ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6

Utilizing ISC Mathematics Class 11 Solutions Chapter 3 Trigonometry Ex 3.6 as a study aid can enhance exam preparation.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6

Very short answer/objective questions (1 to 3) :

Question 1.
Convert the following products into sums or differences:
(i) 2 sin 3x cos 2x
(ii) 2 cos 3x sin 2x
(iii) 2 sin 4x sin 2x
(iv) 2 cos 7x cos 3x.
Solution:
(i) 2 sin 3x cos 2x
= sin (3x + 2x) + sin (3x – 2x)
= sin 5x + sin x
[∵ 2 sin A cos B = sin (A + B) + sin (A – B)]

(ii) 2 cos 3x sin 2x
= sin (3x + 2x) – sin (3x – 2x)
= sin 5x – sin x
[∵ 2 cos A sin B = sin(A + B) – sin (A – B)]

(iii) 2 sin 4x sin 2x
= cos (4x – 2x) – cos (4x + 2x)
= cos 2x – cos 6x
[∵ 2 sin A sin B = cos (A – B) – cos (A + B)]

(iv) 2 cos 7x cos 3x
= cos (7x + 3x) + cos (7x – 3x)
= cos 10x + cos 4x
[∵ 2 cos A cos B = cos (A + B) + cos(A – B)]

Question 2.
Express each of the following as the product of sines and cosines :
(i) sin 10x + sin 6x
(ii) sin 10x – sin 6x
(ill) cos 10x + cos 6x
(iv) cos 10x – cos 6x.
Solution:
(i) sin 10x + sin 6x
= 2 sin \(\left(\frac{10 x+6 x}{2}\right)\) + cos \(\left(\frac{10 x-6 x}{2}\right)\)
= 2 sin 8x cos 2x
[∵ sin C + sin D = 2 sin \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)]

(ii) sin 10x – sin 6x
= 2 cos \(\left(\frac{10 x+6 x}{2}\right)\) sin \(\left(\frac{10 x-6 x}{2}\right)\)
= 2 cos 8x sin 2x
[∵ sin C + sin D = 2 sin \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)]

(iii) cos 10x + cos 6x
= – 2 sin \(\left(\frac{10 x+6 x}{2}\right)\) sin \(\left(\frac{10 x-6 x}{2}\right)\)
= – 2 sin 8x sin 2x
[∵ cos C – cos D = – 2 sin \(\frac{C+D}{2}\) sin \(\frac{C-D}{2}\)]

(iv) cos 10x – cos 6x
= – 2 sin \(\left(\frac{10 x+6 x}{2}\right)\) sin \(\left(\frac{10 x-6 x}{2}\right)\)
= – 2 sin 8x sin 2x
[∵ cos C – cos D = – 2 sin \(\frac{C+D}{2}\) sin \(\frac{C-D}{2}\)]

Question 3.
Prove that :
(i) 2 cos 45° cos 15° = \(\frac{\sqrt{3}+1}{2}\)
(ii) 2 sin 75° sin 15° = \(\frac{1}{2}\)
Solution:
(i) 2 cos 45° cos 15° = cos (45° + 15°) + cos (45° – 15°)
[∵ 2 cos A cos B = cos (A + B) + cos (A – B)]
= cos 60° + cos 30°
= \(\frac{1}{2}+\frac{\sqrt{3}}{2}\)
= \(\frac{1+\sqrt{3}}{2}\)

(ii) 2 sin 75° sin 15°
= cos (75° – 15°) – cos (75° + 15°)
[∵ 2 sin A sin B cos (A – B) – cos (A + B)]
= cos 60° – cos 90°
= \(\frac{1}{2}\)

Short answer questions (4 to 7) :

Question 4.
Prove that :
(i) sin 80° – cos 70° = cos 50°
(ii) cos 5° – sin 25° = sin 35°
(iii) sin 36° + cos 36° = cos 9°
(iv) cos 15°-sin 15° =7
Solution:
(i) L.H.S. = sin 80° – cos 70°
= sin 80° – cos (90° – 20°)
= sin 80° – sin 20°
= 2 cos \(\left(\frac{80^{\circ}+20^{\circ}}{2}\right)\) sin \(\left(\frac{80^{\circ}-20^{\circ}}{2}\right)\)
[∵ sin C – sin D = 2 cos \(\frac{C+D}{2}\) sin \(\frac{C-D}{2}\)]
= 2 cos 50° sin 30°
= 2 cos 50° × \(\frac{1}{2}\)
= cos 50°
= R.H.S.

(ii) cos 5° – sin 25°
= cos 5° – sin (90° – 65°)
= cos 5° – cos 65°
= 2 sin \(\left(\frac{5^{\circ}+65^{\circ}}{2}\right)\) sin \(\left(\frac{65^{\circ}-5^{\circ}}{2}\right)\)
[∵ cos C – cos D = 2 sin \(\frac{C+D}{2}\) sin \(\frac{C-D}{2}\)]
= 2 sin 35° × \(\frac{1}{2}\)
= sin 35°

(iii) sin 36° + cos 36°
= sin (90° – 54°) + cos 36°
= cos 54° + cos 36°
= 2 cos \(\left(\frac{54^{\circ}+36^{\circ}}{2}\right)\) cos \(\left(\frac{54^{\circ}-36^{\circ}}{2}\right)\)
[∵ cos C + cos D = 2 cos \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)]
= 2 cos 45° cos 9°
= \(\frac{2}{\sqrt{2}}\) cos 9°
= √2 cos 9°

(iv) cos 15° – sin 15°
= cos 15° – sin (90° – 75°)
= cos 15° – cos 75°
= 2 sin \(\left(\frac{15^{\circ}+75^{\circ}}{2}\right)\) sin \(\left(\frac{75^{\circ}-15^{\circ}}{2}\right)\)
= 2 sin 45° sin 30°
= 2 × \(\frac{1}{\sqrt{2}} \times \frac{1}{2}\)
= \(\frac{1}{\sqrt{2}}\)

Question 5.
Prove that :
(i) \(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}\) = tan 4x
(ii) \(\frac{\cos 7 x+\cos 5 x}{\sin 7 x-\sin 5 x}\) = cot x
(iii) \(\frac{\sin x+\sin y}{\cos x+\cos y}=\tan \frac{x+y}{2}\)
(iv) \(\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}\)
Solution:
(i) L.H.S. = \(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}\)
= \(\frac{2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)}{2 \cos \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)}\)
[using C – D formulae]
= \(\frac{2 \sin 4 x \cos x}{2 \cos 4 x \cos x}\)
= tan 4x
= R.H.S.

(ii) \(\frac{\cos 7 x+\cos 5 x}{\sin 7 x-\sin 5 x}\)
= \(\frac{2 \cos \left(\frac{7 x+5 x}{2}\right) \cos \left(\frac{7 x-5 x}{2}\right)}{2 \cos \left(\frac{7 x+5 x}{2}\right) \sin \left(\frac{7 x-5 x}{2}\right)}\)
[using C – D formulae]
= \(\frac{2 \cos 6 x \cos x}{2 \cos 6 x \sin x}\)
= cot x
= R.H.S.

(iii) \(\frac{\sin x+\sin y}{\cos x+\cos y}\)
= \(\frac{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}\)
[∵ sin C + sin D = 2 sin \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)
cos C + cos D = 2 cos \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)]
= tan \(\left(\frac{x+y}{2}\right)\)
= R.H.S.

(iv) L.H.S. = \(\frac{\sin x-\sin y}{\cos x+\cos y}\)
= \(\frac{2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}\)
= tan \(\left(\frac{x-y}{2}\right)\)
[using C – D formulae]
= R.H.S.

Question 6.
(i) \(\frac{\cos 20^{\circ}-\cos 70^{\circ}}{\sin 70^{\circ}-\sin 20^{\circ}}\) = 1
(ii) \(\frac{\sin 75^{\circ}-\sin 15^{\circ}}{\cos 75^{\circ}+\cos 15^{\circ}}=\frac{1}{\sqrt{3}}\)
Solution:
(i) L.H.S. = \(\frac{\cos 20^{\circ}-\cos 70^{\circ}}{\sin 70^{\circ}-\sin 20^{\circ}}\)
= \(\frac{2 \sin \left(\frac{20^{\circ}+70^{\circ}}{2}\right) \sin \left(\frac{70^{\circ}-20^{\circ}}{2}\right)}{2 \cos \left(\frac{70^{\circ}+20^{\circ}}{2}\right) \sin \left(\frac{70^{\circ}-20^{\circ}}{2}\right)}\)
= tan 45°
= 1
= R.H.S.

(ii) \(\frac{\sin 75^{\circ}-\sin 15^{\circ}}{\cos 75^{\circ}+\cos 15^{\circ}}\)
= \(\frac{2 \cos \left(\frac{75^{\circ}+15^{\circ}}{2}\right) \sin \left(\frac{75^{\circ}-15^{\circ}}{2}\right)}{2 \cos \left(\frac{75^{\circ}+15^{\circ}}{2}\right) \cos \left(\frac{75^{\circ}-15^{\circ}}{2}\right)}\)
= tan 30°
= \(\frac{1}{\sqrt{3}}\)
= R.H.S.

Question 7.
Prove that :
(i) sin (\(\frac{\pi}{4}\) sin (\(\frac{\pi}{2}\) – x) = \(\frac{1}{2}\) cos 2x
(ii) sec (\(\frac{\pi}{4}\) + x) sec (\(\frac{\pi}{4}\) – x) = 2 sec 2x
(iii) \(\sin \left(\frac{5 \pi}{6}+x\right)+\sin \left(\frac{5 \pi}{6}-x\right)\) = cos x
Solution:
(i) L.H.S. = sin (\(\frac{\pi}{4}\) sin (\(\frac{\pi}{2}\) – x)
= \(\frac{1}{2}\left[2 \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-x\right)\right]\)
= \(\frac{1}{2}\left[\cos \left(\frac{\pi}{4}+x-\frac{\pi}{4}+x\right)\right.\) – \(\left.\cos \left(\frac{\pi}{4}+x+\frac{\pi}{4}-x\right)\right]\)
[∵ 2 sin A sin B = cos (A – B) – cos (A + B)]
= \(\frac{1}{2}\) [cos 2x – cos \(\frac{\pi}{2}\)]
= \(\frac{1}{2}\) cos 2x
= R.H.S.

(ii) L.H.S. = sec (\(\frac{\pi}{4}\) + x) sec (\(\frac{\pi}{4}\) – x)

(iii) L.H.S. = \(\sin \left(\frac{5 \pi}{6}+x\right)+\sin \left(\frac{5 \pi}{6}-x\right)\)

Question 8.
Prove that :
(i) cos 7x + cos 5x + cos 3x + cos x = 4 cos x cos 2x cos 4x
(ii) sin x + sin 2x + sin 4x + sin 5x = 4cos \(\frac{x}{2}\) cos \(\frac{3 x}{2}\) sin 3x
(iii) cos 3x + cos 5x + cos 7x + cos 15x = 4 cos 4x cos 5x cos 6x
(iv) cos x cos \(\frac{x}{2}\) – cos 3x cos \(\frac{9 x}{2}\) = sin 4x sin \(\frac{7 x}{2}\)
Solution:
(i) L.H.S. = cos 7x + cos 5x + cos 3x + cos x
= (cos 7x + cos x) + (cos 5x + cos 3x)
= 2 \(\cos \left(\frac{7 x+x}{2}\right) \cos \left(\frac{7 x-x}{2}\right)\) + 2 \(\cos \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)\)
= 2 cos 4x cos 3x + 2 cos 4x cos x
= 2 cos 4x (cos 3x + cos x)
= 2 cos 4x \(\left[2 \cos \left(\frac{3 x+x}{2}\right) \cos \left(\frac{3 x-x}{2}\right)\right]\)
[∵ cos C + cos D = 2 cos \(\left(\frac{\mathrm{C}+\mathrm{D}}{2}\right)\) cos \(\left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\)]
= 2 cos 4x (2 cos 2x cos x)
= 4 cos x cos 2x cos 4x
= R.H.S.

(ii) L.H.S. = sin x + sin 2x + sin 4x + sin 5x
= (sin 5x + sin x) + sin 4x + sin 2x)
= 2 \(\sin \left(\frac{5 x+x}{2}\right) \cos \left(\frac{5 x-x}{2}\right)\) + 2 \(\sin \left(\frac{4 x+2 x}{2}\right) \cos \left(\frac{4 x-2 x}{2}\right)\)
= 2 sin 3x cos 2x + 2 sin 3x cos x
= 2 sin 3x (cos 2x + cos x)
= 2 sin 3x \(\left[2 \cos \left(\frac{2 x+x}{2}\right) \cos \left(\frac{2 x-x}{2}\right)\right]\)
= 4 cos \(\frac{x}{2}\) cos \(\frac{3 x}{2}\) sin 3x
= R.H.S.

(iii) L.H.S. = cos 3x + cos 5x + cos 7x + cos 15x
= (cos 15x + cos 3x) + (cos 7x + cos 5x)
= \(2 \cos \left(\frac{15 x+3 x}{2}\right) \cos \left(\frac{15 x-3 x}{2}\right)+2 \cos \left(\frac{7 x+5 x}{2}\right) \cos \left(\frac{7 x-5 x}{2}\right)\)
= 2 cos 9x cos 6x + 2 cos 6x cos x
= 2 cos 6x [cos 9x + cos x]
= 2 cos 6x \(\left[2 \cos \left(\frac{9 x+x}{2}\right) \cos \left(\frac{9 x-x}{2}\right)\right]\)
= 2 cos 6x (2 cos 5x cos 4x)
= 4 cos 4x cos 5x cos 6x
= R.H.S.

(iv) L.H.S. = cos x cos \(\frac{x}{2}\) – cos 3x cos \(\frac{9 x}{2}\)

Question 9.
Prove that :
(i) cos 52° + cos 68° + cos 172° = 0
(ii) cos 20° + cos 100° + cos 140° = 0
(iii) \(2 \sin \frac{\pi}{17} \sin \frac{11 \pi}{17}-\cos \frac{5 \pi}{17}+\cos \frac{7 \pi}{17}\) = 0
Solution:
(I) L.H.S. = cos 52° + OS 68° COS 1720
= 2 cos \(\left(\frac{52^{\circ}+68^{\circ}}{2}\right)\) cos \(\left(\frac{52^{\circ}-68^{\circ}}{2}\right)\) + cos (180° – 8°)
[cos C + cos D = 2 cos \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)]
= 2 cos 60° cos 8° – cos 8°
= 2 × \(\frac{1}{2}\) cos 8° – cos 8°
= 0
= R.H.S.

(ii) LH.S. = cos 20° + cos 100° + cos 140°
= 2 cos \(\left(\frac{20^{\circ}+100^{\circ}}{2}\right)\) cos \(\left(\frac{20^{\circ}-100^{\circ}}{2}\right)\) + cos (180° – 40°)
= 2 cos 60° cos 40° – cos 40°
= 2 × \(\frac{1}{2}\) cos 40° – cos 40°
= 0
= R.H.S.

(iii) L.H.S = \(2 \sin \frac{\pi}{17} \sin \frac{11 \pi}{17}-\cos \frac{5 \pi}{17}+\cos \frac{7 \pi}{17}\)

Question 10.
Prove that :
(i) sin 10° sin 50° sin 70° = \(\frac{1}{8}\)
(ii) sin 10° sin 30° sin 50° sin 70° = \(\frac{1}{16}\)
(iii) sin 20° sin 40° sin 60° sin 80° = \(\frac{3}{16}\)
Solution:
(i) L.H.S. = sin 10° sin 50° sin 70°
= \(\frac{1}{2}\) sin 10° [2 sin 70° sin 50°]
= \(\frac{1}{2}\) sin 10° [cos (20°) – cos 120°]
[∵ 2 sin A sin B = cos (A – B) – cos (A + B)]
= \(\frac{1}{2}\) sin 10° [cos 20° – cos (180° – 60°)]
= \(\frac{1}{4}\) (2 cos 20° sin 10°) – \(\frac{1}{2}\) (- cos 60°) sin 10°
= \(\frac{1}{4}\) [sin (20° + 10°) – sin (20° – 10°)] + \(\frac{1}{2}\) × \(\frac{1}{2}\) sin 10°
= \(\frac{1}{4}\) [\(\frac{1}{2}\) – sin 10°] + \(\frac{1}{4}\) sin 10°
= R.H.S.

(ii) L.H.S. = sin 10° sin 30° sin 50° sin 70°
= \(\frac{1}{2}\) [sin 10° sin 50° sin 70°]
= \(\frac{1}{2 \times 2}\) sin 10° [2 sin 70° sin 50°]
= \(\frac{1}{4}\) sin 10° [cos (70° – 50°) – cos (70° + 50°)]
= \(\frac{1}{4}\) sin 10° [cos 20° – cos(180° – 60°)]
= \(\frac{1}{4}\) sin 10° [cos 20° + cos 60°]
= \(\frac{1}{4}\) cos 20° sin 10° sin 10° × \(\frac{1}{2}\)
= \(\frac{1}{2}\) [2 cos 20° sin 10°] + \(\frac{1}{8}\) sin 10°
= \(\frac{1}{8}\) [sin (20° + 10°) – sin (20° – 10°)] + \(\frac{1}{8}\) sin 10°
= \(\frac{1}{8}\left[\frac{1}{2}-\sin 10^{\circ}\right]\)
= \(\frac{1}{16}\)
= R.H.S.

(iii) L.H.S. = sin 20° sin 40° sin 60° sin 80°
= \(\frac{\sqrt{3}}{2}\) [sin 20° sin 40° sin 80°]
= \(\frac{\sqrt{3}}{2}\) sin 20 × \(\frac{1}{2}\) [2 sin 80° sin 40°]
= \(\frac{\sqrt{3}}{4}\) sin 20° [cos (80° – 40°) – cos(80° + 40°)]
[∵ 2 sin A sin B = cos (A – B) – cos(A + B)]
= \(\frac{\sqrt{3}}{4}\) sin 20° [cos 40° – cos 120°]
= \(\frac{\sqrt{3}}{4}\) sin 20° [cos 40° – cos (180° – 60°)]
= \(\frac{\sqrt{3}}{4}\) sin 20° [cos 40° + \(\frac{1}{2}\)]
= \(\frac{\sqrt{3}}{4}\) cos 40° sin 20° + \(\frac{\sqrt{3}}{8}\) sin 20°
= \(\frac{\sqrt{3}}{8}\) (2 cos 40° sin 20°) + \(\frac{\sqrt{3}}{8}\) sin 20°
= \(\frac{\sqrt{3}}{8}\) [sin (40° + 20°) – sin (40° – 20°)] + \(\frac{\sqrt{3}}{8}\) sin 20°
[∵2 cos A sin B = sin (A + B) – sin (A – B)]
= \(\frac{\sqrt{3}}{8}\) [sin 60° – sin 20°] + \(\frac{\sqrt{3}}{8}\) sin 20°
= \(\frac{\sqrt{3}}{8} \times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{8} \sin 20^{\circ}+\frac{\sqrt{3}}{8} \sin 20^{\circ}\)
= \(\frac{3}{16}\)
= R.H.S.

Question 11.
Prove that :
(i) cos 20° cos 40° cos 60° cos 80° = \(\frac{1}{16}\)
(ii) cos 10° cos 30° cos 50° cos 70° = \(\frac{3}{16}\)
Solution:
(i) L.H.S. = cos 20° cos 40° cos 60° cos 80°
= \(\frac{1}{2}\) [cos 20° cos 40° cos 80°]
= \(\frac{1}{4}\) cos 20° [2 cos 80° cos 40°]
= \(\frac{1}{4}\) cos 20° [cos (80° + 40°) + cos (80° – 40°)]
[∵ 2 cos A cos B = cos (A + B) + cos (A – B)]
= \(\frac{1}{4}\) cos 20° [cos 120° + cos 40°]
= \(\frac{1}{4}\) cos 20° [cos (180° – 60°) + cos 40°]
= \(\frac{1}{4}\) cos 20° [- cos 60° + cos 40°]
= \(\frac{1}{4}\) cos 20° [- \(\frac{1}{2}\) + cos 40°]
= \(\frac{1}{8}\) cos 20°+ \(\frac{1}{8}\) (2 cos 40° cos 20°)
= – \(\frac{1}{4}\) cos 20° + \(\frac{1}{8}\) [cos (40° + 20°) + cos (40° – 20°)]
= – \(\frac{1}{8}\) cos 20°+ \(\frac{1}{8}\) [\(\frac{1}{2}\) + cos 20°]
= – \(\frac{1}{4}\) cos 20° + \(\frac{1}{16}\) + \(\frac{1}{8}\) cos 20°
= \(\frac{1}{16}\)
= R.HS.

(ii) L.H.S. = cos 10° cos 30° cos 50° cos 70°
= \(\frac{\sqrt{3}}{2}\) cos 10° (cos 70° cos 50°)
= \(\frac{\sqrt{3}}{4}\) cos 10° (2 cos 70° cos 50°)
= \(\frac{\sqrt{3}}{4}\) cos 10° [cos (70° + 50°) + cos (70° – 50°)]
= \(\frac{\sqrt{3}}{4}\) cos 10° [cos 120° + cos 20°]
= \(\frac{\sqrt{3}}{4}\) cos 10° [cos (180° – 60°) + cos 20°]
= \(\frac{\sqrt{3}}{4}\) cos 10° [- cos 60° + cos 20°]
= \(\frac{\sqrt{3}}{4}\) cos 10° [- \(\frac{1}{2}\) + cos 20°]
= – \(\frac{\sqrt{3}}{8}\) cos 10° + \(\frac{\sqrt{3}}{8}\) (2 cos 20° cos 10°)
= – \(\frac{\sqrt{3}}{8}\) cos 10° + \(\frac{\sqrt{3}}{8}\) [cos 30° + cos 10°]
= – \(\frac{\sqrt{3}}{8}\) cos 10° + \(\frac{\sqrt{3}}{8}\) [\(\frac{\sqrt{3}}{2}\) + cos 10°]
= \(-\frac{\sqrt{3}}{8} \cos 10^{\circ}+\frac{3}{16}+\frac{\sqrt{3}}{8} \cos 10^{\circ}=\frac{3}{16}\)
= R.H.S.

Question 12.
Prove that :
tan 10° tan 50° tan 70° = tan 30°.
Solution:
L.H.S. = tan 10° tan 50° tan 70°
= \(\frac{\sin 10^{\circ} \sin 50^{\circ} \sin 70^{\circ}}{\cos 10^{\circ} \cos 50^{\circ} \cos 70^{\circ}}\) …………………(1)
Now sin 10° sin 50° sin 70° = \(\frac{1}{2}\) sin 10° (2 sin 70° sin 50°)
= \(\frac{1}{2}\) sin 10° [cos 20° – cos 120°]
[∵ 2 sinA sin B = cos (A – B) – cos (A + B)]
= \(\frac{1}{2}\) sin 10° [cos 20° – cos (180° – 60°)]
= \(\frac{1}{2}\) sin 10° [cos 20° + \(\frac{1}{2}\)]
= \(\frac{1}{4}\) (2 cos 20° sin 10°) + \(\frac{1}{4}\) sin 10°
= \(\frac{1}{4}\) [sin (20° + 10°) – sin (20° – 10°)] + \(\frac{1}{4}\) sin 10°
= \(\frac{1}{4}\) [\(\frac{1}{2}\) – sin 10°] + \(\frac{1}{4}\) sin 10°
= \(\frac{1}{8}\) …………………..(2)
= cos 10° cos 50° cos 70°
= \(\frac{1}{2}\) cos 10° (2 cos 70° cos 50°)
= \(\frac{1}{2}\) cos 10° [cos 120° + cos 20°]
= \(\frac{1}{2}\) cos 10° [cos (180° – 60°) + cos 20°]
= \(\frac{1}{2}\) cos 10° [- \(\frac{1}{2}\) + cos 20°]
= – \(\frac{1}{4}\) cos 10° + \(\frac{1}{4}\) (2 cos 20° cos 10°)
= – \(\frac{1}{4}\) cos 10° + \(\frac{1}{4}\) [cos 30° + cos 10°]
= \(\frac{1}{4}\) cos 30°
= \(\frac{1}{4} \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{8}\) …………………..(3)
Using eqn. (2), (3) in eqn. (1) ; we have
L.H.S. = \(\frac{\frac{1}{8}}{\frac{\sqrt{3}}{8}}=\frac{1}{\sqrt{3}}\)
= tan 30°
= R.H.S.

Question 13.
(i) cos x + cos (\(\frac{2 \pi}{3}\) – x) + cos (\(\frac{2 \pi}{3}\) + x) = 0
(ii) \(\cos \frac{\pi}{8}+\cos \frac{3 \pi}{8}+\cos \frac{5 \pi}{8}+\cos \frac{7 \pi}{8}\) = 0.
Solution:
(i) L.H.S. = cos x + cos (\(\frac{2 \pi}{3}\) – x) + cos (\(\frac{2 \pi}{3}\) + x)
= cos x + 2 \(\cos \left(\frac{\frac{2 \pi}{3}-x+\frac{2 \pi}{3}+x}{2}\right) \cos \left(\frac{\frac{2 \pi}{3}-x-\frac{2 \pi}{3}-x}{2}\right)\)
[∵ cos C + cos D = 2 cos \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)]
= cos x + 2 cos \(\frac{2 \pi}{3}\) cos (- x)
= cos x + 2 cos (π – \(\frac{\pi}{3}\)) cos x
= cos x + 2 (- cos \(\frac{\pi}{3}\)) cos x
= cos x – 2 × (\(\frac{1}{2}\)) cos x
= cos x – cos x
= 0
= R.H.S.

(ii) L.H.S. = \(\cos \frac{\pi}{8}+\cos \frac{3 \pi}{8}+\cos \frac{5 \pi}{8}+\cos \frac{7 \pi}{8}\)

Question 14.
Prove that :
(i) 4 sin x sin (\(\frac{\pi}{3}\) – x) sin (\(\frac{\pi}{3}\) + x) = sin 3x
(ii) 4 cos x cos (\(\frac{\pi}{3}\) – x) cos (\(\frac{\pi}{3}\) + x) = cos 3x
Solution:
(i) L.H.S. = 4 sin x sin (\(\frac{\pi}{3}\) – x) sin (\(\frac{\pi}{3}\) + x)

(ii) L.H.S. = 4 cos x cos (\(\frac{\pi}{3}\) – x) cos (\(\frac{\pi}{3}\) + x)

= – cos x + 2 cos 2x cos x
= – cos x + cos (2x + x) + cos (2x – x)
= – cos x + cos 3x + cos x
= cos 3x
= R.H.S.

Question 15.
Prove that :
(i) (cos x + cosy)² + (sin x + siny)² = 4 c0s²
(ii) (cos x + cos y)² + (sin x – sin y)² = 4 cos²
(iii) sin² x + sin² (x – y) – 2 sin x cos y sin (x – y) = sin² y.
Sol.
(i) L.H.S. = (cos x + cos y)² – (sin x + sin y)²

(ii) L.H.S. = (cos x + cos y)² + (sin x – sin y)²

(iii) L.H.S. = sin² x + sin² (x – y) – 2 sin x cos y sin (x – y)
= sin² x + sin (x – y) [sin (x – y) – 2 sin x cos y]
= sin² x + sin (x – y) [sin (x – y) – sin (x + y) – sin (x – y)]
= sin² x + sin (x – y) {- sin (x + y)}
= sin² x – sin (x – y) sin (x + y)
= sin² x – {sin² x – sin² y}
= sin² x – sin² x + sin² y
= sin² y
= R.H.S.

Question 16.
Prove that :
(i) \(\frac{\sin x+\sin y}{\sin x-\sin y}=\tan \frac{x+y}{2} \cot \frac{x-y}{2}\)
(ii) \(\frac{\sin x+\sin 3 x+\sin 5 x}{\cos x+\cos 3 x+\cos 5 x}\) = tan 3x
(iii) \(\frac{\sin 5 x+2 \sin 8 x+\sin 11 x}{\sin 8 x+2 \sin 11 x+\sin 14 x}=\frac{\sin 8 x}{\sin 11 x}\)
(iv) \(\frac{\sin 8 x \cos x-\sin 6 x \cos 3 x}{\cos 2 x \cos x-\sin 3 x \sin 4 x}\) = tan 2x
(v) \(\frac{\cos x+\cos 3 x+\cos 5 x+\cos 7 x}{\sin x+\sin 3 x+\sin 5 x+\sin 7 x}\) = cot 4x
Solution:
(i) L.H.S. = \(\frac{\sin x+\sin y}{\sin x-\sin y}\)
= \(\frac{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)}\)
= \(\tan \left(\frac{x+y}{2}\right) \cot \left(\frac{x-y}{2}\right)\)
= R.H.S

(ii) L.H.S. = \(\frac{\sin x+\sin 3 x+\sin 5 x}{\cos x+\cos 3 x+\cos 5 x}\)

(iii) L.H.S. = \(\frac{\sin 5 x+2 \sin 8 x+\sin 11 x}{\sin 8 x+2 \sin 11 x+\sin 14 x}\)

(iv) L.H.S. = \(\frac{\sin 8 x \cos x-\sin 6 x \cos 3 x}{\cos 2 x \cos x-\sin 3 x \sin 4 x}\)

(v) \(\frac{\cos x+\cos 3 x+\cos 5 x+\cos 7 x}{\sin x+\sin 3 x+\sin 5 x+\sin 7 x}\)
= cot⁴ x

Question 17.
Prove that :
cos α + cos β + cos γ + cos (α + β + γ) = 4 \(\cos \frac{\alpha+\beta}{2} \cos \frac{\beta+\gamma}{2} \cos \frac{\gamma+\alpha}{2}\)
Solution:
L.H.S. = cos α + cos β) + cos γ + cos (α + β + γ)
= (cos α + cos β) + (cos (α + β + γ) + cos γ)

Question 18.
If cos x + cos y = \(\frac{1}{3}\) and sin x + sin y = \(\frac{1}{4}\), prove that \(\frac{x+y}{2}=\frac{3}{4}\).
Solution:
Given cos x + cos y = \(\frac{1}{3}\)

Question 19.
(i) \(\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}\) then show that \(\frac{\tan x}{\tan y}=\frac{a}{b}\).
(ii) If cos (x + 2y) = m cos x, prove that cot y = \(\frac{1+m}{1-m}\) tan (x + y).
Solution:
Given \(\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}\)
Applying componendo and dividendo, we have

(ii) Given, cos (x + 2y) = m cos x
⇒ \(\frac{\cos (x+2 y)}{\cos x}=\frac{m}{1}\)
Applying componendo and dividendo, we have