.(-1/x)5+8c6.x².(-1/x)6
+8c7.x.(-1/x)7+8c8.(-1/x)8
=x8-8.x6+28.x4-56.x²+70-56.(1/x²)+28.(1/x4)
-8.(1/x6)+1/x8
=(x8+1/x8)-8.(x6+1/x6)+28.(x4+1/x4)-56.(x²+1/x²)+70
=2.cos8θ-8.2.cos6θ+28.2.cos4θ-56.2.cos2θ+70
⇒28.i8.sin8θ=2.cos8θ-8.2.cos6θ+28.2.cos4θ-56.2.cos2θ+70
⇒27.sin8θ=cos8θ-8.cos6θ+28.cos4θ-56.cos2θ+35
(4) Let x=cosθ+isinθ; xn=cosnθ+isinnθ
1/x=cosθ-i.sinθ;1/xn=cosnθ-i.sinnθ
x+1/x=2.cosθ––––––(1)
xn+1/xn=2.cosnθ–––––(2)
x-1/x=2i.sinθ;xn-1/xn=2i.sinnθ
(2.cosθ)³.(2i.sinθ)4=(x+1/x)³.(x-1/x)4
=(x+1/x)³.(x-1/x)³.(x-1/x)
=(x²-1/x²)³.(x-1/x)
=(x6-3x²+3/x²-1/x6).(x-1/x)
=x7-x5-3x³+3x+3/x-3/x³-1/x5+1/x7
=(x7+1/x7)-(x5+1/x5)-3(x³+1/x³)+3(x+1/x)
8.cos³θ.16.i4.sin4θ=2cos7θ-2.cos5θ-3.2.cos3θ+3.2.cosθ
⇒64.cos3θ.sin4θ=cos7θ-cos5θ-3.cos3θ+3.cosθ[i4=1]
⇒cos³θ.sin4θ=1/64[cos7θ-cos5θ-3.cos3θ+3.cosθ]
(5) Let x=cosθ+isinθ; xn=cosnθ+isinnθ
1/x=cosθ-i.sinθ;1/xn=cosnθ-i.sinnθ
x+1/x=2.cosθ––––––(1)
xn+1/xn=2.cosnθ–––––(2)
x-1/x=2i.sinθ;xn-1/xn=2i.sinnθ
(2i.sinθ)5=(x-1/x)5
=x5+5c1.x4.(-1/x)+5c2.x³.1/x²+5c3.x²(-1/x)³
+5c4.x.1/x4+5c5.(-1/x)5
=x5-5.x³+10.x-10.1/x+5.1/x³-1/x5
=(x5-1/x5)-5.(x³-1/x³)+10.(x-1/x)
25.i5.sin5θ=2i.sin5θ-5.2isin3θ+10.2i.sinθ
⇒24.i.sin5θ=i.sin5θ-5.i.sin3θ+10.i.sinθ[i5=i4.i=1.i=i]
⇒16.sin5θ=sin5θ-5.sin3θ+10.sinθ––––(3)
Replace θ by π/2-θ in (3)
16.sin5(π/2-θ)=sin5(π/2-θ)-5.sin3(π/2-θ)+10.sin(π/2-θ)
⇒16.cos5θ=sin(5π/2-5θ)-5.sin(3π/2-3θ)+10.cosθ
=sin(2π+π/2-5θ)-5.(-cos3θ)+10.cosθ [sin(3π/2-θ)=-cosθ]
=sin(π/2-5θ)+5.cos3θ+10.cosθ [sin(2π+θ)=sinθ]
⇒16.cos5θ=cos5θ+5.cos3θ+10.cosθ––––(4)
Now to solve cos5θ+5.cos3θ+10.cosθ=1/2
⇒16.cos5θ=1/2 using (4)
⇒cos5θ=1/32=1/25
⇒cosθ=1/2
⇒θ=2nπ±π/3
(6) (i) Let x=cosθ+isinθ; xn=cosnθ+isinnθ
1/x=cosθ-i.sinθ;1/xn=cosnθ-i.sinnθ
x+1/x=2.cosθ––––––(1)
xn+1/xn=2.cosnθ–––––(2)
x-1/x=2i.sinθ;xn-1/xn=2i.sinnθ
(2.cosθ)².(2i.sinθ)5=(x+1/x)².(x-1/x)5
=(x+1/x)².(x-1/x)².(x-1/x)³
=(x²-1/x²)²(x³-3.x².1/x3.x.1/x²-1/x³)
=(x4+1/x4-2)(x³-3x+3/x-1/x³)
=x7-3.x5+3.x³-x+1/x-3/x³+3/x5-1/x7
-2.x³+6x-6/x+2/x³
=(x7-1/x7)-3(x5-1/x5)+(x³-1/x³)+5(x-1/x) ⇒2².cos²θ.25.i5.sin5θ=2.i.sin7θ-3.2i.sin5θ+2i.sin3θ+5.2i.sinθ
⇒26.cos²θ.sin5θ=sin7θ-3.sin5θ+sin3θ+5.sinθ[i5=i4.i=1.i=i]
⇒cos²θ.sin5θ=1/64[sin7θ-3.sin5θ+sin3θ+5.sinθ]
(ii) (2.cosθ)².(2isinθ)6=(x+1/x)².(x-1/x)6
=(x+1/x)²(x-1/x)².(x-1/x)
=(x4+1/x4-2).(x4-4.x²+6-4/x²+1/x4)
=(x8+1/x8)-4(x6+1/x6)+4(x4+1/x4)+4(x²+1/x²)-10
⇒2².cos²θ.26.-1.sin6θ=2.cos8θ-4.2.cos6θ+4.2.cos4θ
+4.2.cos2θ-10
⇒-27.cos²θ.sin6θ=cos8θ-4.cos6θ+4.cos4θ+4.cos2θ-5
⇒sin6θ.cos²θ=-1/27[cos8θ-4.cos6θ+4.cos4θ+4.cos2θ-5]
(7) 24.cos4θ.(2i.sinθ)6
=(x+1/x)4.(x-1/x)6
=(x²-1/x²)4.(x-1/x)²
=(x8-4.x4+6-4/x4+1/x8).(x²+1/x²-2)
=(x10+1/x10)-3.(x6+1/x6)+8(x4+1/x4)+2(x²+1/x²)-12
⇒24.cos4θ.26.i6.sin6θ
=2.cos10θ-3.2.cos6θ+8.2.cos4θ-2.2cos8θ+2.2cos2θ-12
⇒2³.cos4θ.26.-1.sin6θ
=cos10θ-2.cos8θ-3.cos6θ+8.cos4θ+2.cos2θ-6
⇒29.cos4θ.sin6θ=-cos10θ+2.cos8θ+3.cos6θ-8.cos4θ-2.cos2θ+6
(8) (2cosθ).(2i.sinθ)²
=(x+1/x).(x-1/x)²
=(x+1/x).(x²+1/x²-2)
=(x³+1/x³)-(x+1/x)
=2.cos3θ-2.cosθ
⇒(2cosθ).4.i².sin²θ=2.cos3θ-2.cosθ
⇒-8.cosθ.sin²θ=2.cos3θ-2.cosθ
⇒-4.cosθ.sin²θ=cos3θ-cosθ
⇒cosθ.sin²θ=¼(cosθ-cos3θ)
(9) (2i.sinθ)(2cosθ)5
=(x-1/x)(x+1/x)5
=(x-1/x)(x5+5c1.x4.1/x+5c2.x³.1/x²+5c3.x².1/x³+5c4.x.1/x4+1/x5)
=(x-1/x)(x5+5x³+10x+10/x+5/x³+1/x5)
=x6+5x4+10x²+10+5/x²+1/x4-x4-5x²-10-10/x²-5/x4-1/x6
=(x6-1/x6)+4(x4-1/x4)+5(x²-1/x²)
=2i.sin6θ+4.2i.sin4θ+5.2i.sin2θ
⇒sinθ.25.cos5θ=sin6θ+4sin4θ+5sin2θ
⇒sinθ.cos5θ=1/32[sin6θ+4sin4θ+5sin2θ]
(10) (2isinθ)7.(2cosθ)³
=(x-1/x)7(x+1/x)³
=(x-1/x)³(x+1/x)³(x-1/x)4
=(x²-1/x²)³(x-1/x)4
=(x6-3x²+3/x²-1/x6)(x4-4x²+6-4/x²+1/x4)
=x10-4x8+6x6-4x4+x²-3x6+12x4-18x²+12-3/x²+3x²-12+18/x²
-12/x4+3/x6-1/x²+4/x4-6/x6+4/x8-1/x10
=(x10-1/x10)-4(x8-1/x8)+3(x6-1/x6)+8(x4-1/x4)-14(x²-1/x²)
=2i.sin10θ-4.2i.sin8θ+3.2i.sin6θ+8.2i.sin4θ-14.2i.sin2θ
⇒26.i6.sin7θ.2³.cos³θ
=sin10θ-4.sin8θ+3.sin6θ+8.sin4θ-14.sin2θ
⇒29.-1.sin7θ.cos³θ=sin10θ-4.sin8θ+3.sin6θ+8.sin4θ-14.sin2θ
⇒sin7θ.cos³θ=1/512[4.sin8θ-sin10θ-3.sin6θ-8.sin4θ+14.sin2θ]
(11) Given cos5θ=A.cosθ+B.cos3θ+C.cos5θ
Replacing θ by π/2-θ in the above
cos5(π/2-θ)=A.cos(π/2-θ)+B.cos3(π/2-θ)+C.cos5(π/2-θ)
⇒sin5θ=A.sinθ+B.(-sin3θ)+C.cos(2π+π/2-5θ) [cos(270-θ)=-sinθ]
=A.sinθ-B.sin3θ+C.cos(π/2-5θ) [cos(2π+θ)=cosθ]
= A.sinθ-B.sin3θ+C.sin5θ
(12) L.H.S=sin6θ+cos6θ
=1/8[8.sin6θ+8.cos6θ] [multiplying nr and dr by 8]
=1/8[(2.sin²θ)³+(2.cos²θ)³]
=1/8[(1-cos2θ)³+(1+cos2θ)³]
=1/8[2+6.cos²2θ] –––(1) [(a-b)³+(a+b)³=2a.(a²+3b²)]
Now cos4θ=cos2(2θ)=2.cos²(2θ)-1
⇒1+cos4θ=2.cos²(2θ)
⇒6.cos²(2θ)=3+3.cos4θ [multiplying both sides by 3]
Using the above result in (1)
sin6θ+cos6θ=1/8[2+3+3.cos4θ]=1/8[5+3.cos4θ]
Exercise XII
-––––––––––
(1) (a) Ltx↦0(tanx-sinx)/sin³x
tanx-sinx=(x-x³/3+2x5/15-...)-(x-x³/3!+x5/5!-...)
= x³(1/3+1/6)+terms containing higher power of x––––(1)
sin³x=(x-x³/3!+x5/5!-...)³
=x³(1-x²/3!+x4/5!-...)³-–––(2)
(1)/(2)⇒(tanx-sinx)/sin³x=1∕3+1/6
Ltx↦0
(b) cotθ+cot2θ=cosθ/sinθ + cos2θ/sin2θ
=[sinθ.cos2θ+cosθ.sin2θ]/sinθ.sin2θ
=sin3θ/sinθ.sin2θ
[cotθ+cot2θ]/cot3θ=[sin3θ/sinθ.sin2θ]/cot3θ
=[sin3θ/sinθ.sin2θ]/[cos3θ/sin3θ]
=sin²3θ/sinθ.sin2θ.cos3θ
sin²3θ=[3θ-(3θ)³/3!+...]²
=[3θ-27.θ³/6]²
=[3θ-9θ³/2]²
=9.θ²+81.θ6/4-27.θ4
sinθ.sin2θ=[θ-θ³/3!+...].[2θ-(2θ)³/3!+...]
=2.θ²-8.θ4-2.θ4/6
Ltθ↦0sin²3θ/sinθ.sin2θ
=Ltθ↦0 9.θ²/2.θ² (omitting higher powers of θ)
=9/2
(c) tanx-sinx
=(x+x³/3+2x5/15+...)-(x-x³/3!+x5/5!-...)
=x³/3+2x5/15+x³/6-x5/120
=x³/3+x³/6(omitting higher powers of x)
=x³/2
(tanx-sinx)/x³=x³/2.x³=½
Ltx↦0(tanx-sinx)/x³=½
(d) cos²θ-2.cos²3θ
= cos²θ-2.[4.cos³θ-3.cosθ]²
= cos²θ-2.[cosθ(4.cos²θ-3)]²
= cos²θ-2.cos²θ(4cos²θ-3)²
= cos²θ[1-2(4cos²θ-3)²]
= cos²θ[1-2(4(1-sin²θ)-3)²]
= cos²θ[1-2(4-4sin²θ-3)²]
= cos²θ[1-2(1-4sin²θ)²]
= cos²θ[1-2(1+16sin4θ-8sin²θ)]
= cos²θ[16sin²θ-32sin4θ-1]
= (1+sinθ)(1-sinθ)(16sin²θ-32sin4θ-1)
(1-sinθ)/(cos²θ-2cos²3θ)=(1-sinθ)/ (1+sinθ)(1-sinθ)(16sin²θ-32sin4θ-1)
=1/(1+sinθ)(16sin²θ-32sin4θ-1)
Ltθ↦π/2(1-sinθ)/(cos²θ-2cos²3θ)=1/(1+1)(16-32-1)
=-1/34
(e) (cosθ-sin2θ)/cos3θ
=(cosθ-2sinθ.cosθ)/(4cos³θ-3cosθ)
=cosθ[1-2sinθ]/cosθ[4cos²θ-3]
=(1-2sinθ)/(4cos²θ-3)
Ltθ↦π/2 (cosθ-sin2θ)/cos3θ
= Ltθ↦π/2(1-2sinθ)/(4cos²θ-3)
=(1-2)/(0-3)=1/3
(f) (sin2x-2sinx)/x³
= (2sinx.cosx-2sinx)/x³
= 2sinx(cosx-1)/x³
= 2.sinx∕x.(cosx-1)/x²
cosx-1=1-x²/2+x4/24-...-1
= -x²/2 (omitting higher powers)
(cosx-1)/x²=-½
Ltx↦0(sin2x-2sinx)/x³
=Ltx↦02.sinx/x.(cosx-1)/x²
= 2.Ltx↦0sinx/x.Ltx↦0(cosx-1)/x²
= 2.1.-½
=-1
(g) cos1-cos(cosx)
= cos1-cos(1-x²/2+x4/24-....)
= cos1-cos(1-x²/2) (omitting higher powers of x)
= 2.sin([4-x²]/4).sin(-x²/4)
= -2.sin(1-x²/4).sin(x²/4)
sin(1-x²/4)=(1-x²/4)-(1-x²/4)³/3!+....
= 1-x²/4 (omitting higher powers of x)
sin(x²/4)= x²/4-(x²/4)³/3!+....
= x²/4 (omitting higher powers of x)
Hence cos1-cos(cosx)
= -2.sin(1-x²/4).sin(x²/4)
= -2.(1-x²/4).(x²/4)
= (-2+x²/2).(x²/4)
[cos1-cos(cosx)]/x²
= (-2+x²/2)/4
Ltx↦0[cos1-cos(cosx)]/x²
=-2/4=-½
(h) ex-e-x-2tan-1x
=(1+x+x²/2+x³/6+....)-(1-x+x²/2-x³/6+....)
-2(x-x³/3+x5/5-....)
=1+x+x²/2+x³/6-1+x-x²/2+x³/6-2x+2x³/3
(omitting higher powers)
=2x³/6+2x³/3
= x³
log(1+x)-log(1-x)-2x
=(x-x²/2+x³/3-....)-(-x-x²/2-x³/3-....)-2x
=2x³/3
Ltx↦0[ex-e-x-2tan-1x]/[log(1+x)-log(1-x)-2x]
Ltx↦0[x³÷2x³/3]=3/2
(2) x-sinx
= x-(x-x³/3!+x5/5!-....)
= x³/6 (omitting higher powers)
1-cosx
= 1-(1-x²/2!+x4/4!-....)
= x²/2(omitting higher powers)
sinx(1-cosx)
=(x-x³/3!+x5/5!-....)(x²/2)
=x³/2 (omitting higher powers)
Ltx↦0[x-sinx]/sinx(1-cosx)
Ltx↦0[x³/6]÷[x³/2]
= 1/3
(3) logcosx
=log[1-x²/2!+x4/4!-....]
=log[1-(x²/2!-x4/4!)]
=log[1-(x²/2-x4/24)]
Let y=x²/2-x4/24
log(1-y)
=-y-y²/2-....
=-(x²/2-x4/24)-(x²/2-x4/24)²/2-....
[x²/2-x4/24]²
=x4/4+x8/24.24-2x6/48
=x4/4 (omitting higher powers)
[x²/2-x4/24]²/2=x4/8
logcosx=log(1-y)
=-[x²/2-x4/24]-x4/8
= -x²/2-x4/12
¼.cos2x
=¼[1-(2x)²/2!+(2x)4/4!-....]
=¼[1-4x²/2+16x4/24] (omitting higher powers)
=¼-x²/2+x4/6
Hence logcosx+¼-¼.cos2x
=-x²/2-x4/12+¼-¼+x²/2-x4/6
=-x4/4
[logcosx+¼-¼.cos2x]/x4= -¼
Lt ↦0[logcosx+¼-¼.cos2x]/x4= -¼
(4) sinx=x-x³/6+x5/120-....
sin(sinx)=sin(x-x³/6+x5/120-....)
= (x-x³/6+x5/120) -1∕6.(x-x³/6+x5/120)3
+1∕120.(x-x³/6+x5/120)5
(x-x³/6+x5/120)³= x³-x5/2 (omitting powers higher than x5)
(x-x³/6+x5/120)5 = x5 (omitting powers higher than x5)
Hence sin(sinx) = (x-x³/6+x5/120) -1∕6.(x-x³/6+x5/120)3
+ 1∕120.(x-x³/6+x5/120)5
= x-x³/6+x5/120 - 1∕6.(x³-x5/2) + 1/120.x5
= x-x³/3+x5/10
sin(sinx)-sinx+x³/6
= x - x³/3 + x5/10 - (x-x³/6+x5/120) + x³/6
= 11x5/120
Lt x↦0[sin(sinx)-sinx+x³/6]/x5
= 11/120
[Note: As per text book last term in the Nr is x²/6. It should be
x³/6 to get the above solution.]
(5) ex-1+log(1-x)
= 1+x/1!+x²/2!+x³/3!+....-1-x-x²/2-x³/3-....
= -x³/6
Ltx↦0[ex-1+log(1-x)]/x³
= -x³/6÷x³
= -1/6
(6) a - θ.sinθ-b.cosθ
= a - θ[θ-θ³/3!+θ5/5!-....]-b[1-θ²/2!+θ4/4!-θ6/6!+....]
= a - θ² + θ4/6 - θ6/120 - b +b.θ²/2 - b.θ4/24 + b.θ6/720
[a - θ.sinθ-b.cosθ]/θ4 = 1/12
⇒ (a-b)/θ4 + (b/2-1)/θ² + (1/6 - b/24) + θ²(b/720-1/120)
= 1/12
1/6 - b/24 = 1/12 ⇒ b= 2
a-b=0 ⇒ a = 2
(7) θ[a+bcosθ]-c.sinθ
= θ[a+b(1-θ²/2!+θ4/4!-θ6/6!+....)]
-c.[θ-θ³/3!+θ5/5!-θ7/7!+....]
= aθ + bθ[1-θ²/2!+θ4/4!-θ6/6!+....]
-c.[θ-θ³/3!+θ5/5!-θ7/7!+....]
= aθ + bθ[1-θ²/2!+θ4/4!-θ6/6!+....]
-cθ[1-θ²/3!+θ4/5!-θ6/7!+....]
1/θ5[ θ[a+bcosθ]-c.sinθ]
= a/θ4 + b/θ4[1-θ²/2!+θ4/4!-θ6/6!+....]
-c/θ4[1-θ²/3!+θ4/5!-θ6/7!+....]
= 1/θ4(a+b-c)-b/θ².2! + b/4!-b.θ²/6! +c/θ².3! -c/5! +c.θ²/7!
= (a+b-c)/θ4 + [-1/2! + c/3!]/θ²
+ b/4! - c/5! + θ².[c/7! - b/6!]
If Ltθ↦0[θ(a+b.cosθ)-c.sinθ]/θ5 = 1
then b/4! - c/5! = 1 ⇒5b-c=120 ––– (1)
Also c/3! - b/2! = 0 ⇒ 3b - c = 0 –––(2)
Solving (1) and (2) b = 60
and c = 180
a + b - c = 0 ⇒ a = 120
(8) Let θ = π/2 - α
when θ = π/2 nearly, α = 0
(sinθ)1/n = [sin(π/2-α)]1/n
= (cosα)1/n
= [(1-tan²(α/2))/(1+tan²(α/2))]1/n
= [1-tan²(α/2)/n]/[1+tan²(α/2)/n]
[expanding and neglecting higher powers]
= [n-tan²(α/2)]/[n+tan²(α/2)]
Nr = n-tan²(α/2)
Nr.cos²(α/2) = ncos²(α/2)-sin²(α/2)
= n(1+cosα)/2-(1-cosα)/2–––(1)
Dr = n+tan²(α/2)
Dr.cos²(α/2) = n.cos²(α/2) + sin²(α/2)
= n(1+cosα)/2 + (1-cosα)/2 ––– (2)
(1)÷(2) gives
[n-tan²(α/2)]/[n+tan²(α/2)]
= [n(1+cosα)-(1-cosα)]/[n(1+cosα)+(1-cosα)]
=( n+n.cosα-1+cosα)/(n+n.cosα+1-cosα)
= [(n-1)+cosα.(n+1)]/[(n+1)+cosα.(n-1)]
= [(n-1)+(n+1).sinθ]/[(n+1)+(n-1).sinθ] since α=π/2-θ
It is possible only when n>1
(9) y=perimeter of a regular n-gon inscribed in a circle of radius r
= 2nr.sin(π/n)
x= perimeter of a regular n-gon circumscribed to the circle
= 2nr.tan(π/n)
(x+2y)/3 = [2nr.tan(π/n)+2.2nr.sin(π/n)]/3
Ltn↦∞[2nr.tan(π/n)+2.2nr.sin(π/n)]/3
Let n=1/h
As n↦∞, h↦0
Lth↦0[2.r/h.tan(hπ)+4.r/h.sin(hπ)]/3
= 1∕3[2r.Lth↦0tan(hπ)/h] + 1/3[4r.Lth↦0sin(hπ)/h]
Lth↦0tan(hπ)/h
Lth↦0[hπ-(hπ)³/3+2(hπ)5/15+...]/h=Lth↦0[π-h².π³/3+2.h4.π5/15= πLth↦0sin(hπ)/h=Lth↦0[hπ-(hπ)³/6+(hπ)5/120]/h= Lth↦0[π-h²π³/6+h4π5/120]= π(x+2y)/3=1/3(2rπ)+1/3(4rπ)= 2πr
(10) 8.sin(θ/2)
= 8[θ/2-(θ/2)³/3!+(θ/5)5/5!]= 8[θ/2 - θ³/48 + θ5/32.120]= 4θ - θ³/6 + θ5/480sinθ = θ - θ³/3! + θ5/5! = θ - θ³/6 + θ5/1208.sin(θ/2) - sinθ= 4θ - θ³/6 + θ5/480 - θ + θ³/6 - θ5/120= 3θ - 3θ5/4801/3[8.sin(θ/2) - sinθ]= θ - θ5/480Hence the error in taking 1/3[8.sin(θ/2) - sinθ] as θ is θ5/480
(11) 3.sinθ/(2+cosθ)= 3.sinθ/(2+1-θ²/2! + θ4/4!)= 3(θ-θ³/3!+θ5/5!)/(3-θ²/2!+θ4/4!)= 3(θ-θ³/6+θ5/120)/3(1-θ²/6+θ4/72)= (θ-θ³/6+θ5/120).(1-θ²/6+θ4/72)-1= (θ-θ³/6+θ5/120).(1-[θ²/6-θ4/72])-1= (θ-θ³/6+θ5/120).[1+(θ²/6-θ4/72)+(θ²/6-θ4/72)²]= (θ-θ³/6+θ5/120)(1+θ²/6-θ4/72+θ4/36) [omitting powers higher than θ5]= (θ-θ³/6+θ5/120)(1+θ²/6+θ4/72)= θ + θ5/72 - θ5/36 + θ5/120= θ + θ5/120 - θ5/72= θ - θ5/180
(12) a.sin2θ+b.sin3θ= a[2θ-(2θ)³/3!+(2θ)5/5!]+ b[3θ - (3θ)³/3! + (3θ)5/5!]= θ[2a+3b] - 8a.θ³/6 - 27b.θ³/6+ higher powers of θ= θ[2a+3b] - θ³/6[8a+27b] + higher powers of θAs θ is small and the expression is equal to θ,2a + 3b = 1–––(1)8a + 27b = 0–––(2)Solving (1) and (2) we geta = 9/10 and b = -4/15Sub a, b values in a.sin2θ+b.sin3θwe get9/10[2θ-(2θ)³/3!+(2θ)5/5!]-4/15[3θ - (3θ)³/3! + (3θ)5/5!]=θ-3θ5/10Hence proved
(13) sin(α+δ) - sinα= sinα.cosδ + cosα.sinδ - sinα= sinα.1 + cosα.δ - sinα[δ is the radian measure of 1’]= sinα +δ.cosα-sinα= δ.cosαsin(α+δ) - sinα = δ.cosα⇒ sinα.cosδ + cosα.sinδ - sinα = δ.cosα⇒ sinα.[1-δ²/2!+δ4/4!]+cosα[δ-δ³/3!+δ5/5]- sinα = δ.cosα⇒ sinα - δ²/2.sinα - sinα + δ.cosα - δ³/6.cosα-δ.cosα=0⇒ δ²/2.sinα + δ³/6.cosα = 0
(14) cos(α+x) = cosα.cosx - sinα.sinx= cosα[1-x²/2!] - sinα[x - x³/3!]= cosα - x²/2.cosα - x.sinα + x³/6.sinα= cosα - x.sinα - x²/2.cosα + x³/6.sinα
(15) Since sinθ/θ tends to 1, when θ tends to zeroso, sinθ/θ = 1 nearly, when θ is very small.Here, given that sinθ/θ = 2165/2166⇒sinθ/θ is nearly equal to 1 and hence θ isvery small.Expanding sinθ in powers of θ,we havesinθ/θ = [θ - θ³/3! + θ5/5!-...]/θ⇒2165/2166 = 1 - θ²/6, neglecting θ³ and higher powers of θ⇒ θ²/6 = 1 - 2165/2166⇒ θ² = 1/2166⇒ θ² = 6/2166 = 1/361⇒ θ= 1/√361 radians =180/π.√361⇒ θ = 180.7/22.19 degrees = 630/209 degrees = 3.01 degrees = nearly 3°
(16) Given tanθ/θ = 2524/2523⇒ [θ + θ³/3 +2θ5/15 + ...]/θ = 2524/2523⇒ 1 + θ²/3 = 2524/2523 [omitting higher powers of θ]⇒ θ²/3 = 1/2523⇒ θ²=3/2523 = 1/841⇒ θ = 1/29 radians = 180/π.29 degrees = 180.7/22.29 = 1260/638 degrees = 1.97492163 = 1°58’ approximately [1 deg = 60′]
(17) Given sin(π/6 + θ) = 0.51⇒ sin(π/6 + θ) = 0.5 nearly = ½ nearly = sin(π/6)⇒ θ is very small so that cosθ=1and sinθ=θNow, sin(π/6 + θ) = 0.51⇒ sin(π/6).cosθ + cos(π/6).sinθ = 0.51⇒ 0.5.1 + 0.87.θ = 0.51⇒ 0.87θ = 0.01⇒ θ = 1/87 radians = 0.012 radians
(18) Since tan(π/4) = 1, it follows thatθ differes from π/4 by a very small quantity, say x.Assume that θ = π/4 + x, where x is small.Since x is small, we take sinx=x and cosx=1Hence tanx = xHere tanθ = 1.0024⇒tan(π/4 + x) = 1.0024⇒[tan(π/4) + tanx]/[1 - tan(π/4).tanx] = 1.0024⇒ (1 + x)/(1 - x) = 1.0024⇒ 1 + x = (1 - x)1.0024⇒ x = 0.0024/2.0024 = 0.0012 (approximately)
(19) Given tan(π/4 - θ) = 1.001⇒ tan(π/4-θ) = 1 nearly = tan(π/4)⇒ θ is very small so that tanθ = θNow, tan(π/4-θ) = 1.001⇒ [tan(π/4) - tanθ]/[1 + tan(π/4).tanθ] = 1.001⇒ (1-θ)/(1+θ) = 1.001⇒ 1 - θ = (1 + θ)1.001⇒ θ = 0.001/2.001 = 0.0005 radians
(20) 
(21)
(22) 
(23) 
(24)
(25) Same as Q.No (22)
(26)
(27)
(28)
(29)
Exercises XIII
(1)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(2)
(3)
(4)
(5)
(6) (i)
(6)(ii)
(7a)
Therefore θ = 1/3[2nπ ± π/2] or nπ or nπ ± π¼
2. cos3θ + cos2θ + cosθ = 0
⇒ cos3θ + cosθ + cos2θ = 0
⇒ 2cos2θ.cosθ + cos2θ = 0
⇒ cos2θ[2cosθ + 1] = 0
⇒ cos2θ = 0 or 2cosθ + 1 = 0
Now cos2θ = 0 ⇒ 2θ = 2nπ ± π½ ⇒θ = nπ ± π¼
2cosθ + 1 = 0 ⇒ 2cosθ = -1 ⇒ cosθ = -½
⇒ cosθ = cos(π-π/3)
⇒ cosθ = cos(2π/3)
⇒ θ = 2nπ ± 2π/3
∴ θ = nπ ± π/4 or 2nπ ± 2π/3
3. cos2θ + 5cosθ = 2
⇒ 2cos²θ - 1 + 5cosθ -2 = 0
⇒ 2cos²θ + 5cosθ - 3 = 0
⇒ cosθ = (-5 ± √49)/4 = (-5±7)/4
⇒ cosθ = 1/2 or -3
But -3 is numerically > 1.
Hence there is no real value of θ.
∴ cosθ = 1/2 ⇒ θ = 2nπ ± π/3
4. cos3x - cos4x = cos5x - cos6x
⇒ 2sin(7x/2).sin(x/2) = 2 sin(11x/2).sin(x/2)
⇒ sin(x/2)[sin(7x/2) - sin(11x/2)] = 0
⇒ sin(x/2) = 0 ⇒x/2 = nπ ⇒ x = 2nπ
or sin(7x/2) - sin(11x/2) = 0
⇒ 2cos(9x/2).sin(-x) = 0
⇒ -2cos(9x/2).sinx = 0
⇒ cos(9x/2) = 0 ⇒9x/2 = 2nπ ± π/2 ⇒ x = (4nπ ± π)/9
sinx=0 ⇒ x = nπ
∴ x = 2nπ or nπ or (4nπ ± π)/9
5. sin2θ - sinθ = cos2θ - cosθ
⇒ 2cos(3θ/2).sin(θ/2) = 2sin(3θ/2).sin(-θ/2)
⇒ cos(3θ/2).sin(θ/2) = -sin(3θ/2).sin(θ/2)
⇒ sin(θ/2).[cos(3θ/2) + sin(3θ/2)] = 0
⇒ sin(θ/2) = 0 or cos(3θ/2) + sin(3θ/2) = 0
sin(θ/2) = 0 ⇒ θ/2 = nπ ⇒ θ = 2nπ
cos(3θ/2) + sin(3θ/2) = 0
⇒ cos(3θ/2) + cos(π/2-3θ/2) = 0
⇒ 2 cos (π/4). cos(3θ/2-π/4) = 0
⇒ (1/√2).cos(3θ/2 - π/4) = 0
⇒ cos(3θ/2 - π/4) = 0
⇒ sin(π/2 - (3θ/2 - π/4)) = 0
⇒ π/2 - 3θ/2 + π/4 = nπ
⇒ 3π/4 - 3θ/2 = nπ
⇒ θ = π/2 - 2nπ/3
∴ θ = 2nπ or π/2 - 2nπ/3
6. 2sin²x + 3cosx - 3 = 0
⇒ 2(1-cos²x) + 3 cosx - 3 = 0
⇒ 2 -2cos²x + 3 cosx - 3 = 0
⇒ -2cos²x + 3cosx - 1 = 0
⇒ 2cos²x - 3cosx + 1 = 0
⇒ cosx = (3±1)/4 = 1 or ½
cosx = 1 ⇒ x = 2nπ
cosx = ½ ⇒ x = 2nπ ± π/3
∴ x = 2nπ or 2nπ ± π/3
7. sinx + cos2x = 0.2
⇒ sinx + 1 - 2sin²x = 0.2
⇒ -2sin²x + sinx + 1 - 0.2 = 0
⇒ -2sin²x + sinx + 0.8 = 0
⇒ 2sin²x - sinx - 0.8 = 0
⇒ 20sin²x - 10sinx - 8 = 0 (multiplying by 10)
⇒ 10sin²x - 5sinx - 4 = 0 (dividing by 2)
⇒ sinx = (5 ± √185)/20 = 0.93 or -0.43
⇒ sinx = 0.93 = 68°26′
⇒ x = (-1)ⁿ.68°26′ + nπ
sinx = -0.43 = 25°28′ ⇒ x = (-1)^(n+1).25°28′ + nπ
∴ x = (-1)^n.68°26′ + n180° or x = (-1)^(n+1).25°28′ + n180°
8. 2cos3θ + 6cosθ + 1 = 0
⇒ 2cos3θ + 6cosθ = -1
⇒ cos3θ + 3cosθ = -½
⇒ 4cos³θ - 3cosθ + 3cosθ = -½
⇒ 4cos³θ = -½
⇒ 4cos³θ = -½
⇒ cos³θ = -1/8
⇒ cosθ = -½ = cos(2π/3)
⇒ θ = 2nπ ± 2π/3
9. 2sinx + cosecx = 3
⇒ 2sin²x + 1 = 3sinx [multiplying both sides by sinx]
⇒ 2sin²x - 3sinx + 1 = 0
⇒ sinx = (3±1)/4 = 1 or ½
⇒ x = (-1)ⁿπ/2 + nπ or (-1)ⁿπ/6 + nπ
10. 5sin²x + 3sinx - 2 = 0
10. 5sin²x + 3sinx - 2 = 0
⇒ sinx = (-3 ±√49)/10 = -1 or 0.4
sinx = -1 ⇒ x = (-1)^(n+1).π½ + nπ or sinx = (-1)^n.23° 34′ + n180°
11. sinθ + sin2θ = sin3θ
11. sinθ + sin2θ = sin3θ
⇒ sinθ + 2sinθ.cosθ = 3sinθ - 4sin³θ
⇒ 3sinθ - 4sin³θ - sinθ - 2sinθ.cosθ = 0
⇒ -4sin³θ + 2sinθ -2sinθ.cosθ = 0
⇒ 4sin³θ - 2sinθ + 2sinθ.cosθ = 0
⇒ 2sin³θ - sinθ + sinθ.cosθ = 0
⇒ sinθ[2sin²θ - 1 + cosθ] = 0
⇒ sinθ = 0 or 2sin²θ - 1 + cosθ = 0
sinθ = 0 ⇒ θ = nπ
2sin²θ - 1 + cosθ = 0
⇒ 2(1 - cos²θ) - 1 + cosθ = 0
⇒ 2 -2cos²θ - 1 + cosθ = 0
⇒ -2cos²θ + cosθ + 1 = 0
⇒ 2cos²θ - cosθ - 1 = 0
⇒ cosθ = (1 ± 3)/4 = 1 or -½
⇒ cosθ = 1 ⇒ θ = 2nπ
cosθ = -½ ⇒ θ = 2nπ ± 2π/3
Therefore θ = nπ or 2nπ or 2nπ ± 2π/3
12. cos2θ + 5cosθ = 2
12. cos2θ + 5cosθ = 2
⇒ 2cos²θ - 1 + 5cosθ - 2 = 0
⇒ 2cos²θ + 5cosθ - 3 = 0
⇒ cosθ = (-5±7)/4 = ½ or -3
But -3 is numerically > 1. Hence there is no real value of θ.
cosθ = ½ ⇒ θ = 2nπ ± π/3
13. 3sinx + 4cosx = 2
13. 3sinx + 4cosx = 2
Let rcosα = 3 and rsinα = 4 ⇒ r² = 25 ⇒ r = 5
cosα = 3/5 and sinα = 4/5 ⇒ α = 53°7′
The equation takes the form rcosα.sinx + rsinα.cosx = 2
⇒ r[sinx.cosα + cosx.sinα] = 2
⇒ 5.sin(x + α) = 2 ⇒ sin(x + α) = 2/5 = 0.4
⇒ sin(x + 53°7′) = 0.4
Let θ = x + 53°7′ ⇒ sinθ = 0.4 ⇒ θ = (-1)^n.23°34′ + n180°
⇒ x + 53°7′ = (-1)^n.23°34′ + n.180°
14. 2sinx + 5cosx = 3
Let rcosα = 2 and rsinα = 5 ⇒ r² = 4 + 25 = 29 ⇒ r = 5.385164807
therefore α = 68°11′
∴ The equation takes the form rcosα.sinx + rsinα.cosx = 3
⇒ r[sinx.cosα + cosx.sinα] = 3
⇒ r.sin(x + α) = 3
⇒ sin(x + α) = 3/r =3/5.385164807
⇒ x + α = (-1)^n.33°51′ n.180°
⇒ x + 68°11′ = (-1)^n.33°.51′ + n.180°
15. sinx + √3.cosx = 1
½.sinx + ½√3.cosx = ½ (multiplying both sides by ½)
⇒ sin(π/6).sinx + cos(π/6).cosx = ½
⇒ cos(x - π/6) = ½
⇒ x - π/6 = 2nπ ± π/3
⇒ x = 2nπ + π/3 + π/6 = 2nπ + π/2 or
x = 2nπ - π/3 + π/6 = 2nπ - π/6
16. 2cosx - 3sinx = 2
16. 2cosx - 3sinx = 2
⇒ 2cosx - 3√1-cos²x = 2
⇒ -3√1-cos²x = 2 - 2cosx
Squaring, 9(1-cos²x) = 4 + 4cos²x - 8cosx
⇒ 9 - 9cos²x - 4 - 4cos²x + 8cosx = 0
⇒ -13cos²x + 8cosx + 5 = 0
⇒ 13cos²x - 8cosx - 5 = 0
⇒ cosx = (8±18)/26 = (8+18)/26 =1 ⇒x = 2nπ
or cosx = -10/26 = -5/13 = -0.39 ⇒ x = ±67°2′ + n360°
∴ x = nπ or n360° ± 67°2′
17. 5cosx - 4sinx = 6
⇒ Let rcosα = 5 and rsinα = 4
∴ r² = 41 ⇒ r = 6.4
∴ α = 38°39′
The given equation takes the form rcosα.cosx - rsinα.sinx = 6
⇒ r[cosα.cosx - sinα.sinx] =6
⇒ 6.4cos(α + x) = 6 ⇒ cos(α+ x) = 6/6.4 = 0.9375
⇒ cos(38°39′ + x) = 0.9375
Let θ = x + 38°39′ ⇒ cosθ = 0.9375 ⇒ θ = 2nπ ± 20°21′
∴ x + 38°39′ =n.360° ± 20°21′
18. cosx + sinx = 2cos2x
⇒ cosx + sinx = 2(cos²x - sin²x) = 2(cosx - sinx)(cosx + sinx)
⇒ cosx + sinx - 2(cosx - sinx)(cosx + sinx) = 0
⇒ (cosx + sinx)[1 -2(cosx - sinx)] = 0
⇒ cosx + sinx = 0 ⇒1/√2.cosx + 1/√2.sinx =0 ⇒ sin(π/4).cosx + cos(π/4).sinx = 0
⇒ sin(π/4 + x) = 0 ⇒ π/4 + x = nπ ⇒ x = nπ - π/4
(or) 1 - 2cosx + 2sinx = 0
⇒ -2cosx + 2sinx = -1
⇒ 2cosx - 2sinx = 1
⇒ cosx - sinx = ½
Let rcosα = 1 and rsinα = -1
∴ tanα = -1 and α = -π/4, r = √2 = 1.414
∴ rcosα.cosx - rsinα.sinx = ½
⇒ rcos(α + x) = ½
⇒ √2.cos(x - π/4) = ½
⇒ cos(x - π/4) = 1/2√2 = 0.35
⇒ x - π/4 = 2nπ ± 69°19′
⇒ x = 2nπ ± 69°19′ + π/4
∴ x = nπ - π/4 or 2nπ ± 69°19′ + π/4
19. tanx + cotx = 3
⇒sinx/cosx + cosx/sinx = 3
⇒ (sin²x + cos²x)/sinx.cosx = 3
⇒ 1/sinx.cosx = 3
⇒ 2/2sinx.cosx = 3
⇒ 2/sin2x = 3
⇒ sin2x = ⅔ = 0.67
∴ x = 20°54′ + nπ
20. 1 + sin²θ = 3sinθ.cosθ
(1 + sin²θ)² = 9sin²θ.cos²θ (squaring both sides)
⇒ 1 + sin⁴θ + 2sin²θ = 9sin²θ.cos²θ
⇒ 1 + sin⁴θ + 2sin²θ = 9sin²θ.(1 - sin²θ)
⇒ 1 + sin⁴θ + 2sin²θ = 9sin²θ - 9sin⁴θ
⇒ 1 + sin⁴θ + 2sin²θ - 9sin²θ + 9sin⁴θ = 0
⇒ 1 + 10sin⁴θ _ 7sin²θ = 0
Let t= sin²θ . ∴ t² = sin⁴θ
∴ 1 + 10t² - 7t = 0
⇒ 10t² -5t -2t + 1 = 0
⇒ 5t(2t - 1) -1(2t - 1) = 0
⇒ (2t - 1)(5t - 1) = 0
⇒ t = ½ or t = ⅕
∴ sin²θ = ½ ⇒ sinθ = 1/√2 = 0.707213578
∴ θ = n.180° + 45°
sin²θ = ⅕ ⇒ sinθ = 1/√5 = 0.447227191 ⇒ θ = 26.56 ⇒ θ = n.180° + 26°34′
21. tanx + tan2x + tan3x = 0
Let t = tanx
∴ t + 2t/(1 - t²) + (3t - t³)/(1 - 3t²) = 0
⇒ [t(1 - t²)(1 - 3t²) + 2t(1 - 3t²) + (3t - t³)(1 - t²)]/(1 - t²)(1 - 3t²) = 0
⇒ t(1 - t²)((1 - 3t²) + 2t - 6t³ + 3t - 3t³ - t³ + t⁵ = 0
⇒ t - 3t³ - t³ + 3t⁵ + t⁵ - 10t³ + 5t = 0
⇒ 4t⁵ - 14t³ + 6t = 0
⇒ 2t⁵ - 7t³ + 3t = 0
⇒ t(2t⁴ - 7t² + 3) = 0
⇒ t = 0 (or) 2t⁴ - 7t² + 3 = 0
put t² = k ∴ 2k² - 7k + 3 = 0
⇒ k = (7±5)/4 = 3 or ½
∴ t² = 3 or t² = ½ = 0.5
⇒ t = √3 or t = 0.707106781
⇒ tanx = √3 or tanx = 0.707106781
⇒ x = 60° ± n.180° or x = 35°15′ ± n.180°
22. tanθ = 1 - sec2θ
22. tanθ = 1 - sec2θ
= 1 - (1 + tan²θ)/(1 - tan²θ) ∵ cos2θ = (1 - tan²θ)/(1 + tan²θ)
= (-2tan²θ)/(1 - tan²θ)
tanθ - tan³θ = -2tan²θ
-tan³θ + 2tan²θ + tanθ = 0
tan³θ -2tan²θ - tanθ = 0
tanθ[tan²θ - 2tanθ - 1] = 0
tanθ = 0 ⇒ θ = n.180°
tan²θ - 2tanθ - 1 = 0
⇒ tanθ = 1 ± √2 ⇒ tanθ = 2.4144 or 0.4144
⇒ θ = 67°30′+ n180° or θ = 22°30′ + n180°
23. cosec⁴θ + 9.cosec²θ -6cosec²θ.cotθ = 10
cosec⁴θ = (cosec²θ)² = (1 + cot²θ)² = 1 + cot⁴θ + 2cot²θ
9 cosec²θ = 9 + 9cot²θ
6.cosec²θ.cotθ = 6cotθ + 6cot³θ
Gn eqn.becomes
1 + cot⁴θ + 2cot²θ + 9 + 9cot²θ - 6cotθ - 6cot³θ = 10
⇒ cot⁴θ - 6cot³θ + 11cot²θ - 6cotθ = 0
⇒ cotθ[cot³θ - 6cot²θ + 11cotθ -6] = 0
⇒ cotθ = 0 ⇒ θ =n180° + 90
Let Z = cotθ
The above eqn becomes z³ - 6z²+ 11z - 6 = 0
⇒(z-1)(z² -5z + 6) = 0 ⇒z = 1,3,2
z = 1 ⇒ cotθ = 1 ⇒ tanθ = 1 ⇒θ = n180° + 45
z = 3 ⇒ cotθ = 3 ⇒ tanθ = ⅓ ⇒ θ = 18.43494865 ⇒θ = n180° + 18°26′
z = 2 ⇒ cotθ = 2 ⇒ tanθ = ½ ⇒ θ = 26.56505118 ⇒ θ = 26°33′ + n180°
24. tanA = -cot2A
24. tanA = -cot2A
⇒ sinA/cosA = -cos2A/sin2A
⇒ sinA.sin2A = -cosA.cos2A
⇒ cos2A.cosA + sin2A.sinA = 0
⇒ cos(2A - A) = 0 ⇒ cosA = 0 ⇒ A =2nπ ± π½
25 tanθ + 4cos2θ + 1 = 0
25 tanθ + 4cos2θ + 1 = 0
⇒ tanθ + 4(1 - tan²θ)/(1 + tan²θ) + 1 = 0
⇒ tanθ( 1 + tan²θ) + 4(1-tan²θ) + 1 + tan²θ = 0
⇒ tanθ + tan³θ + 4 - 4tan²θ + 1 + tan²θ = 0
⇒ tan³θ - 3.tan²θ + tanθ + 5 = 0
⇒ (tanθ + 1)(tan²θ - 4tanθ + 5) = 0
⇒ tanθ = -1 ⇒ θ = n.180° - 45
tan²θ - 4tanθ + 5 = 0
⇒ tanθ = (4±2i)/2 = 2 ± i
considering the real value
tanθ = 2 ⇒ θ = tan⁻2 = 63.43494882 = 63°26′
∴ θ = n.180° + 63°26′
26. √3.(tanθ + secθ) = 4
⇒ secθ + tanθ = 4/√3 —(1)
Multiplying Nr and Dr by secθ - tanθ
(sec²θ - tan²θ)/(secθ - tanθ) = 4/√3
⇒ 1/(secθ - tanθ) = 4/√3
⇒ secθ - tanθ = √3/4 — (2)
(1) - (2) ⇒ 2tanθ = 4/√3 - √3/4 = 13/4√3
⇒ tanθ = 13/8√3 = 0938194187
⇒ θ = 43.1735511 + nπ
⇒ θ = 43° 10′ + n180°
27 Sec²θ - (√3 + 1)tanθ + √3 - 1 = 0
⇒ 1 + tan²θ - √3tanθ - tanθ + √3 - 1 = 0
⇒ tan²θ - tanθ[√3 + 1] + √3 = 0
⇒ tanθ = (√3 + 1 + sqrt(4-2√3))/2 = 1.732050807
⇒ θ = nπ + π/3
Also tanθ = (√3 + 1 - sqrt(4-2√3))/2 = 1
⇒ θ = nπ + π/4
28. tanpx = cotqx
⇒ tanpx = tan(π/2 - qx)
⇒ px = π/2 - qx + nπ
⇒ px + qx = π/2 + nπ
⇒ x(p + q) = π(2n + 1)/2
⇒ x = π(2n + 1)/2(p + q)
29. If cos(A - B) = 1/2 and sin(A + B) = 1/2, find the smallest positive values of A and B and their most general values.
Solution : cos(A - B) = 1/2 ⇒ A - B = 60°
sin(A + B) = 1/2 ⇒ A + B = 30° or 150°
A - B = 60°
A + B = 30°
⇒ A = 45°
When A = 45, B =-15 not +ve.
∴ A - B = 60
A + B = 150
⇒ A = 105
∴ B = 45
General value of A - B = 2nπ ± π/3––(1)
General value of A + B = (-1)^m.π/6 + mπ - - (2)
(1) + (2) ⇒ A = nπ ± π/6 + (-1)^m.π/12 + mπ/2
⇒ A = π[n ± 1/6 + (-1)^m/12 + m/2]
(1) - (2) ⇒ B = π[-n ± 1/6 + (-1)^m/12 + m/2]
30. Explain why the same series of two angles are given by the equations.
30. Explain why the same series of two angles are given by the equations.
θ + π/4 = nπ + (-1)^n.π/6 and θ - π/4 = 2nπ ± π/3
Solution : Given θ + π/4 = nπ + (-1)^n.π/6 — (a)
sin(θ + π/4) = sin(nπ + (-1)^n.π/6) = ½
⇒sinθ.cosπ/4 + cosθ.sinπ/4 = ½
⇒ 1/√2 . cosπ/4 + 1/√2.cosθ = ½
⇒ sinθ + cosθ = 1/√2
Also given that θ - π/4 = 2nπ ± π/3 —(b)
cos(θ - π/4) = cos(2nπ ± π/3)
⇒ cos(θ - π/4) = ½
⇒ cosθ.cosπ/4 + sinθ.sinπ/4 = ½
⇒ cosθ.1/√2 + sinθ.1/√2 = ½
⇒ cosθ + sinθ = 1/√2
Equations (a) and (b) are general solution to the equation cosθ + sinθ = 1/√2.
The general solution can frequently be obtained in many ways. The various
forms which the result takes are merely different modes of expressing the same series of angles.
Exercises IX
(1)
Exercises IX
(1)
(i) sin6θ
cos6θ + isin6θ = (cosθ+isinθ)6
= cos6θ + 6c1 cos5θ.(isinθ) + 6c2.cos4θ.(isinθ)2 + 6c3.cos3θ(isinθ)3
+ 6c4.cos2θ.(isinθ)4 + 6c5.cosθ.(isinθ)5 + 6c6.(isinθ)6
= cos6θ + 6i.cos5θ.sinθ - 15.cos4θ.sin2θ-20i.cos3θ.sin3θ + 15.cos2θ.sin4θ + 6i.cosθ.sin5θ - sin6θ
Equating the imaginary parts, we have
sin6θ = 6.cos5θ.sinθ - 20.cos3θ.sin3θ + 6.cosθ.sin5θ
= sinθ.(6.cos5θ - 20.cos3θ.sin2θ + 6.cosθ.sin4θ)
= sinθ.[6.cos5θ - 20.cos3θ.(1 - cos2θ) + 6.cosθ.(sin2θ)2]
= sinθ.[6.cos5θ - 20.cos3θ +20.cos5 + 6.cosθ.(1 - cos2θ)2]
= sinθ.[26.cos5θ - 20.cos3θ + 6.cosθ.(1 + cos4θ - 2.cos2θ)
= sinθ.[26.cos5θ - 20.cos3θ + 6.cosθ + 6.cos5θ - 12.cos3θ]
= sinθ.[32.cos5θ - 32.cos3θ + 6.cosθ]
(ii) cos5θ
cos5θ + isin5θ = (cosθ + isinθ)5 = cos5θ + 5c1.cos4θ.(isinθ) + 5c2.cos3θ.(isinθ)2 + 5c3.cos2θ.(isinθ)3 + 5c4.cosθ.(isinθ)4 + 5c5.(isinθ)5
= cos5θ + 5.cos4
Equating the real parts, we have
cos5θ = cos
= cos
= cos
= 11.cos
= 11.cos
= 16.cos
(iii) cos9θ
cos9θ + isin9θ = (cosθ + isinθ)9
= cos9θ + 9c1.cos8θ.(isinθ) + 9c2.cos7θ.(isinθ)2 + 9c3.cos6θ(isinθ)3
+ 9c4.cos5θ.(isinθ)4 + 9c5.cos4θ.(isinθ)5 + 9c6.cos3θ.(isinθ)6
+ 9c7.cos2θ.(isinθ)7 + 9c8.cosθ.(isinθ)8 + 9c9.(isinθ)9
= cos9θ + 9i.cos8θ.sinθ - 36.cos7θ.sin2θ - 84i.cos6θ.sin3θ
+ 126.cos5θ.sin4θ + 126i.cos4θ.sin5θ - 84.cos3θ.sin6θ
- 36i.cos2θ sin7θ + 9.cosθ.sin8θ + i.sin9θ
Equating the real parts, we have
cos9θ = cos9θ - 36.cos7θ.sin2θ + 126.cos5θ.sin4θ - 84.cos3θ.sin6θ + 9.cosθ.sin8θ
= cos9θ - 36.cos7θ.(1 - cos2θ) + 126.cos5θ.(sin2θ)2 - 84.cos3θ.(sin2θ)3 + 9.cosθ.(sin2θ)4
= cos
= 37.cos
= 37.cos
= 247.cos
(vii) tan9θ
= 247.cos9θ - 540.cos7θ + 378.cos5θ - 84.cos3θ + 9.cosθ + 9.cos9θ + 36.cos5θ + 18.cos5θ - 36.cos7θ - 36.cos3θ
= 256.cos9θ - 576.cos7θ + 432.cos5θ - 120.cos3θ + 9.cosθ
(iv) cos4θ
(iv) cos4θ
cos4θ + i.sin4θ = (cosθ + isinθ)4
= cos4θ + 4c1.cos3θ.(isinθ) + 4c2.cos2θ.(isinθ)2 + 4c3.cosθ.(isinθ)3 + 4c4.(isinθ)4
= cos4θ + 4.cos3θ.(isinθ) - 6.cos2θ.sin2θ - 4i.cosθ.sin3θ + sin4θ
Equating the real parts, we have
cos4θ = cos4θ - 6.cos2θ.sin2θ + sin4θ
= cos4θ - 6.cos2θ.(1 - cos2θ) + (sin2θ)2
= cos4θ - 6.cos2θ + 6.cos4θ + (1 - cos2θ)2
= 7.cos4θ - 6.cos2θ + 1+ cos4θ - 2.cos2θ
= 8.cos4θ - 8.cos2θ + 1
(v) tan4θ
we know that tannθ = (nc1.tanθ - nc3.tan3θ + ...)/(1 - nc2.tan2θ + nc4.tan4θ - ...)
Put n=4 in the above
tan4θ = (4.tanθ - 4.tan3θ)/(1 - 6.tan2θ + tan4θ)
= 4.tanθ(1 - tan2θ)/(1 - 6.tan2θ + tan4θ)
(vi) tan5θ
(vi) tan5θ
= (5c1.tanθ - 5c3.tan3θ + 5c5.tan5θ)/(1 - 5c2.tan2θ + 5c4.tan4θ)
= (5.tanθ - 10.tan3θ + tan5θ)/(1 - 10.tan2θ + 5.tan4θ)
(vii) tan9θ
Nr = 9c1.tanθ - 9c3.tan3θ + 9c5.tan5θ - 9c7.tan7θ + tan9θ
= 9.tanθ - 84.tan3θ + 126.tan5θ - 36.tan7θ + tan9θ
Dr = 1 - 9c2.tan2θ + 9c4.tan4θ - 9c6.tan6θ + 9c8.tan8θ
= 1 - 36.tan2θ + 126.tan4θ - 84.tan6θ + 9.tan8θ
tan9θ
=(9.tanθ - 84.tan3θ + 126.tan5θ - 36.tan7θ + tan9θ)/(1 - 36.tan2θ + 126.tan4θ - 84.tan6θ + 9.tan8θ)
(2) (cosθ + isinθ)11 = cos11θ + 11c1.cos10θ.(isinθ) + 11c2.cos9θ.(isinθ)2 + 11c3.cos8θ.(isinθ)3 + ...+ 11c10.cosθ.(isinθ)10 + (isinθ)11
= cos
Equating the real parts, last term in cos11θ is -11.cosθ.sin10θ.
(cosθ + isinθ)9 = cos9θ + 9c1.cos8θ.(isinθ) + 9c2.cos7θ.(isinθ)2 + 9c3.cos6θ.(isinθ)3 + .... + 9c8.cosθ.(isinθ)8 + (isinθ)9
= cos9θ + i.9.cos8θ.sinθ - 36.cos7θ.sin2θ - i.84.cos6θ.sin3θ + .... 9.cosθ.sin8θ + i.sin9θ
Equating the imaginary parts, last term in the expansion of sin9θ is sin9θ.
(3) (a) cos7θ + isin7θ = (cosθ + isinθ)7
= cos
Equating the imaginary parts
sin7θ = 7.cos
sin7θ/sinθ = 7.cos
cos
cos
sin7θ/sinθ = 7(1 - 3.sin
= 7 - 21.sin
= 7 - 56.sin
3 (b) From (1) (iii), Equating the imaginary parts
3 (b) From (1) (iii), Equating the imaginary parts
sin9θ = 9.cos8θ.sinθ - 84.cos6θ.sin3θ + 126.cos4θ.sin5θ - 36.cos2θ.sin7θ + sin9θ
sin9θ/sinθ = 9.cos8θ - 84.cos6θ.sin2θ + 126.cos4θ.sin4θ - 36.cos2θ.sin6θ + sin8θ – (1)
sin4θ = (sin2θ)2 = (1 - cos2θ)2 = 1 + cos4θ - 2.cos2θ
sin6θ = sin4θ.sin2θ = (1 + cos4θ - 2.cos2θ).(1 - cos2θ) = 1 - cos2θ + cos4θ - cos6θ - 2.cos2θ + 2.cos4θ = 1 - 3.cos2θ + 3.cos4θ - cos6θ
sin8θ = sin6θ.sin2θ = (1 - 3.cos2θ + 3.cos4θ - cos6θ).(1 - cos2θ)
= 1 - cos2θ - 3.cos2θ + 3.cos4θ + 3.cos4θ - 3.cos6θ - cos6θ + cos8θ = 1 - 4.cos2θ + 6.cos4θ - 4.cos6θ + cos8θ
84.cos6θ.sin2θ = 84.cos6θ.(1 - cos2θ) = 84.cos6θ - 84.cos8θ
126.cos4θ.sin4θ = 126.cos4θ.(1 + cos4θ - 2.cos2θ)
= 126.cos4θ + 126.cos8θ - 252.cos6θ
36.cos2θ.sin6θ = 36.cos2θ.(1 - 3.cos2θ + 3.cos4θ - cos6θ)
= 36.cos2θ - 108.cos4θ + 108.cos6θ - 36.cos8θ
From (1),
sin9θ/sinθ = 9.cos8θ - 84.cos6θ + 84.cos8θ + 126.cos4θ - 252.cos6θ + 126.cos8θ - 36.cos2θ + 108.cos4θ - 108.cos6θ + 36.cos8θ + 1 - 4.cos2θ + 6.cos4θ - 4.cos6θ + cos8θ
= 256.cos8θ - 448.cos6θ + 240.cos4θ + 1 - 40.cos2θ
(4) (cosθ + isinθ)
case 1 : When n is even
Let n = 2m, where m is a positive integer
Then the last term in the binomial expression (1) is
(isinθ)n = in.sinnθ = (i)2m.sinnθ = (i2)m.sinnθ = (-1)m.sinnθ
= (-1)n/2.sinnθ.which is real and the last but one term is
ncn-1.cosθ.(isinθ)n-1 = n.cosθ.in-1.(sinθ)n-1
= n.cosθ.i2m-1.(sinθ)n-1
= n.cosθ.i2m-2.i.(sinθ)n-1(multiplying nr and dr by i)
= i.i2(m-1).n.cosθ.(sinθ)n-1
= i.(i2)m-1.n.cosθ.(sinθ)n-1
= i.(-1)m-1.n.cosθ.(sinθ)n-1
= i.(-1)n/2-1.n.cosθ.(sinθ)n-1 which is imaginary
Hence the last term in the expansion of cosnθ is (-1)n/2.sinnθ and the last term in the expansion of sinnθ is (-1)n/2-1.n.cosθ.(sinθ)n-1
Case (2) : When n is odd
Let n = 2m + 1, where m is a positive integer
Then (i)n = (i)2m+1 = i2m.i = (i2)m.i = (-1)m.i = (-1)(n-1)/2
(i)
(isinθ)
n
= n.cosθ.(sinθ)
Hence the last term in the expansion of cosnθ is n.cosθ.(sinθ)
and sinnθ is (-1)
(5) Given equation is tan2θ = λ.tan(θ+α)
2.tanθ/(1 - tan2θ) = λ.(tanθ + tanα)/(1 - tanθ.tanα)
⇒ 2.tanθ(1 - tanθ.tanα) = λ.(tanθ + tanα).(1 - tan²θ)
⇒ 2.tanθ - 2.tan²θ.tanα = λ.(tanθ - tan³θ + tanα - tanα.tan²θ)
⇒ 2.tanθ - 2.tan²θ.tanα = λ.tanθ - λ.tan³θ + λ.tanα - λ.tanα.tan²θ
⇒ 2.tanθ - 2.tan²θ.tanα - λ.tanθ + λ.tan³θ - λ.tanα + λ.tanα.tan²θ = 0
⇒ λ.tan³θ + λ.tanα.tan²θ - 2.tanα.tan²θ + 2.tanθ - λ.tanθ - λ.tanα = 0
⇒ λ.tan³θ + tan²θ.(λ.tanα - 2.tanα) + tanθ.(2 - λ) - λ.tanα = 0
⇒ λ.tan³θ - tanα.tan²θ.(2 - λ) + tanθ.(2 - λ) - λ.tanα = 0
This is a cubic equation is tanθ.
Its roots are given as tanθ1,tanθ2 and tanθ3.
S1 =∑tanθ1 = (2 - λ).tanα/λ
S2 = ∑tanθ1.tanθ2 = (2 - λ)/λ and S3 = tanθ1.tanθ2.tanθ3 = tanα
tan(θ1 + θ2 + θ3) = (S1 - S3)/( 1 - S2)
S1 - S3 = (2 - λ).tanα/λ - tanα
= [(2 - λ).tanα - λ.yanα]/λ
= (2.tanα - λ.tanα - λ.tanα)/λ
= (2.tanα - 2λ.tanα)/λ
= 2.tanα.(1 - λ)/λ
1 - S2 = 1 - (2 - λ)/λ
= 2.(λ-1)/λ
tan(θ1 + θ2 + θ3) = (S1 - S3)/(1-S2)
= 2.tanα.(1-λ).λ/λ.2.(λ-1)
= -tanα except when λ=1
when λ=1, fraction takes indeterminate form.
Thus tan(θ1+θ2+θ3) = -tanα = tan(-α)
⇒ θ1+θ2+θ3 = nπ-α, where n is any integer
or θ1+θ2+θ3
Thus θ1+θ2+θ3 + α is a multiple of π
(6) Given tan(θ+π∕4) = 3.tan3θ
⇒ [tanθ+tan(π/4)]/[1-tanθ.tan(π/4)] = 3.[3.tanθ-tan³θ]/[1-3.tan²θ]
⇒ (1+tanθ)/(1-tanθ) = 3.[3.tanθ-tan³θ]/[1-3.tan²θ]
Let tanθ=x
∴ (1+x)/(1-x) = 3.(3x-x³)/(1-3x²)
⇒ (1+x)(1-3x²)=3(1-x)(3x-x³)
⇒ 1-3x²+x-3x³ = 3(3x-x³-3x²+x4)
⇒ 1-3x³-3x²+x = 9x-3x³-9x²+3x4
⇒ 3x4-3x³-9x²+9x-1+3x³+3x²-x=0
⇒ 3x4+0.x³-6x²+8x-1=0
It is a 4th degree equation. Hence 4 roots
Coefficient of x³ is zero. α,β,γ and δ are the roots of the equation.
∑tanα = -(coeff. of x³)/coeff.of x4 =0
⇒ tanα+tanβ+tanγ+tanδ=0
(7) Given sin(θ+α)=a.sin2θ+b
⇒sinθ.cosα+cosθ.sinα = a.2sinθ.cosθ+b
⇒2t/(1+t²).cosα +(1-t²)/(1+t²).sinα = 2a.2t/(1+t²).(1-t²)/(1+t²) + b
where t = tan(θ/2)
⇒2t.cosα+(1-t²).sinα=4at.(1-t²)/(1+t²)+b(1+t²)
⇒2t.(1+t²).cosα+(1-t4).sinα=4at.(1-t²)+b.(1+t²)²
⇒2t(1+t²)cosα+(1-t4)sinα=4at(1-t²)+b(1+t4+2t²)
⇒2t(1+t²)cosα+(1-t4)sinα=4at(1-t²)+b(1+t4+2t²)
⇒2tcosα+2t³cosα+sinα-t4sinα=4at-4at³+b+bt4+2bt²
⇒t4(b+sinα)-2t³(cosα+2a)+2bt²+2t(2a-cosα)+b-sinα=0
Let the four roots of the above equation be t1,t2,t3 and t4.
S1=∑t1=2(2a+cosα)/(b+sinα)
S2=∑t1t2=2b/(b+sinα)
S3=∑t1t2t3=-2(2a-cosα)/(b+sinα)
S4=t1t2t3t4=(b-sinα)/(b+sinα)
S1-S3=(4a+2cosα+4a-2cosα)/(b+sinα)=8a/(b+sinα)
1-S2+S4=1-2b/(b+sinα)+(b-sinα)/(b+sinα)
=(b+sinα-2b++b-sinα)/(b+sinα)
=0/(b+sinα)=0
tan[(θ1+θ2+θ3+θ4)/2]=(S1-S3)/(1-S2+S4)
= 8a/0=∞
⇒(θ1+θ2+θ3+θ4)/2=kπ+π/2
⇒θ1+θ2+θ3+θ4=2kπ+π=(2k+1)π
(8) cos2θ+a.cosθ+b.sinθ+c=0
Let tan(θ/2)=t
sinθ=[2tan(θ/2)]/[1+tan²(θ/2)]
cosθ=[1-tan²(θ/2)]/[1+tan²(θ/2)]
Let tan(θ/2)=t
Then sinθ=2t/(1+t²) and cosθ=(1-t²)/(1+t²)
sin²θ= 4t²/(1+t²)²
cos2θ=1-2.sin²θ = 1-2.4t²/(1+t²)²
= [(1+t²)²-8.t²]/(1+t²)²
= (1+t4+2.t²-8.t²)/(1+t²)²
= (1+t4 -6.t²)/(1+t²)²
Using the above values in the given equation we have
(1+t4-6.t²)/(1+t²)² + a.(1-t²)/(1+t²)+b.2t/(1+t²) + c=0
⇒1+t4-6.t²+a(1-t²)(1+t²)+b.2t(1+t²)+c.(1+t²)²=0
⇒1+t4-6.t²+a(1-t4)+2bt+2bt³+c(1+t4+2t²)=0
⇒1+t4-6t²+a-at4+2bt+2bt³+c+ct4
⇒t4(1-a+c)+t³.2b+t².(2c-6)+t.2b+1+a+c=0-–––––(1)
This is a fourth degree equation in t
ie., tan(θ/2) giving four roots α,β,γ and δ.
ie.,tan(α/2),tan(β/2),tan(γ/2) and tan(δ/2).
By theory of equations (1) gives
S1=∑α=∑tan(α/2) = -2b/(1-a+c)
S2 = ∑αβ = ∑tan(α/2).tan(β/2) = (2c-6)/(1-a+c)
S3 = ∑αβγ = ∑tan(α/2).tan(β/2).tan(γ/2)
= -2b/((1-a+c) = S1
S4 = αβγδ = tan(α/2)tan(β/2)tan(γ/2)tan(δ/2)
= (1+a+c)/(1-a+c)
tan(α+β+γ+δ/2) = (S1-S3)/(1-S2+S4)
= (S1-S1)/(1-S2+S4)––––––(2)
1-S2+S4 = 1 - (2c-6)/(1-a+c) + (1+a+c)/(1-a+c)
= (1-a+c-2c+6+1+a+c)/(1-a+c)
= 8/(1-a+c) ≠ zero
(2)⇒ tan(α+β+γ+δ/2) = 0 = tan0
⇒(α+β+γ+δ)/2=0=nπ,n is an integer
⇒α+β+γ+δ = 2nπ
(9) Given asin2θ+bsinθ+c=0
⇒a.2sinθ.cosθ+b.sinθ+c=0
⇒2a.2t/(1+t²).(1-t²)/(1+t²)+b.2t/(1+t²)+c=0 where t =tan(θ/2)
⇒4at(1-t²)+2b(1+t²)+c(1+t²)²=0
⇒4at-4at³+2b+2bt²+c(1+t4 +2t²)=0
⇒ct4-4at³+t²(2b+2c)+4at+2b+c=0–––––(1)
This is a fourth degree equation in t.
ie., tan(θ/2) giving four roots
Let t1=tan(θ1/2),t2=tan(θ2/2),t3=tan(θ3/3) and t4=tan(θ4/4)
By theory of equations (1) gives
S1=∑t1=∑tan(θ1/2)=4a/c
S2=∑t1t2=(2b+2c)/c
S3=∑t1t2t3=-4a/c
S
1-S
S
tan[(θS1-S3)/(1-S2+S4)
=8a/c÷0=∞=tan(π/2)
⇒(θ1+θ2+θ3+θ4)/2=π/2 + nπ
⇒θ1+θ2+θ3+θ4 = 2nπ+π=π(2n+1) is an odd multiple of π
(10) a.cos2θ+b.sin2θ+c.cosθ+d.sinθ=0
⇒a.(1-2sin²θ)+b.2sinθ.cosθ+c.cosθ+d.sinθ=0––––(1)
Let t=tan(θ/2)
1-2sin²θ=1-2(2t/1+t²)²
=1-2.4t²(1+t²)²
=[(1+t²)²-8t²]/(1+t²)²
=(1+t4+2t²-8t²)/(1+t²)²
=(1+t4-6t²)/(1+t²)²
2sinθcosθ=2.2t/(1+t²).(1-t²)/(1+t²)=4t(1-t²)/(1+t²)²
c.cosθ=c.(1-t²)/(1+t²)
d.sinθ=d.2t/(1+t²)
Substituting the above values in (1) we get
a.(1+t4-6t²)+4bt.(1-t²)+c.(1-t4)+2dt.(1+t²)=0
⇒t4(a-c)+t³(2d-4b)-6at²+t(2d+4b)+a+c=0
This is a fourth degree equation in t
ie.,tan(θ/2) giving four roots t1,t2,t3 and t4.
ie., tan(α/2),tan(β/2),tan(γ/2) and tan(δ/2)
S1=∑t1=∑tan(α/2)=(4b-2d)/(a-c)
S2=∑t1t2=∑tan(α/2).tan(β/2)=-6a/(a-c)
S3=∑t1t2t3=∑tan(α/2).tan(β/2).tan(γ/2)
=-(2d+4b)/(a-c)
S4=tan(α/2).tan(β/2).tan(γ/2).tan(δ/2)
=(a+c)/(a-c)
S1-S3=(4b-2d)/(a-c)+(2d+4b)/(a-c)
=8b/(a-c)
1-S2+S4=1+6a/(a-c)+(a+c)/(a-c)=8a/(a-c)
tan(α+β+γ+δ/2)=(S1-S3)/(1-S2+S4)
=8b/(a-c)÷8a/(a-c)
= b/a
(11) Given a²cos²θ+b²sin²θ+2gacosθ+2fb.sinθ+c=0
Let t=tan(θ/2)
sinθ=2.tan(θ/2)/(1+tan²(θ/2))=2t/(1+t²)
cosθ=(1-tan²(θ/2))/(1+tan²θ/2)=(1-t²)/(1+t²)
Using the above values in the given equation we get
a²[(1-t²)/(1+t²)]²+b².[2t/(1+t²)]²+2ga.(1-t²)/(1+t²)+2fb.2t/(1+t²)+c=0
⇒t4(a²-2ga+c)+t³.4fb+t²(4b²-2a²+2c)+4fb.t+a²+2ga+c=0
This is a fourth degree equation in t
ie., tan(θ/2) giving four roots t1,t2,t3 and t4.
ie., tan(θ1/2),tan(θ2/2),tan(θ3/2) and tan(θ4/2)
By theory of equations we have
S1=∑t1=∑tan(θ1/2)=-4bf/(a²-2ga+c)
S2=∑t1t2=∑tan(θ1/2)tan(θ2/2)=(4b²-2a²+2c)/(a²-2ga+c)
S3=∑t1t2t3=∑tan(θ1/2).tan(θ2/2).tan(θ3/2)
=-4fb(a²-2ga+c)=S1
S4=∑t1t2t3t4=tan(θ1/2).tan(θ2/2).tan(θ3/2).tan(θ4/2)
=(a²+2ga+c)/(a²-2ga+c)
tan(θ1+θ2+θ3+θ4/2)=(S1-S3)/(1-S2+S4)
=(S1-S1)/(1-S2+S4)
1-S2+S4=1-(4b²-2a²+2c)/(a²-2ga+c)+(a²+2ga+c)/(a²-2ga+c)
=(4a²-4b²)/(a²-2ga+c)≠0
Hence tan(θ1+θ2+θ3+θ4/2)=0=tan0
⇒θ1+θ2+θ3+θ4/2=0=nπ,n being an integer
⇒θ1+θ2+θ3+θ4=2nπ,an even multiplier of π
(12) tanx/tan3x=tanx÷(3tanx-tan³x)/(1-3.tan²x)
=tanx.(1-3.tan²x)/(3tanx-tan³x)
=(1-3tan²x)/(3-tan²x)=n say
⇒1-3tan²x=n(3-tan²x)
⇒1-3tan²x=3n-ntan²x
⇒n.tan²x-3.tan²x=3n-1
⇒tan²x.(n-3)=3n-1
⇒tan²x=(3n-1)/(n-3)=(1-3n)/(3-n)
These two values of tan²x must be positive, and therefore
n must be greater than 3 or less than 1/3.
(13) Let x=cosθ+isinθ
1/x=cosθ-i.sinθ
x-1/x=2i.sinθ; x+1/x=2.cosθ
xn=cosnθ+i.sinnθ
1/xn=cosnθ-i.sinnθ
xn+1/xn=2.cosnθ––––––(1)
(x-1/x)8=(2i.sinθ)8
(x-1/x)8=x8+8c1.x7.(-1/x)+8c2.x6.(-1/x)²+8c3.x5.(-1/x)³
+8c4.x4.(-1/x)4+8c5.x³.(-1/x)5+8c6.x².(-1/x)6
+8c7.x.(-1/x)7+8c8.(-1/x)8
= x8-8.x6+28.x4-56.x²+70-56.(1/x²)+28.(1/x4)-8.(1/x6)+1/x8
(15)(ii) (1+cos10θ)/(1+cos2θ)
= x8-8.x6+28.x4-56.x²+70-56.(1/x²)+28.(1/x4)-8.(1/x6)+1/x8
= x8+1/x8 - 8.(x6+1/x6)+28.(x4+1/x4)-56.(x²+1/x²)+70
28.i8.sin8θ=2.cos8θ-8.2.cos6θ+28.2.cos4θ-56.2.cos2θ+70 using (1)
⇒27.sin8θ=cos8θ-8.cos6θ+28.cos4θ-56.cos2θ+35
⇒sin8θ=(1/128)[cos8θ-8.cos6θ+28.cos4θ-56.cos2θ+35]–––––(2)
(x+1/x)8=28.cos8θ
(x+1/x)8=x8+8c1.x7(1/x)+8c2.x6.(1/x)²+8c3.x5.(1/x)³+8c4.x4.(1/x)4
+8c5.x³(1/x)5+8c6.x².(1/x)6+8c7.x.(1/x)7+8c8(1/x)8
=x8+8.x6+28.x4+56.x²+70+56.(1/x²)+28.(1/x4)+8.(1/x6)+1/x8
= x8+1/x8 + 8.(x6+1/x6)+28.(x4+1/x4)+56.(x²+1/x²)+70
= 2.cos8θ+8.2.cos6θ+28.2.cos4θ+56.2.cos2θ+70
28.cos8θ= 2.cos8θ+8.2.cos6θ+28.2.cos4θ+56.2.cos2θ+70
⇒27.cos8θ=cos8θ+8.cos6θ+28.cos4θ+56.cos2θ+35
⇒cos8θ=(1/128).[cos8θ+8.cos6θ+28.cos4θ+56.cos2θ+35]––––(3)
(2) +(3)⇒128.cos8θ+128.sin8θ=2.cos8θ+56.cos4θ+70
⇒64.(cos8θ+sin8θ)=cos8θ+28.cos4θ+35
(14) cos3θ+sin3θ=0
(14) cos3θ+sin3θ=0
⇒4.cos³θ-3.cosθ+3.sinθ-4.sin³θ=0
⇒4(cos³θ-sin³θ)-3(cosθ-sinθ)=0
⇒4(cosθ-sinθ)(cos²θ+cosθ.sinθ+sin²θ)-3(cosθ-sinθ)=0
⇒4(cosθ-sinθ)(1+cosθ.sinθ)-3(cosθ-sinθ)=0
⇒4(1+cosθ.sinθ)-3=0
⇒4(1+cosθ.sinθ)-3=0
⇒4+4cosθ.sinθ-3=0
⇒1+2.2sinθ.cosθ=0
⇒1+2.sin2θ=0
⇒2.sin2θ=-1
⇒sin2θ=-1/2
⇒2.tanθ/(1+tan²θ)=-1/2
⇒-4.tanθ=1+tan²θ
⇒tan²θ+4.tanθ+1=0–––––(1)
⇒tanθ=(-4±√12)/2=-2+√3 or -2-√3
Let x=tanθ
(1)⇒x²+4x+1=0
⇒x=-2±√3
tan(π/6)=1/√3
⇒tan(2π/12)=1/√3
⇒2.tan(π/12)/(1-tan²π/12)=1/√3
⇒2t/(1-t²)=1/√3 where t = tan(π/12)
⇒t²+2t.√3-1=0
⇒t=-√3±2
since tan(π/12)>0,tan(π/12)=-√3+2
-tan(π/12)=-2+√3
tan(5π/12)=tan(π/4+π/6)
=[tan(π/4)+tan(π/6)]/[1-tan(π/4).tan(π/6)]
=[1+1/√3]/[1-1.1/√3]
=(√3+1)/(√3-1)
=2+√3
-tan(5π/12)=-2-√3
Hence -tan(π/12) and -tan(5π/12) are the roots of the equation x²+4x+1=0
(15) (i) 2(1+cos8θ)=2(1+cos2(4θ))
(15) (i) 2(1+cos8θ)=2(1+cos2(4θ))
=2.2cos²4θ
=(2.cos4θ)²––––(1)
From Q.no.(1)(iv)
cos4θ=8.cos4θ-8.cos²θ+1
Using the above value in (1)
2(1+cos8θ)=[2.(8cos4θ-8cos²θ+1)]²
=(16.cos4θ-16.cos²θ+2)²
= [(2cosθ)4-4.4cos²θ+2]²
=[(2cosθ)4-4.(2cosθ)²+2]²
=(x4-4x²+2)² where x = 2cosθ
(15)(ii) (1+cos10θ)/(1+cos2θ)
1+cos10θ=1+cos2(5θ)=2cos²5θ
From (1)(ii) ,cos5θ=16cos5θ-20cos³θ+5cosθ
cos²5θ=(16cos5θ-20cos³θ+5cosθ)²
=[cosθ(16cos4θ-20cos²θ+5)]²
=cos²θ(16cos4θ-20.cos²θ+5)²
2.cos²5θ=2.cos²θ(16.cos4θ-20.cos²θ+5)²
(1+cos10θ)/(1+cos2θ)=(16.cos4θ-20.cos²θ+5)²
=[(2cosθ)4-5.(2cosθ)²+5]²
=(x4-5x²+5)² where x = 2cosθ
(15)(iii) (1+cos9θ)/(1+cosθ)=2.cos²(9θ/2)/2cos²(θ/2)
(15)(iii) (1+cos9θ)/(1+cosθ)=2.cos²(9θ/2)/2cos²(θ/2)
=2cos²(9θ/2).2sin²(θ/2)/2cos²(θ/2).2sin²(θ/2) [multiplying Nr and Dr by sin²(θ/2)]
=[2cos(9θ/2).sin(θ/2)]²/[2cos(θ/2).sin(θ/2)]²
=[(sin5θ-sin4θ)/sinθ]²
sin5θ=5.cos4θ.sinθ-10.cos²θ.sin³θ+sin5θ
sin4θ=4.cos³θ.sinθ-4.cosθ.sin³θ
sin5θ-sin4θ=5.cos4θ.sinθ-10.cos²θ.sin³θ+sin5θ-4.cos³θ.sinθ+4.cosθ.sin³θ
[sin5θ-sin4θ]/sinθ=5.cos4θ-10.cos²θ.sin²θ+sin4θ-4.cos³θ+4.cosθ.sin²θ
=5.cos4θ-10.cos²θ.(1-cos²θ)+(sin²θ)²-4.cos³θ+4.cosθ(1-cos²θ)
=5.cos4θ-10.cos²θ+10.cos4θ+(1-cos²θ)²-4.cos³θ+4.cosθ-4.cos³θ
=15.cos4θ-10.cos²θ+1+cos4θ-2.cos²θ-8.cos³θ+4.cosθ
=16.cos4θ-8.cos³θ-12.cos²θ+4.cosθ+1
=(2.cosθ)4-(2.cosθ)³-3.(2.cosθ)²+2.2cos+1
(1+cos9θ)/(1+cosθ)=([sin5θ-sin4θ]/sinθ)²
=(x4-x³-3.x²+2x+1)², where x=2cosθ
(15)(iv) (1+cos7θ)/(1+cosθ)
(15)(iv) (1+cos7θ)/(1+cosθ)
=2.cos²(7θ/2)/2.cos²(θ/2)
=2.cos²(7θ/2).2.sin²(θ/2)/2.cos²(θ/2).2.sin²(θ/2)
[Multiplying Nr. and Dr by 2.sin²(θ/2)]
=[2.cos(7θ/2).sin(θ/2)]²/[2.cos(θ/2).sin(θ/2)]²
=[(sin4θ-sin3θ)/sinθ]²–––––(1)
sin4θ=4.cos³θ.sinθ-4.cosθ.sin³θ
sin3θ=3.sinθ-4.sin³θ
(sin4θ-sin3θ)/sinθ=4.cos³θ-4.cosθ.sin²θ-3+4.sin²θ
=4.cos³θ-4.cosθ.(1-cos²θ)-3+4.(1-cos²θ)
=4.cos³θ-4.cosθ+4.cos³θ-3+4-4.cos²θ
=8.cos³θ-4.cos²θ-4.cosθ+1
=(2.cosθ)³-(2.cosθ)²-2.(2cosθ)+1
=x³-x²-2x+1 , where x=2cosθ
Using the above value in (1) we get
(1+cos7θ)/(1+cosθ)=(x³-x²-2x+1)²
Exercise X
(4) Given cos4θ=½
Exercise X
-–––––––––
(1)We know that sin18°=(√5-1)/4 cos36°=(√5+1)/4 LHS=sin(π/5).sin(2π/5).sin(3π/5).sin(4π/5)
=sin36°.sin72°.sin108°.sin144°
=sin36°.sin72°.sin(180-72).sin(180-36) [sin(180-θ)=sinθ]
=sin36°.sin72°.sin72°.sin36°
=sin²36°.sin²72°
=(sin36°.sin72°)²
=(2.sin36°.sin72°/2)²
=¼(2.sin36°.sin72°)²
=¼[cos(36-72)-cos(36+72)]² [2.sinA.sinB=cos(A-B)-cos(A+B)]
=¼[cos(-36)-cos108]²
=¼[cos36-cos(90+18)]² [cos(-θ)=cosθ]
=¼[cos36-(-sin18)]² [cos(90+θ)=-sinθ]
=¼[cos36+sin18]²
=¼[(√5+1)/4+(√5-1)/4]²
=¼[(2√5/4)]²=5∕16=RHS
(2) Let θ=nπ/7
(2) Let θ=nπ/7
⇒7θ=nπ
⇒sin7θ=sin(nπ)=0
From Page no.68 Worked example 1
sin7θ=7.sinθ-56.sin³θ+112.sin5θ-64.sin7θ
sin7θ=0
⇒7.sinθ-56.sin³θ+112.sin5θ-64.sin7θ=0
⇒sinθ(7-56.sin²θ+112.sin4θ-64.sin6θ)=0
⇒sinθ=0 or 7-56.sin²θ+112.sin4θ-64.sin6θ=0
⇒64.sin6θ-112.sin4θ+56.sin²θ-7=0
⇒64.x6-112.x4+56.x²-7=0 where x=sinθ––––(1)
sinθ=0 corresponds to n=0
Therefore the roots of the equation (1) are
sin(π/7),sin(2π/7),sin(3π/7),sin(4π/7),sin(5π/7)
and sin(6π/7).
Since sin(π-θ)=sinθ
sin(6π/7)=sin(π-π/7)=sin(π/7)
sin(5π/7)=sin(π-2π/7)=sin(2π/7)
sin(4π/7)=sin(π-3π/7)=sin(3π/7)
Product of the roots
=sin(π/7).sin(2π/7).sin(3π/7).sin(4π/7).sin(5π/7).sin(6π/7)
=sin²(π/7).sin²(2π/7).sin²(3π/7)=7/64
[Negattive sign is discarded since all the terms of the
expression on the left side are positive]
(3) Let θ denote any of the angles
(3) Let θ denote any of the angles
π/10,5π/10,9π/10,13π/10 and 17π/10.
5θ=π/2,5π/2,9π/2,13π/2 and 17π/2.
=(2nπ+π/2),n=0,1,2,3,4
sin5θ=sin(2nπ+π/2)=1
⇒5.sinθ-20.sin³θ+16.sin5θ=1
⇒16.sin5θ-20.sin³θ+5.sinθ=1––––(1)
Let x=sinθ
(1)⇒16.x5-20.x³+5x=1 has five roots viz.,
sin(π/10),sin(5π/10),sin(9π/10),sin(13π/10) and sin(17π/10)
(i) Sum of the roots
=sin(π/10)+sin(5π/10)+sin(9π/10)+sin(13π/10)+sin(17π/10)
=-0/16=0––––(2)
(ii) From (2)
sin18+sin90++sin162+sin234+sin306=0 [π/10=18]
⇒sin18+1+sin(180-18)+sin(180+54)+sin(360-54)=0
⇒sin18+1+sin18-sin54-sin54=0[sin(180-θ)=sinθ,sin(180+θ)=-sinθ, sin(360-θ)=-sinθ]
⇒2.sin18+1-2.sin54=0
⇒sin18+½-sin54=0
⇒sin18+½=sin54
(4) Given cos4θ=½
⇒8.cos4θ+1-8.cos²θ=½
⇒16.cos4θ-16.cos²θ+1=0
⇒16c4-16c²+1=0 where c=cosθ
cos4θ=½
⇒4θ=cos-(½)
⇒θ=¼.cos-(½)
=¼[(π/3)+2nπ,(5π/3)+2nπ]
when n=0,θ=(π/12),(5π/12)
when n=1,θ=(7π/12),(11π/12)
After this cosine value repeats.
Hence roots of the given equation are cos(π/12),cos(5π/12),cos(7π/12) and cos(11π/12).
Let x=c²
Then the original eqn becomes 16x²-16x+1=0
Then cos²(π/12) and cos²(5π/12) are the roots of the equation
it is not necessary to mention cos2(7π/12) and cos2(11π/12)
they differ only in sign, so their squares are the same roots of the quadratic.
⇒θ=2nπ/7,n=1,2,3,4,5,6,7
⇒7θ=2nπ
⇒4θ+3θ=2nπ
⇒4θ=2nπ-3θ
⇒cos4θ=cos(2nπ-3θ)
⇒cos4θ=cos3θ
⇒2.cosθ²2θ-1=4.cos³θ-3.cosθ [cos2θ=2.cos²θ-1]
⇒2.cosθ²2θ-1=4.cos³θ-3.cosθ [cos2θ=2.cos²θ-1]
⇒2.[cos2θ]²-1=4.cos³θ-3.cosθ
⇒2.[2.cos²θ-1]²-1=4.cos³θ-3.cosθ
⇒2[4cos4θ+1-4cos²θ]-1=4cos³θ-3cosθ
⇒8cos4θ+2-8cos²θ-1-4cos³θ+3cosθ=0
⇒8cos4θ-4cos³θ-8cos²θ+3cosθ+1=0
Let x=cosθ
8x4-4x³-8x²+3x+1=0
⇒(x-1)(8x³+4x²-4x-1)=0––––––(1)
⇒x-1=0
⇒x=1
⇒cosθ=1
⇒θ=14π/7=2π
Hence the roots of the equation 8x³+4x²-4x-1=0 are the
remaining six roots namely
cos(2π/7),cos(4π/7),cos(6π/7),cos(8π/7),cos(10π/7) and cos(12π/7)
But cos(12π/7)=cos(2π-2π/7)=cos(2π/7)
cos(10π/7)=cos(2π-4π/7)=cos(4π/7)
cos(8π/7)=cos(2π-6π/7)=cos(6π/7)
Hence cos(2π/7),cos(4π/7) and cos(6π/7) are the roots
of the equation 8x³+4x²-4x-1=0
Sum of the roots=cos(2π/7)+cos(4π/7)+cos(6π/7)=-4/8
⇒1-2.sin²(π/7)+1-2.sin²(2π/7)+1-2.sin²(3π/7)=-1/2
⇒3-2[sin²(π/7)+sin²(2π/7)+sin²(3π/7)]=-1/2
⇒-2[sin²(π-π/7)+sin²(2π/7)+sin²(π-3π/7)]=-½-3 [sin(π-θ)=sinθ]
⇒-2[sin²(6π/7)+sin²(2π/7)+sin²(4π/7)]=-7∕2
⇒sin²(2π/7)+sin²(4π/7)+sin²(6π/7)=7∕4
(6) Let θ=2π/7 or 4π/7 or 6π/7 or 8π/7 or 10π/7 or 12π/7 or 14π/7
⇒θ=2nπ/7,n=1,2,3,4,5,6,7
⇒7θ=2nπ
⇒4θ+3θ=2nπ
⇒4θ=2nπ-3θ
⇒cos4θ=cos(2nπ-3θ)
⇒cos4θ=cos3θ
⇒2.cosθ²2θ-1=4.cos³θ-3.cosθ [cos2θ=2.cos²θ-1]
⇒2.cosθ²2θ-1=4.cos³θ-3.cosθ [cos2θ=2.cos²θ-1]
⇒2.[cos2θ]²-1=4.cos³θ-3.cosθ
⇒2.[2.cos²θ-1]²-1=4.cos³θ-3.cosθ
⇒2[4cos4θ+1-4cos²θ]-1=4cos³θ-3cosθ
⇒8cos4θ+2-8cos²θ-1-4cos³θ+3cosθ=0
⇒8cos4θ-4cos³θ-8cos²θ+3cosθ+1=0
Let x=cosθ
8x4-4x³-8x²+3x+1=0
⇒(x-1)(8x³+4x²-4x-1)=0––––––(1)
⇒x-1=0
⇒x=1
⇒cosθ=1
⇒θ=14π/7=2π
Hence the roots of the equation 8x³+4x²-4x-1=0 are the
remaining six roots namely
cos(2π/7),cos(4π/7),cos(6π/7),cos(8π/7),cos(10π/7) and cos(12π/7)
But cos(12π/7)=cos(2π-2π/7)=cos(2π/7)
cos(10π/7)=cos(2π-4π/7)=cos(4π/7)
cos(8π/7)=cos(2π-6π/7)=cos(6π/7)
Hence cos(2π/7),cos(4π/7) and cos(6π/7) are the roots
of the equation 8x³+4x²-4x-1=0
Put 2x=y in the above equation
y³+y²-2y-1=0 has roots 2cos(2π/7),2cos(4π/7),2cos(6π/7)
(7) Let θ=π/11,3π/11,5π/11,7π∕11,....19π/11,21π/11
(11 values)
Let z=cosθ+isinθ
z11=cos11θ+isin11θ=-1
⇒z11+1=0
⇒(z+1)(z10-z9+z8-z7+z6-z5+z4
⇒z+1=0⇒z=-1⇒θ=11π/11=π
z+1/z=2.cosθ
Let x=cosθ
Therefore z+1/z=2x
z10-z9+z8-z7+z6-z5+z4
Dividing the above equation by z
(z
z²+1/z²=(z+1/z)²-2=4x²-2
z³+1/z³=(z+1/z)³-3(z+1/z)=8x³-3.2x=8x³-6x
now
(z²+1/z²)(z³+1/z³)=z
(4x²-2)(8x³-6x)=z
32x
z
Substituting these values in (1)
32x
⇒32x
where x=cosθ has 10 values π/11,3π/11,...19π/11,21π/11
cos(21π/11)=cos(2π-π/11)=cos(π/11)
cos(19π/11)=cos(2π-3π/11)=cos(3π/11)
cos(17π/11)=cos(2π-5π/11)=cos(5π/11)
cos(15π/11)=cos(2π-7π/11)=cos(7π/11)
cos(13π/11)=cos(2π-9π/11)=cos(9π/11)
Hence the equation has five roots
cos(π/11),cos(3π/11),cos(5π/11),cos(7π/11) and cos(9π/11)
cos(π/11)+cos(3π/11)+cos(5π/11)+cos(7π/11)+cos(9π/11)=-(-16/32)=½
(sum of the roots)
(8) (i) When n is odd
nc1.tanθ-nc3.tan³θ+nc5.tan5θ...+(-1)(n-1)/2.tannθ
tan nθ= ––––––––––––––––––––––––––––––––––––––––––––––––
1-nc2.tan²θ+nc4.tan4θ...+n(-1)(n-1)/2.tann-1θ
When n is even
nc1.tanθ-nc3.tan³θ+nc5.tan5θ...+n(-1)(n-2)/2.tann-1θ
tan nθ= –––––––––––––––––––––––––––––––––––––––––––––––––––
1-nc2.tan²θ+nc4.tan4θ...+(-1)(n/2).tannθ
Let θ=nπ/7,n=0,1,2,3,4,5,6
⇒7θ=nπ
⇒tan7θ=tannπ=0
7c1.tanθ-7c3.tan³θ+7c5.tan5θ-7c7.tan7θ
⇒ –––––––––––––––––––––––––––––––––––––––=0
1-7c2.tan²θ+7c4.tan4θ-7c6.tan6θ
⇒7c1.tanθ-7c3.tan³θ+7c5.tan5θ-7c7.tan7θ=0––––(1)
7c1=7,7c3=35,7c5=21,7c7=1
(1)⇒7.tanθ-35.tan³θ+21.tan5θ-tan7θ=0
⇒tanθ(tan6θ-21.tan4θ+35.tan²θ-7)=0
Let tanθ=x
x(x6-21x4+35x²-7)=0
omitting x=tanθ=0 we have,
x6-21x4+35x²-7=0 whose roots are
tan(π/7),tan(2π/7),tan(3π/7),tan(4π/7),tan(5π/7) and tan(6π/7).
But tan(4π/7)=tan(π-3π/7)=-tan(3π/7)
tan(5π/7)=tan(π-2π/7)=-tan(2π/7) and
tan(6π/7)=tan(π-π/7)=-tan(π/7).
Product of the roots
=tan(π/7).tan(2π/7).tan(3π/7).tan(4π/7).tan(5π/7).tan(6π/7)=-7
⇒[-tan(6π/7)].tan(2π/7).[-tan(4π/7)].tan(4π/7).[-tan(2π/7)].tan(6π/7)=-7
⇒tan²(2π/7).tan²(4π/7).tan²(6π/7)=7
⇒tan(2π/7).tan(4π/7).tan(6π/7)=√7
In Q.No.5 we have proved
cos(2π/7),cos(4π/7) and cos(6π/7) are the roots of
8x³+4x²-4x-1=0
Product of the roots:
cos(2π/7).cos(4π/7).cos(6π/7)=-(-1/8)=1/8–––––(2)
To prove:
sin(2π/7).sin(4π/7).sin(6π/7)=√7/8
tan(2π/7).tan(4π/7).tan(6π/7)=√7(proved already)
⇒sin(2π/7).sin(4π/7).sin(6π/7)/cos(2π/7).cos(4π/7).cos(6π/7)=√7
⇒sin(2π/7).sin(4π/7).sin(6π/7)÷1/8=√7(using (2))
⇒sin(2π/7).sin(4π/7).sin(6π/7)=√7/8
(ii) When n is odd,
tannθ=[nc1.t-...+(-1)(n-1)/2.ncn.tn]/[1-nc2.t²+...],where t=tanθ
and is zero for θ=rπ/n
Hence the values of tan(rπ/n) for r=0 to (n-1) are the roots of
nc1.t-...-(-1)(n-1)/2.nc(n-2).t(n-2)+(-1)(n-1)/2.ncn.tn=0
⇒tn-nc2.t(n-2)+...=0
Sum of the products two together = coeff. of t(n-2)=-½n(n-1)=n(1-n)/2
(9) tan13θ=0 is satisfied by θ=nπ/13 where n is
any integer or zero.
considering the nr for tan13θ
13t-286.t³+1287.t
where t = tanθ
t(t
The factor t corresponds to θ=0
It follows that ±tan(π/13),±tan(2π/13),±tan(3π/13),±tan(4π/13),
±tan(5π/13) and ±tan(6π/13) are the roots of
t12-78.t10+715.t8-1716.t6+1287.t4-286.t²+13=0
Product of the roots
tan²(π/13).tan²(2π/13).tan²(3π/13).tan²(4π/13).tan²(5π/13).tan²(6π/13)=13
⇒tan(π/13).tan(2π/13).tan(3π/13).tan(4π/13).tan(5π/13).tan(6π/13)=√13
(10) Let θ=π/7,2π/7,3π/7,4π/7,5π/7,6π/7
(10) Let θ=π/7,2π/7,3π/7,4π/7,5π/7,6π/7
7θ=any multiple of π and hence tan7θ=0
⇒[7.tanθ-35.tan³θ+21.tan5θ-tan7θ]/[1-21.tan²θ+35.tan4θ-7.tan6θ]=0
⇒7.tanθ-35.tan³θ+21.tan5θ-tan7θ=0
⇒tanθ[7-35.tan²θ+21.tan4θ-tan6θ]=0
If tanθ=0 then θ=π=7π/7 which gives none of the
above angles.
Therefore tan6θ-21.tan4θ+35.tan²θ-7=0–––––(1)
Roots of the above equation are tan(π/7),tan(2π/7),
tan(3π/7),tan(4π/7),tan(5π/7) and tan(6π/7).
tan(π-4π/7)=tan(4π/7)=tan(3π/7) [tan(π-θ)=tanθ]
tan(5π/7)=tan(π-5π/7)=tan(2π/7)
tan(6π/7)=tan(π-6π/7)=tan(π/7)
Hence roots of (1) are tan²(π/7),tan²(2π/7) & tan²(3π/7)
Put tan²θ=x in (1)
Then x³-21.x²+35x-7=0 whose roots are
tan²(π/7),tan²(2π/7) & tan²(3π/7).
(11) Let θ denote any of the angles 2π/7,4π/7,...14π/7
(ii) cos6θ=cos6θ-6c2.cos4θ.sin²θ+6c4.cos²θ.sin4θ-6c6.sin6θ
Put x=1/z in the above equation
(11) Let θ denote any of the angles 2π/7,4π/7,...14π/7
so that 7θ=an even multiple of π=2nπ
⇒4θ+3θ=2nπ
⇒4θ=2nπ-3θ
⇒sin4θ=sin(2nπ-3θ)=-sin3θ
⇒2.sin2θ.cos2θ=-(3.sinθ-4.sin³θ)
⇒2.2sinθ.cosθ.((1-2sin²θ)=4sin³θ-3sinθ
⇒4cosθ(1-2sin²θ)=4sin²θ-3 if sinθ≠0 ie., if θ≠14π/7
⇒4√1-sin²θ(1-2sin²θ)=4sin²θ-3
Lets sinθ=x
4.√1-x²(1-2x²)=4x²-3
⇒16(1-x²)(1-2x²)²=(4x²-3)² [squaring both sides]
⇒16(1-x²)(1+4x4-4x²)=16x4+9-24x²
⇒16(1+4x4-4x²-x²-4x6+4x4)=16x4+9-24x²
⇒16(1-4x6+8x4-5x²)=16x4+9-24x²
⇒16-64x6+128x4-80x²=16x4+9-24x²
⇒16x4+9-24x²-16+64x6-128x4+80x²=0
⇒64x6-112x4+56x²-7=0
⇒(8x³)²-(4√7x²-√7)²=0
⇒(8x³+4√7x²-√7)(8x³-4√7x²+√7)=0
⇒(x³+√7x²/2-√7/8)(x³-√7x²/2+√7/8)=0––––(1)
The roots of the equation (1) are
sin(2π/7),sin(4π/7),...sin(12π/7)
sin(8π/7)=sin(2π-6π/7)=-sin(6π/7)
sin(10π/7)=sin(2π-4π/7)=-sin(4π/7)
sin(12π/7)=sin(2π-2π/7)=-sin(2π/7)
Equation (1) resolved into two equations
x³-(√7/2)x²+√7/8=0–––––(2) and
x³+(√7/2)x²-√7/8=0–––––(3)
If we replace x by -x in equation (2), we get equation (3).
So the roots of the equation (3) are the roots of the equation (2) with their signs changed.
Thus if α,β,γ are the roots of (2), then -α,-β,-γ are the roots of (3).
So to write the roots of the equation (2) we have to select one out of
sin(2π/7),sin(12π/7),one out of sin(4π/7),sin(10π/7) and one out of
sin(6π/7),sin(8π/7).
For equation (2) sum of the roots is is positive,sum of the product
of the roots taken two at a time is zero and the product of the roots
is negative.
To satisfy these requirements one root of the equation (2) should be negative and two positive.
The values of sin(2π/7),sin(4π/7),sin(6π/7) are positive
while those of sin(8π/7),sin(10π/7),sin(12π/7) are negative.
The following are the three possibilities to select the roots of (2):
sin(2π/7),sin(4π/7),sin(8π/7)––––(4)
(or) sin(2π/7),sin(6π/7),sin(10π/7)–––––(5)
(or) sin(4π/7),sin(6π/7),sin(12π/7)–––––(6)
If we select the values (4) as the roots,the sum of
If we select the values (4) as the roots,the sum of
the products of the roots taken two at a time
= sin(2π/7).sin(4π/7)+sin(2π/7).sin(8π/7)+sin(4π/7).sin(8π/7)
=½[cos(2π/7)-cos(6π/7)+cos(6π/7)-cos(10π/7)+cos(4π/7)-cos(12π/7)]
=½[cos(2π/7)-cos(10π/7)+cos(4π/7)-cos(12π/7)]
=½[cos(2π-2π/7)-cos(10π/7)+cos(2π-4π/7)-cos(12π/7)]
=½[cos(12π/7)-cos(10π/7)+cos(10π/7)-cos(12π/7)]
=½*0=0
But when we consider the sum of the products of the roots taken two
at a time by taking the values (5) and (6), none of these comes out to
be zero.
Hence set (4) represents the roots of (2).
Hence sin(2π/7),sin(4π/7),sin(8π/7) are the roots of
x³-(√7/2)x²+√/8=0
∴ sin(2π/7)+sin(4π/7)+sin(8π/7)=√7/2 and
sin(2π/7).sin(4π/7).sin(8π/7)=√7/8
(12) (i) sin9θ/sinθ
(12) (i) sin9θ/sinθ
=256cos8θ-448cos6θ+240cos4θ-40cos²θ+1
[Ref Exercise IX Q.No.3(b)]
Let θ=π/9,2π/9,3π/9 and 4π/9
in each of these cases sin9θ=0
256x8 - 448x6 + 240x4 - 40x2 + 1 = 0 where x= cosθ
Let x²=1/z
then the above equation reduces to
z
hence sec²(π/9)+sec²(2π/9)+sec²(3π/9)+sec²(4π/9)=40
⇒sec²(π/9)+sec²(2π/9)+4+sec²(4π/9)=40
⇒sec²(π/9)+sec²(2π/9)+sec²(4π/9)=36
Prod of the roots:
sec²(π/9).sec²(2π/9).sec²(3π/9).sec²(4π/9)=256
sec²(π/9).sec²(2π/9).4.sec²(4π/9)=256
⇒sec(π/9).sec(2π/9).sec(4π/9)=8
(ii) cos6θ=cos6θ-6c2.cos4θ.sin²θ+6c4.cos²θ.sin4θ-6c6.sin6θ
=cos6θ-15.cos4θ.sin²θ+15.cos²θ.sin4θ-sin6θ
=cos6θ-15.cos4θ.(1-cos²θ)+15.cos²θ.(sin²θ)²-(sin²θ)³
=cos6θ-15.cos4θ+15.cos6θ+15.cos²θ.(1-cos²θ)²-(1-cos²θ)³
=16.cos6θ-15.cos4θ+15.cos²θ.(1+cos4θ-2.cos²θ)-(1-3.cos²θ+3.cos4θ-cos6θ)
=16.cos6θ-15.cos4θ+15.cos²θ+15.cos6θ-30.cos4θ-1+3.cos²θ-3.cos4θ+cos6θ
=32.cos6θ-48.cos4θ+18.cos²θ-1
Let cos6θ=½
then θ=π/18,5π/18,7π/18,11π/18,13π/18 and 17π/18.
cos6θ=½
⇒2.cos6θ=1
⇒2.[32.cos6θ-48.cos4θ+18.cos²θ-1]=1
⇒64.cos
⇒64.cos6θ-96.cos4θ+36.cos²θ-3=0-–––(1)
-cos(11π/18)=cos(π-11π/18)=cos(7π/18)
-cos(13π/18)=cos(π-13π/18)=cos(5π/18)
-cos(17π/18)=cos(π-17π/18)=cos(π/18)
Put cos²θ=x in (1)
Then (1) becomes 64x³-96x²+36x-3=0
Hence cos²(π/18),cos²(5π/18) and cos²(7π/18)
are the roots of the above equation.
Put x=1/z in the above equation
Then 64.(1/z³)-96.(1/z²)+36.(1/z)-3=0
⇒64-96.z+36.z²-3z³=0
⇒3z³-36z²+96z-64=0
Hence sec²(π/18)+sec²(5π/18)+sec²(7π/18)=-(-36)/3=12
[x=cos²θ=1/z ⇒z=sec²θ & sum of the roots=12]
Product of the roots
=sec²(π/18).sec²(5π/18).sec²(7π/18)=-(-64)/3=64/3
⇒sec(π/18).sec(5π/18).sec(7π/18)=8/√3
Exercise XI
Exercise XI
––––––––––
(1) Let x=cosθ+isinθ; xn=cosnθ+isinnθ
1/x=cosθ-i.sinθ;1/xn=cosnθ-i.sinnθ
x+1/x=2.cosθ––––––(1)
xn+1/xn=2.cosnθ–––––(2)
Raising both sides of (1) to the power of 7, we have
(2.cosθ)7=(x+1/x)7
=x7+7c1.x6.(1/x)+7c2.x5.(1/x)²+7c3.x4.(1/x)³
+7c4.x3.(1/x)4+7c5.x².(1/x)5+7c6.x.(1/x)6+7c7.(1/x)7
=x7+7.x5+21.x³+35.x+35.1/x+21.(1/x³)+7.(1/x5)+1/x7
⇒27.cos7θ=(x7+1/x7)+7.(x5+1/x5)+21.(x³+1/x³)+35.(x+1/x)
=2.cos7θ+7.2.cos5θ+21.2.cos3θ+35.2.cosθ
⇒26.cos7θ=cos7θ+7.cos5θ+21.cos3θ+35.cosθ
(2) Let x=cosθ+isinθ; xn=cosnθ+isinnθ
(2) Let x=cosθ+isinθ; xn=cosnθ+isinnθ
1/x=cosθ-i.sinθ;1/xn=cosnθ-i.sinnθ
x+1/x=2.cosθ––––––(1)
xn+1/xn=2.cosnθ–––––(2)
Raising both sides of (1) to the power of 8, we have
(2.cosθ)
=x8+8c1.x7.(1/x)+8c2.x6.(1/x)²+8c3.x5.(1/x)³+8c4.x4.(1/x)4
+8c5.x³.(1/x)5+8c6.x².(1/x)6+8c7.x(1/x)7+8c8.(1/x)8
=x8+8.x6+28.x4+56.x²+70+56.(1/x²)+28.(1/x4)+8.(1/x6)+1/x8
=x8+1/x8+8.(x6+1/x6)+28(x4+1/x4)+56(x²+1/x²)+70
28.cos8θ=2.cos8θ+8.2.cos6θ+28.2.cos4θ+56.2.cos2θ+70
⇒27.cos8θ=cos8θ+8.cos6θ+28.cos4θ+56.cos2θ+35
(3) Let x=cosθ+isinθ; xn=cosnθ+isinnθ
1/x=cosθ-i.sinθ;1/xn=cosnθ-i.sinnθ
x+1/x=2.cosθ––––––(1)
xn+1/xn=2.cosnθ–––––(2)
(2i.sinθ)8=(x-1/x)8
=x8+8c1.x7.(-1/x)+8c2.x6.(-1/x)²+8c3.x5.(-1/x)³
+8c4.x4.(-1/x)4+8c5.x³.(-1/x)5+8c6.x².(-1/x)6
+8c7.x.(-1/x)7+8c8.(-1/x)8
=x8-8.x6+28.x4-56.x²+70-56.(1/x²)+28.(1/x4)
-8.(1/x6)+1/x8
=(x8+1/x8)-8.(x6+1/x6)+28.(x4+1/x4)-56.(x²+1/x²)+70
=2.cos8θ-8.2.cos6θ+28.2.cos4θ-56.2.cos2θ+70
⇒28.i8.sin8θ=2.cos8θ-8.2.cos6θ+28.2.cos4θ-56.2.cos2θ+70
⇒27.sin8θ=cos8θ-8.cos6θ+28.cos4θ-56.cos2θ+35
(4) Let x=cosθ+isinθ; xn=cosnθ+isinnθ
1/x=cosθ-i.sinθ;1/xn=cosnθ-i.sinnθ
x+1/x=2.cosθ––––––(1)
xn+1/xn=2.cosnθ–––––(2)
x-1/x=2i.sinθ;xn-1/xn=2i.sinnθ
(2.cosθ)³.(2i.sinθ)4=(x+1/x)³.(x-1/x)4
=(x+1/x)³.(x-1/x)³.(x-1/x)
=(x²-1/x²)³.(x-1/x)
=(x6-3x²+3/x²-1/x6).(x-1/x)
=x7-x5-3x³+3x+3/x-3/x³-1/x5+1/x7
=(x7+1/x7)-(x5+1/x5)-3(x³+1/x³)+3(x+1/x)
8.cos³θ.16.i4.sin4θ=2cos7θ-2.cos5θ-3.2.cos3θ+3.2.cosθ
⇒64.cos3θ.sin4θ=cos7θ-cos5θ-3.cos3θ+3.cosθ[i4=1]
⇒cos³θ.sin4θ=1/64[cos7θ-cos5θ-3.cos3θ+3.cosθ]
(5) Let x=cosθ+isinθ; xn=cosnθ+isinnθ
1/x=cosθ-i.sinθ;1/xn=cosnθ-i.sinnθ
x+1/x=2.cosθ––––––(1)
xn+1/xn=2.cosnθ–––––(2)
x-1/x=2i.sinθ;xn-1/xn=2i.sinnθ
(2i.sinθ)5=(x-1/x)5
=x5+5c1.x4.(-1/x)+5c2.x³.1/x²+5c3.x²(-1/x)³
+5c4.x.1/x4+5c5.(-1/x)5
=x5-5.x³+10.x-10.1/x+5.1/x³-1/x5
=(x5-1/x5)-5.(x³-1/x³)+10.(x-1/x)
25.i5.sin5θ=2i.sin5θ-5.2isin3θ+10.2i.sinθ
⇒24.i.sin5θ=i.sin5θ-5.i.sin3θ+10.i.sinθ[i5=i4.i=1.i=i]
⇒16.sin5θ=sin5θ-5.sin3θ+10.sinθ––––(3)
Replace θ by π/2-θ in (3)
16.sin5(π/2-θ)=sin5(π/2-θ)-5.sin3(π/2-θ)+10.sin(π/2-θ)
⇒16.cos5θ=sin(5π/2-5θ)-5.sin(3π/2-3θ)+10.cosθ
=sin(2π+π/2-5θ)-5.(-cos3θ)+10.cosθ [sin(3π/2-θ)=-cosθ]
=sin(π/2-5θ)+5.cos3θ+10.cosθ [sin(2π+θ)=sinθ]
⇒16.cos5θ=cos5θ+5.cos3θ+10.cosθ––––(4)
Now to solve cos5θ+5.cos3θ+10.cosθ=1/2
⇒16.cos5θ=1/2 using (4)
⇒cos5θ=1/32=1/25
⇒cosθ=1/2
⇒θ=2nπ±π/3
(6) (i) Let x=cosθ+isinθ; xn=cosnθ+isinnθ
1/x=cosθ-i.sinθ;1/xn=cosnθ-i.sinnθ
x+1/x=2.cosθ––––––(1)
xn+1/xn=2.cosnθ–––––(2)
x-1/x=2i.sinθ;xn-1/xn=2i.sinnθ
(2.cosθ)².(2i.sinθ)5=(x+1/x)².(x-1/x)5
=(x+1/x)².(x-1/x)².(x-1/x)³
=(x²-1/x²)²(x³-3.x².1/x3.x.1/x²-1/x³)
=(x4+1/x4-2)(x³-3x+3/x-1/x³)
=x7-3.x5+3.x³-x+1/x-3/x³+3/x5-1/x7
-2.x³+6x-6/x+2/x³
=(x7-1/x7)-3(x5-1/x5)+(x³-1/x³)+5(x-1/x) ⇒2².cos²θ.25.i5.sin5θ=2.i.sin7θ-3.2i.sin5θ+2i.sin3θ+5.2i.sinθ
⇒26.cos²θ.sin5θ=sin7θ-3.sin5θ+sin3θ+5.sinθ[i5=i4.i=1.i=i]
⇒cos²θ.sin5θ=1/64[sin7θ-3.sin5θ+sin3θ+5.sinθ]
(ii) (2.cosθ)².(2isinθ)6=(x+1/x)².(x-1/x)6
=(x+1/x)²(x-1/x)².(x-1/x)
=(x4+1/x4-2).(x4-4.x²+6-4/x²+1/x4)
=(x8+1/x8)-4(x6+1/x6)+4(x4+1/x4)+4(x²+1/x²)-10
⇒2².cos²θ.26.-1.sin6θ=2.cos8θ-4.2.cos6θ+4.2.cos4θ
+4.2.cos2θ-10
⇒-27.cos²θ.sin6θ=cos8θ-4.cos6θ+4.cos4θ+4.cos2θ-5
⇒sin6θ.cos²θ=-1/27[cos8θ-4.cos6θ+4.cos4θ+4.cos2θ-5]
(7) 24.cos4θ.(2i.sinθ)6
=(x+1/x)4.(x-1/x)6
=(x²-1/x²)4.(x-1/x)²
=(x8-4.x4+6-4/x4+1/x8).(x²+1/x²-2)
=(x10+1/x10)-3.(x6+1/x6)+8(x4+1/x4)+2(x²+1/x²)-12
⇒24.cos4θ.26.i6.sin6θ
=2.cos10θ-3.2.cos6θ+8.2.cos4θ-2.2cos8θ+2.2cos2θ-12
⇒2³.cos4θ.26.-1.sin6θ
=cos10θ-2.cos8θ-3.cos6θ+8.cos4θ+2.cos2θ-6
⇒29.cos4θ.sin6θ=-cos10θ+2.cos8θ+3.cos6θ-8.cos4θ-2.cos2θ+6
(8) (2cosθ).(2i.sinθ)²
=(x+1/x).(x-1/x)²
=(x+1/x).(x²+1/x²-2)
=(x³+1/x³)-(x+1/x)
=2.cos3θ-2.cosθ
⇒(2cosθ).4.i².sin²θ=2.cos3θ-2.cosθ
⇒-8.cosθ.sin²θ=2.cos3θ-2.cosθ
⇒-4.cosθ.sin²θ=cos3θ-cosθ
⇒cosθ.sin²θ=¼(cosθ-cos3θ)
(9) (2i.sinθ)(2cosθ)5
=(x-1/x)(x+1/x)5
=(x-1/x)(x5+5c1.x4.1/x+5c2.x³.1/x²+5c3.x².1/x³+5c4.x.1/x4+1/x5)
=(x-1/x)(x5+5x³+10x+10/x+5/x³+1/x5)
=x6+5x4+10x²+10+5/x²+1/x4-x4-5x²-10-10/x²-5/x4-1/x6
=(x6-1/x6)+4(x4-1/x4)+5(x²-1/x²)
=2i.sin6θ+4.2i.sin4θ+5.2i.sin2θ
⇒sinθ.25.cos5θ=sin6θ+4sin4θ+5sin2θ
⇒sinθ.cos5θ=1/32[sin6θ+4sin4θ+5sin2θ]
(10) (2isinθ)7.(2cosθ)³
=(x-1/x)7(x+1/x)³
=(x-1/x)³(x+1/x)³(x-1/x)4
=(x²-1/x²)³(x-1/x)4
=(x6-3x²+3/x²-1/x6)(x4-4x²+6-4/x²+1/x4)
=x10-4x8+6x6-4x4+x²-3x6+12x4-18x²+12-3/x²+3x²-12+18/x²
-12/x4+3/x6-1/x²+4/x4-6/x6+4/x8-1/x10
=(x10-1/x10)-4(x8-1/x8)+3(x6-1/x6)+8(x4-1/x4)-14(x²-1/x²)
=2i.sin10θ-4.2i.sin8θ+3.2i.sin6θ+8.2i.sin4θ-14.2i.sin2θ
⇒26.i6.sin7θ.2³.cos³θ
=sin10θ-4.sin8θ+3.sin6θ+8.sin4θ-14.sin2θ
⇒29.-1.sin7θ.cos³θ=sin10θ-4.sin8θ+3.sin6θ+8.sin4θ-14.sin2θ
⇒sin7θ.cos³θ=1/512[4.sin8θ-sin10θ-3.sin6θ-8.sin4θ+14.sin2θ]
(11) Given cos5θ=A.cosθ+B.cos3θ+C.cos5θ
Replacing θ by π/2-θ in the above
cos5(π/2-θ)=A.cos(π/2-θ)+B.cos3(π/2-θ)+C.cos5(π/2-θ)
⇒sin5θ=A.sinθ+B.(-sin3θ)+C.cos(2π+π/2-5θ) [cos(270-θ)=-sinθ]
=A.sinθ-B.sin3θ+C.cos(π/2-5θ) [cos(2π+θ)=cosθ]
= A.sinθ-B.sin3θ+C.sin5θ
(12) L.H.S=sin6θ+cos6θ
=1/8[8.sin6θ+8.cos6θ] [multiplying nr and dr by 8]
=1/8[(2.sin²θ)³+(2.cos²θ)³]
=1/8[(1-cos2θ)³+(1+cos2θ)³]
=1/8[2+6.cos²2θ] –––(1) [(a-b)³+(a+b)³=2a.(a²+3b²)]
Now cos4θ=cos2(2θ)=2.cos²(2θ)-1
⇒1+cos4θ=2.cos²(2θ)
⇒6.cos²(2θ)=3+3.cos4θ [multiplying both sides by 3]
Using the above result in (1)
sin6θ+cos6θ=1/8[2+3+3.cos4θ]=1/8[5+3.cos4θ]
Exercise XII
Exercise XII
-––––––––––
(1) (a) Ltx↦0(tanx-sinx)/sin³x
tanx-sinx=(x-x³/3+2x5/15-...)-(x-x³/3!+x5/5!-...)
= x³(1/3+1/6)+terms containing higher power of x––––(1)
sin³x=(x-x³/3!+x5/5!-...)³
=x³(1-x²/3!+x4/5!-...)³-–––(2)
(1)/(2)⇒(tanx-sinx)/sin³x=1∕3+1/6
Ltx↦0
(b) cotθ+cot2θ=cosθ/sinθ + cos2θ/sin2θ
(b) cotθ+cot2θ=cosθ/sinθ + cos2θ/sin2θ
=[sinθ.cos2θ+cosθ.sin2θ]/sinθ.sin2θ
=sin3θ/sinθ.sin2θ
[cotθ+cot2θ]/cot3θ=[sin3θ/sinθ.sin2θ]/cot3θ
=[sin3θ/sinθ.sin2θ]/[cos3θ/sin3θ]
=sin²3θ/sinθ.sin2θ.cos3θ
sin²3θ=[3θ-(3θ)³/3!+...]²
=[3θ-27.θ³/6]²
=[3θ-9θ³/2]²
=9.θ²+81.θ6/4-27.θ4
sinθ.sin2θ=[θ-θ³/3!+...].[2θ-(2θ)³/3!+...]
=2.θ²-8.θ4-2.θ4/6
Ltθ↦0sin²3θ/sinθ.sin2θ
=Ltθ↦0 9.θ²/2.θ² (omitting higher powers of θ)
=9/2
(c) tanx-sinx
(c) tanx-sinx
=(x+x³/3+2x5/15+...)-(x-x³/3!+x5/5!-...)
=x³/3+2x5/15+x³/6-x5/120
=x³/3+x³/6(omitting higher powers of x)
=x³/2
(tanx-sinx)/x³=x³/2.x³=½
Ltx↦0(tanx-sinx)/x³=½
(d) cos²θ-2.cos²3θ
(d) cos²θ-2.cos²3θ
= cos²θ-2.[4.cos³θ-3.cosθ]²
= cos²θ-2.[cosθ(4.cos²θ-3)]²
= cos²θ-2.cos²θ(4cos²θ-3)²
= cos²θ[1-2(4cos²θ-3)²]
= cos²θ[1-2(4(1-sin²θ)-3)²]
= cos²θ[1-2(4-4sin²θ-3)²]
= cos²θ[1-2(1-4sin²θ)²]
= cos²θ[1-2(1+16sin4θ-8sin²θ)]
= cos²θ[16sin²θ-32sin4θ-1]
= (1+sinθ)(1-sinθ)(16sin²θ-32sin4θ-1)
(1-sinθ)/(cos²θ-2cos²3θ)=(1-sinθ)/ (1+sinθ)(1-sinθ)(16sin²θ-32sin4θ-1)
=1/(1+sinθ)(16sin²θ-32sin4θ-1)
Ltθ↦π/2(1-sinθ)/(cos²θ-2cos²3θ)=1/(1+1)(16-32-1)
=-1/34
(e) (cosθ-sin2θ)/cos3θ
(e) (cosθ-sin2θ)/cos3θ
=(cosθ-2sinθ.cosθ)/(4cos³θ-3cosθ)
=cosθ[1-2sinθ]/cosθ[4cos²θ-3]
=(1-2sinθ)/(4cos²θ-3)
Ltθ↦π/2 (cosθ-sin2θ)/cos3θ
= Ltθ↦π/2(1-2sinθ)/(4cos²θ-3)
=(1-2)/(0-3)=1/3
(f) (sin2x-2sinx)/x³
(f) (sin2x-2sinx)/x³
= (2sinx.cosx-2sinx)/x³
= 2sinx(cosx-1)/x³
= 2.sinx∕x.(cosx-1)/x²
cosx-1=1-x²/2+x4/24-...-1
= -x²/2 (omitting higher powers)
(cosx-1)/x²=-½
Ltx↦0(sin2x-2sinx)/x³
=Ltx↦02.sinx/x.(cosx-1)/x²
= 2.Ltx↦0sinx/x.Ltx↦0(cosx-1)/x²
= 2.1.-½
=-1
(g) cos1-cos(cosx)
(g) cos1-cos(cosx)
= cos1-cos(1-x²/2+x4/24-....)
= cos1-cos(1-x²/2) (omitting higher powers of x)
= 2.sin([4-x²]/4).sin(-x²/4)
= -2.sin(1-x²/4).sin(x²/4)
sin(1-x²/4)=(1-x²/4)-(1-x²/4)³/3!+....
= 1-x²/4 (omitting higher powers of x)
sin(x²/4)= x²/4-(x²/4)³/3!+....
= x²/4 (omitting higher powers of x)
Hence cos1-cos(cosx)
= -2.sin(1-x²/4).sin(x²/4)
= -2.(1-x²/4).(x²/4)
= (-2+x²/2).(x²/4)
[cos1-cos(cosx)]/x²
= (-2+x²/2)/4
Ltx↦0[cos1-cos(cosx)]/x²
=-2/4=-½
(h) ex-e-x-2tan-1x
(h) ex-e-x-2tan-1x
=(1+x+x²/2+x³/6+....)-(1-x+x²/2-x³/6+....)
-2(x-x³/3+x5/5-....)
=1+x+x²/2+x³/6-1+x-x²/2+x³/6-2x+2x³/3
(omitting higher powers)
=2x³/6+2x³/3
= x³
log(1+x)-log(1-x)-2x
=(x-x²/2+x³/3-....)-(-x-x²/2-x³/3-....)-2x
=2x³/3
Ltx↦0[ex-e-x-2tan-1x]/[log(1+x)-log(1-x)-2x]
Ltx↦0[x³÷2x³/3]=3/2
(2) x-sinx
(2) x-sinx
= x-(x-x³/3!+x5/5!-....)
= x³/6 (omitting higher powers)
1-cosx
= 1-(1-x²/2!+x4/4!-....)
= x²/2(omitting higher powers)
sinx(1-cosx)
=(x-x³/3!+x5/5!-....)(x²/2)
=x³/2 (omitting higher powers)
Ltx↦0[x-sinx]/sinx(1-cosx)
Ltx↦0[x³/6]÷[x³/2]
= 1/3
(3) logcosx
(3) logcosx
=log[1-x²/2!+x4/4!-....]
=log[1-(x²/2!-x4/4!)]
=log[1-(x²/2-x4/24)]
Let y=x²/2-x4/24
log(1-y)
=-y-y²/2-....
=-(x²/2-x4/24)-(x²/2-x4/24)²/2-....
[x²/2-x4/24]²
=x4/4+x8/24.24-2x6/48
=x4/4 (omitting higher powers)
[x²/2-x4/24]²/2=x4/8
logcosx=log(1-y)
=-[x²/2-x4/24]-x4/8
= -x²/2-x4/12
¼.cos2x
=¼[1-(2x)²/2!+(2x)4/4!-....]
=¼[1-4x²/2+16x4/24] (omitting higher powers)
=¼-x²/2+x4/6
Hence logcosx+¼-¼.cos2x
=-x²/2-x4/12+¼-¼+x²/2-x4/6
=-x4/4
[logcosx+¼-¼.cos2x]/x4= -¼
Lt ↦0[logcosx+¼-¼.cos2x]/x4= -¼
(4) sinx=x-x³/6+x5/120-....
(5) ex-1+log(1-x)
(4) sinx=x-x³/6+x5/120-....
sin(sinx)=sin(x-x³/6+x5/120-....)
= (x-x³/6+x5/120) -1∕6.(x-x³/6+x5/120)3
+1∕120.(x-x³/6+x5/120)5
(x-x³/6+x5/120)³= x³-x5/2 (omitting powers higher than x5)
(x-x³/6+x5/120)5 = x5 (omitting powers higher than x5)
Hence sin(sinx) = (x-x³/6+x5/120) -1∕6.(x-x³/6+x5/120)3
+ 1∕120.(x-x³/6+x5/120)5
= x-x³/6+x5/120 - 1∕6.(x³-x5/2) + 1/120.x5
= x-x³/3+x5/10
sin(sinx)-sinx+x³/6
= x - x³/3 + x5/10 - (x-x³/6+x5/120) + x³/6
= 11x5/120
Lt x↦0[sin(sinx)-sinx+x³/6]/x5
= 11/120
[Note: As per text book last term in the Nr is x²/6. It should be
x³/6 to get the above solution.]
(5) ex-1+log(1-x)
= 1+x/1!+x²/2!+x³/3!+....-1-x-x²/2-x³/3-....
= -x³/6
Ltx↦0[ex-1+log(1-x)]/x³
= -x³/6÷x³
= -1/6
(6) a - θ.sinθ-b.cosθ
(6) a - θ.sinθ-b.cosθ
= a - θ[θ-θ³/3!+θ5/5!-....]-b[1-θ²/2!+θ4/4!-θ6/6!+....]
= a - θ² + θ4/6 - θ6/120 - b +b.θ²/2 - b.θ4/24 + b.θ6/720
[a - θ.sinθ-b.cosθ]/θ4 = 1/12
⇒ (a-b)/θ4 + (b/2-1)/θ² + (1/6 - b/24) + θ²(b/720-1/120)
= 1/12
1/6 - b/24 = 1/12 ⇒ b= 2
a-b=0 ⇒ a = 2
(7) θ[a+bcosθ]-c.sinθ
(7) θ[a+bcosθ]-c.sinθ
= θ[a+b(1-θ²/2!+θ4/4!-θ6/6!+....)]
-c.[θ-θ³/3!+θ5/5!-θ7/7!+....]
= aθ + bθ[1-θ²/2!+θ4/4!-θ6/6!+....]
-c.[θ-θ³/3!+θ5/5!-θ7/7!+....]
= aθ + bθ[1-θ²/2!+θ4/4!-θ6/6!+....]
-cθ[1-θ²/3!+θ4/5!-θ6/7!+....]
1/θ5[ θ[a+bcosθ]-c.sinθ]
= a/θ4 + b/θ4[1-θ²/2!+θ4/4!-θ6/6!+....]
-c/θ4[1-θ²/3!+θ4/5!-θ6/7!+....]
= 1/θ4(a+b-c)-b/θ².2! + b/4!-b.θ²/6! +c/θ².3! -c/5! +c.θ²/7!
= (a+b-c)/θ4 + [-1/2! + c/3!]/θ²
+ b/4! - c/5! + θ².[c/7! - b/6!]
If Ltθ↦0[θ(a+b.cosθ)-c.sinθ]/θ5 = 1
then b/4! - c/5! = 1 ⇒5b-c=120 ––– (1)
Also c/3! - b/2! = 0 ⇒ 3b - c = 0 –––(2)
Solving (1) and (2) b = 60
and c = 180
a + b - c = 0 ⇒ a = 120
(8) Let θ = π/2 - α
(8) Let θ = π/2 - α
when θ = π/2 nearly, α = 0
(sinθ)1/n = [sin(π/2-α)]1/n
= (cosα)1/n
= [(1-tan²(α/2))/(1+tan²(α/2))]1/n
= [1-tan²(α/2)/n]/[1+tan²(α/2)/n]
[expanding and neglecting higher powers]
= [n-tan²(α/2)]/[n+tan²(α/2)]
Nr = n-tan²(α/2)
Nr.cos²(α/2) = ncos²(α/2)-sin²(α/2)
= n(1+cosα)/2-(1-cosα)/2–––(1)
Dr = n+tan²(α/2)
Dr.cos²(α/2) = n.cos²(α/2) + sin²(α/2)
= n(1+cosα)/2 + (1-cosα)/2 ––– (2)
(1)÷(2) gives
[n-tan²(α/2)]/[n+tan²(α/2)]
= [n(1+cosα)-(1-cosα)]/[n(1+cosα)+(1-cosα)]
=( n+n.cosα-1+cosα)/(n+n.cosα+1-cosα)
= [(n-1)+cosα.(n+1)]/[(n+1)+cosα.(n-1)]
= [(n-1)+(n+1).sinθ]/[(n+1)+(n-1).sinθ] since α=π/2-θ
It is possible only when n>1
(9) y=perimeter of a regular n-gon inscribed in a circle of radius r
(9) y=perimeter of a regular n-gon inscribed in a circle of radius r
= 2nr.sin(π/n)
x= perimeter of a regular n-gon circumscribed to the circle
= 2nr.tan(π/n)
(x+2y)/3 = [2nr.tan(π/n)+2.2nr.sin(π/n)]/3
Ltn↦∞[2nr.tan(π/n)+2.2nr.sin(π/n)]/3
Let n=1/h
As n↦∞, h↦0
Lth↦0[2.r/h.tan(hπ)+4.r/h.sin(hπ)]/3
= 1∕3[2r.Lth↦0tan(hπ)/h] + 1/3[4r.Lth↦0sin(hπ)/h]
Lth↦0tan(hπ)/h
Lth↦0[hπ-(hπ)³/3+2(hπ)5/15+...]/h
=Lth↦0[π-h².π³/3+2.h4.π5/15
= π
Lth↦0sin(hπ)/h
=Lth↦0[hπ-(hπ)³/6+(hπ)5/120]/h
= Lth↦0[π-h²π³/6+h4π5/120]
= π
(x+2y)/3
=1/3(2rπ)+1/3(4rπ)
= 2πr
(10) 8.sin(θ/2)
= 8[θ/2-(θ/2)³/3!+(θ/5)5/5!]
= 8[θ/2 - θ³/48 + θ5/32.120]
= 4θ - θ³/6 + θ5/480
sinθ = θ - θ³/3! + θ5/5!
= θ - θ³/6 + θ5/120
8.sin(θ/2) - sinθ
= 4θ - θ³/6 + θ5/480 - θ + θ³/6 - θ5/120
= 3θ - 3θ5/480
1/3[8.sin(θ/2) - sinθ]
= θ - θ5/480
Hence the error in taking 1/3[8.sin(θ/2) - sinθ]
as θ is θ5/480
(11) 3.sinθ/(2+cosθ)
= 3.sinθ/(2+1-θ²/2! + θ4/4!)
= 3(θ-θ³/3!+θ5/5!)/(3-θ²/2!+θ4/4!)
= 3(θ-θ³/6+θ5/120)/3(1-θ²/6+θ4/72)
= (θ-θ³/6+θ5/120).(1-θ²/6+θ4/72)-1
= (θ-θ³/6+θ5/120).(1-[θ²/6-θ4/72])-1
= (θ-θ³/6+θ5/120).[1+(θ²/6-θ4/72)+(θ²/6-θ4/72)²]
= (θ-θ³/6+θ5/120)(1+θ²/6-θ4/72+θ4/36)
[omitting powers higher than θ5]
= (θ-θ³/6+θ5/120)(1+θ²/6+θ4/72)
= θ + θ5/72 - θ5/36 + θ5/120
= θ + θ5/120 - θ5/72
= θ - θ5/180
(12) a.sin2θ+b.sin3θ
= a[2θ-(2θ)³/3!+(2θ)5/5!]
+ b[3θ - (3θ)³/3! + (3θ)5/5!]
= θ[2a+3b] - 8a.θ³/6 - 27b.θ³/6
+ higher powers of θ
= θ[2a+3b] - θ³/6[8a+27b] + higher powers of θ
As θ is small and the expression is equal to θ,
2a + 3b = 1–––(1)
8a + 27b = 0–––(2)
Solving (1) and (2) we get
a = 9/10 and b = -4/15
Sub a, b values in a.sin2θ+b.sin3θ
we get
9/10[2θ-(2θ)³/3!+(2θ)5/5!]
-4/15[3θ - (3θ)³/3! + (3θ)5/5!]
=θ-3θ5/10
Hence proved
(13) sin(α+δ) - sinα
= sinα.cosδ + cosα.sinδ - sinα
= sinα.1 + cosα.δ - sinα
[δ is the radian measure of 1’]
= sinα +δ.cosα-sinα
= δ.cosα
sin(α+δ) - sinα = δ.cosα
⇒ sinα.cosδ + cosα.sinδ - sinα = δ.cosα
⇒ sinα.[1-δ²/2!+δ4/4!]+cosα[δ-δ³/3!+δ5/5]
- sinα = δ.cosα
⇒ sinα - δ²/2.sinα - sinα + δ.cosα - δ³/6.cosα-δ.cosα=0
⇒ δ²/2.sinα + δ³/6.cosα = 0
(14) cos(α+x) = cosα.cosx - sinα.sinx
= cosα[1-x²/2!] - sinα[x - x³/3!]
= cosα - x²/2.cosα - x.sinα + x³/6.sinα
= cosα - x.sinα - x²/2.cosα + x³/6.sinα
(15) Since sinθ/θ tends to 1, when θ tends to zero
so, sinθ/θ = 1 nearly, when θ is very small.
Here, given that sinθ/θ = 2165/2166
⇒sinθ/θ is nearly equal to 1 and hence θ is
very small.
Expanding sinθ in powers of θ,
we have
sinθ/θ = [θ - θ³/3! + θ5/5!-...]/θ
⇒2165/2166 = 1 - θ²/6, neglecting θ³ and
higher powers of θ
⇒ θ²/6 = 1 - 2165/2166
⇒ θ² = 1/2166
⇒ θ² = 6/2166 = 1/361
⇒ θ= 1/√361 radians =180/π.√361
⇒ θ = 180.7/22.19 degrees
= 630/209 degrees
= 3.01 degrees
= nearly 3°
(16) Given tanθ/θ = 2524/2523
⇒ [θ + θ³/3 +2θ5/15 + ...]/θ = 2524/2523
⇒ 1 + θ²/3 = 2524/2523 [omitting higher powers of θ]
⇒ θ²/3 = 1/2523
⇒ θ²=3/2523 = 1/841
⇒ θ = 1/29 radians
= 180/π.29 degrees
= 180.7/22.29
= 1260/638 degrees
= 1.97492163
= 1°58’ approximately [1 deg = 60′]
(17) Given sin(π/6 + θ) = 0.51
⇒ sin(π/6 + θ) = 0.5 nearly
= ½ nearly
= sin(π/6)
⇒ θ is very small so that cosθ=1and sinθ=θ
Now, sin(π/6 + θ) = 0.51
⇒ sin(π/6).cosθ + cos(π/6).sinθ = 0.51
⇒ 0.5.1 + 0.87.θ = 0.51
⇒ 0.87θ = 0.01
⇒ θ = 1/87 radians = 0.012 radians
(18) Since tan(π/4) = 1, it follows that
θ differes from π/4 by a very small quantity, say x.
Assume that θ = π/4 + x, where x is small.
Since x is small, we take sinx=x and cosx=1
Hence tanx = x
Here tanθ = 1.0024
⇒tan(π/4 + x) = 1.0024
⇒[tan(π/4) + tanx]/[1 - tan(π/4).tanx] = 1.0024
⇒ (1 + x)/(1 - x) = 1.0024
⇒ 1 + x = (1 - x)1.0024
⇒ x = 0.0024/2.0024
= 0.0012 (approximately)
(19) Given tan(π/4 - θ) = 1.001
⇒ tan(π/4-θ) = 1 nearly = tan(π/4)
⇒ θ is very small so that tanθ = θ
Now, tan(π/4-θ) = 1.001
⇒ [tan(π/4) - tanθ]/[1 + tan(π/4).tanθ] = 1.001
⇒ (1-θ)/(1+θ) = 1.001
⇒ 1 - θ = (1 + θ)1.001
⇒ θ = 0.001/2.001 = 0.0005 radians
(20)
(21)
(22)
(23)
(24)
(25) Same as Q.No (22)
(26)
(27)
(28)
(29)
Exercises XIII
(1)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(2)
(3)
(4)
(5)
(6) (i)
(6)(ii)
(7a)
