Open In App

Class 11 RD Sharma Solutions – Chapter 8 Transformation Formulae – Exercise 8.1

Last Updated : 30 Apr, 2021
Improve
Improve
Like Article
Like
Save
Share
Report

Question 1. Express each of the following as the sum or difference of sines and cosines:

(i) 2 sin 3θ cos θ

Solution:

By using the trigonometric identity,

2 sin A cos B = sin (A+B) + sin (A-B)

Taking A = 3θ and B = θ

2 sin 3θ cos θ = sin (3θ+θ) + sin (3θ-θ)

= sin 4θ + sin 2θ

(ii) 2 cos 3θ sin 2θ

Solution:

By using the trigonometric identity,

2 sin A cos B = sin (A+B) + sin (A-B)

Taking A = 2θ and B = 3θ

2 cos 3θ sin 2θ = sin (3θ+2θ) + sin (2θ-3θ)

= sin 5θ + sin (-θ)

= sin 5θ – sin θ

(iii) 2 sin 4θ sin 3θ

Solution:

By using the trigonometric identity,

2 sin A sin B = cos (A-B) – cos (A+B)

Taking A = 4θ and B = 3θ

2 sin 4θ sin 3θ = cos (4θ-3θ) – cos (4θ+3θ)

= cos θ – cos 7θ

(iv) 2 cos 7θ cos 3θ

Solution:

By using the trigonometric identity,

2 cos A cos B = cos (A+B) + cos (A-B)

Taking A = 7θ and B = 3θ

2 cos 7θ cos 3θ = cos (7θ+3θ) + cos (7θ-3θ)

= cos 10θ – cos 4θ

Question 2. Prove that:

(i) 2 \hspace{0.1cm}sin \frac{5\pi}{12}\hspace{0.1cm} sin \frac{\pi}{12} = \frac{1}{2}

Solution:

By using the trigonometric identity,

2 sin A sin B = cos (A-B) – cos (A+B)

Taking A = \frac{5\pi}{12}  and B = \frac{\pi}{12}

2 \hspace{0.1cm}sin \frac{5\pi}{12}\hspace{0.1cm} sin \frac{\pi}{12} = cos (\frac{5\pi}{12}-\frac{\pi}{12}) - cos (\frac{5\pi}{12}+\frac{\pi}{12})\\ = cos (\frac{4\pi}{12}) - cos (\frac{6\pi}{12})\\ = cos (\frac{\pi}{3}) - cos (\frac{\pi}{2})\\ = \frac{1}{2} - 0\\ = \frac{1}{2}

Hence, LHS = RHS

(ii) 2\hspace{0.1cm} cos \frac{5\pi}{12} \hspace{0.1cm}cos \frac{\pi}{12} = \frac{1}{2}

Solution:

By using the trigonometric identity,

2 cos A cos B = cos (A+B) + cos (A-B)

Taking A = \frac{5\pi}{12}  and B = \frac{\pi}{12}

2\hspace{0.1cm} cos \frac{5\pi}{12} \hspace{0.1cm}cos \frac{\pi}{12} = cos (\frac{5\pi}{12}+\frac{\pi}{12}) + cos (\frac{5\pi}{12}-\frac{\pi}{12})

= cos (\frac{6\pi}{12}) + cos (\frac{4\pi}{12})\\ = cos (\frac{\pi}{2}) + cos (\frac{\pi}{3})\\ = 0 + \frac{1}{2}\\ = \frac{1}{2}

Hence, LHS = RHS

(iii) 2 \hspace{0.1cm}sin \frac{5\pi}{12}\hspace{0.1cm} cos \frac{\pi}{12} = \frac{(\sqrt{3} + 2)}{2}

Solution:

By using the trigonometric identity,

2 sin A cos B = sin (A+B) + sin (A-B)

Taking A = \frac{5\pi}{12}  and B = \frac{\pi}{12}

2 \hspace{0.1cm}sin \frac{5\pi}{12}\hspace{0.1cm} cos \frac{\pi}{12} = sin (\frac{5\pi}{12}+\frac{\pi}{12}) + sin (\frac{5\pi}{12}-\frac{\pi}{12})

= sin (\frac{6\pi}{12}) + sin (\frac{4\pi}{12})

= sin (\frac{\pi}{2})  + sin (\frac{\pi}{3})

= 1 + \frac{\sqrt{3}}{2}

= \frac{\sqrt{3}+2}{2}

Hence, LHS = RHS

Question 3. Show that:

(i) sin 50° cos 85° = \frac{(1 - \sqrt{2}sin 35\degree)}{2\sqrt{2}}

Solution:

By using the trigonometric identity,

2 sin A cos B = sin (A+B) + sin (A-B)

sin A cos B = \frac{1}{2} (sin (A+B) + sin (A-B))

Taking A = 50° and B = 85°

sin 50° cos 85° = \frac{1}{2} (sin (50°+85°) + sin (50°-85°))

= \frac{1}{2} (sin (135°) + sin (-35°))

= \frac{1}{2} (sin (180°-45°) – sin (35°))     (sin(-θ)=-sin θ)

= \frac{1}{2} (sin (45°) – sin (35°))     (sin(Ï€-θ)=sin θ)

= \frac{1}{2}(\frac{1}{\sqrt{2}} - sin (35\degree))\\ = \frac{1}{2}(\frac{(1 - \sqrt{2}sin 35\degree)}{\sqrt{2}})\\ = \frac{(1 - \sqrt{2}sin 35\degree)}{2\sqrt{2}}

Hence, LHS = RHS

(ii) sin 25° cos 115° = \frac{1}{2}(sin 40 \degree - 1)

Solution:

By using the trigonometric identity,

2 sin A cos B = sin (A+B) + sin (A-B)

sin A cos B = \frac{1}{2} (sin (A+B) + sin (A-B))

Taking A = 25° and B = 115°

sin 25° cos 115° = \frac{1}{2} (sin (25°+85°) + sin (25°-115°))

= \frac{1}{2} (sin (140°) + sin (-90°))

= \frac{1}{2} (sin (180-40°) – sin (90°))     (sin(-θ)=-sin θ)

= \frac{1}{2} (sin (40°) – sin (90°))     (sin(Ï€-θ)=sin θ)

= \frac{1}{2} (sin (40°) – 1)

Hence, LHS = RHS

Question 4. Prove that : 4 \hspace{0.1cm}cos \hspace{0.1cm}\theta \hspace{0.1cm}cos\hspace{0.1cm} (\frac{\pi}{3} + \theta) \hspace{0.1cm}cos\hspace{0.1cm} (\frac{\pi}{3} - \theta) = cos 3\theta

Solution:

4 \hspace{0.1cm}cos \hspace{0.1cm}\theta \hspace{0.1cm}cos\hspace{0.1cm} (\frac{\pi}{3} + \theta) \hspace{0.1cm}cos\hspace{0.1cm} (\frac{\pi}{3} - \theta) = 2 \hspace{0.1cm}cos \hspace{0.1cm}\theta(2 \hspace{0.1cm}cos\hspace{0.1cm} (\frac{\pi}{3} + \theta) \hspace{0.1cm}cos\hspace{0.1cm} (\frac{\pi}{3} - \theta))

By using the trigonometric identity,

2 cos A cos B = cos (A+B) + cos (A-B)

Taking A = \frac{\pi}{3} +θ and B = \frac{\pi}{3} -θ

= 2 \hspace{0.1cm}cos \hspace{0.1cm}\theta(cos\hspace{0.1cm} (\frac{\pi}{3} + \theta+\frac{\pi}{3} - \theta) +cos\hspace{0.1cm} (\frac{\pi}{3} + \theta)-(\frac{\pi}{3} - \theta))\\ = 2 \hspace{0.1cm}cos \hspace{0.1cm}\theta(cos\hspace{0.1cm} (\frac{2\pi}{3}) +cos\hspace{0.1cm} (2\theta))\\ = 2 \hspace{0.1cm}cos \hspace{0.1cm}\theta(cos\hspace{0.1cm} (\pi+\frac{\pi}{3}) +cos\hspace{0.1cm} (2\theta))\\ = 2 \hspace{0.1cm}cos \hspace{0.1cm}\theta(- cos\hspace{0.1cm} (\frac{\pi}{3}) +cos\hspace{0.1cm} (2\theta))\\ = 2 \hspace{0.1cm}cos \hspace{0.1cm}\theta(- \frac{1}{2} +cos\hspace{0.1cm} (2\theta))\\ = - cos \hspace{0.1cm}\theta + [2 \hspace{0.1cm}cos \hspace{0.1cm}\theta\hspace{0.1cm} cos\hspace{0.1cm} (2\theta))]

Using the identity again, we have

Taking A = 2θ and B = θ

= - cos \hspace{0.1cm}\theta + [cos\hspace{0.1cm} (2\theta + \theta) +cos\hspace{0.1cm} (2\theta - \theta)]\\ = - cos \hspace{0.1cm}\theta + cos\hspace{0.1cm} (3\theta) +cos\hspace{0.1cm} (\theta)\\ = cos\hspace{0.1cm} (3\theta)

Hence, LHS = RHS

Question 5. Prove that :

(i) cos 10° cos 30° cos 50° cos 70° = \frac{3}{16}

Solution:

cos 10° cos 30° cos 50° cos 70° = cos 30° cos 10° cos 50° cos 70°

= \frac{\sqrt{3}}{2}  (cos 10° cos 50°) cos 70°

By using the trigonometric identity,

2 cos A cos B = cos (A+B) + cos (A-B)

cos A cos B = \frac{1}{2} [cos (A+B) + cos (A-B)]

Taking A = 10° and B = 50°

= \frac{\sqrt{3}}{2}  (\frac{1}{2} [cos (10°+50°) + cos (10°-50°)]) cos 70°

= \frac{\sqrt{3}}{4}  (cos (60°) + cos (-40°)) cos 70°

= \frac{\sqrt{3}}{4}  (\frac{1}{2}  + cos (40°)) cos 70°

= \frac{\sqrt{3}}{8}  cos 70° + \frac{\sqrt{3}}{4}  (cos 70° cos (40°))

Again using the identity, we get

= \frac{\sqrt{3}}{8}  cos 70° + \frac{\sqrt{3}}{4}  (\frac{1}{2} [cos (70°+40°) + cos (70°-40°)])

= \frac{\sqrt{3}}{8}  cos 70° + \frac{\sqrt{3}}{8}  [cos (110°) + cos (30°)]

= \frac{\sqrt{3}}{8}  cos 70° + \frac{\sqrt{3}}{8}  [cos (110°) + \frac{\sqrt{3}}{2} ]

= \frac{\sqrt{3}}{8}  cos 70° + \frac{\sqrt{3}}{8}  cos (110°) + \frac{3}{16}

= \frac{\sqrt{3}}{8}  (cos 70° + cos (110°)) + \frac{3}{16}

= \frac{\sqrt{3}}{8}  (cos 70° + cos (180°-70°)) + \frac{3}{16}

= \frac{\sqrt{3}}{8}  (cos 70° – cos (70°)) + \frac{3}{16}

= \frac{3}{16}

Hence, LHS = RHS

(ii) cos 40° cos 80° cos 160° = -\frac{1}{8}

Solution:

cos 40° cos 80° cos 160° = cos 80° (cos 40° cos 160°)

By using the trigonometric identity,

2 cos A cos B = cos (A+B) + cos (A-B)

cos A cos B = \frac{1}{2} [cos (A+B) + cos (A-B)]

Taking A = 160° and B = 40°

= cos 80° (\frac{1}{2} [cos (160°+40°) + cos (160°-40°)])

= cos 80° (\frac{1}{2} [cos (200°) + cos (120°)])

= cos 80° (\frac{1}{2} [cos (180°+20°) + cos (180°-60°)])

= cos 80° (\frac{1}{2} [- cos (20°) + (-cos (60°))])

= cos 80° (\frac{1}{2} [- cos (20°) – cos (60°)])

= cos 80° (\frac{1}{2} [- cos (20°) – \frac{1}{2} ])

= -\frac{1}{2}  (cos 80° cos (20°) + \frac{1}{2}  cos 80°])

Again using the identity, we get

= -\frac{1}{2} ((\frac{1}{2}[cos (80\degree+20\degree) + cos (80\degree-20\degree)]) + \frac{1}{2} cos 80\degree)\\ = -\frac{1}{4} ((cos (100\degree) + cos (60\degree)) + cos 80\degree)\\ = -\frac{1}{4} (cos (180\degree-80\degree) + cos (60\degree) + cos 80\degree)\\ = -\frac{1}{4} (- cos (80\degree) + cos (60\degree) + cos 80\degree)\\ = -\frac{1}{4} (cos (60\degree))\\ = -\frac{1}{4} (\frac{1}{2})\\ = -\frac{1}{8}

Hence, LHS = RHS

(iii) sin 20° sin 40° sin 80° = \frac{\sqrt{3}}{8}

Solution:

sin 20° sin 40° sin 80° = (sin 20° sin 40°) sin 80°

By using the trigonometric identity,

2 sin A sin B = cos (A-B) – cos (A+B)

sin A sin B = \frac{1}{2} [cos (A-B) – cos (A+B)]

Taking A = 40° and B = 20°

= (\frac{1}{2} [cos (40°-20°) – cos (40°+20°)]) sin 80°

= \frac{1}{2}  sin 80° [cos (20°) – cos (60°)]

= \frac{1}{2}  sin 80° [cos (20°) – \frac{1}{2} ]

= \frac{1}{2}  [sin 80° cos (20°) – \frac{1}{2}  sin 80°]

By using the trigonometric identity,

2 sin A cos B = sin (A+B) + sin (A-B)

sin A cos B = \frac{1}{2} [sin (A+B) + sin (A-B)]

Taking A = 80° and B = 20°

= \frac{1}{2} [(\frac{1}{2}[sin (80\degree+20\degree) + sin (80\degree-20\degree)]) - \frac{1}{2} sin 80\degree]\\ = \frac{1}{4} [(sin (80\degree+20\degree) + sin (80\degree-20\degree)) - sin 80\degree]\\ = \frac{1}{4} [sin (100\degree) + sin (60\degree) - sin 80\degree]\\ = \frac{1}{4} [sin (180\degree-80\degree) + \frac{\sqrt{3}}{2} - sin 80\degree]\\ = \frac{1}{4} [sin (80\degree) + \frac{\sqrt{3}}{2} - sin 80\degree]\\ = \frac{1}{4} [\frac{\sqrt{3}}{2}]\\ = \frac{\sqrt{3}}{8}

Hence, LHS = RHS

(iv) cos 20° cos 40° cos 80° = \frac{1}{8}

Solution:

cos 20° cos 40° cos 80° = cos 40° (cos 20° cos 80°)

By using the trigonometric identity,

2 cos A cos B = cos (A+B) + cos (A-B)

cos A cos B = \frac{1}{2} [cos (A+B) + cos (A-B)]

Taking A = 80° and B = 20°

= cos 40\degree (\frac{1}{2}[cos (80\degree+20\degree) + cos (80\degree-20\degree)])\\ = \frac{1}{2} cos 40\degree [cos (100\degree) + cos (60\degree)]\\ = \frac{1}{2} cos 40\degree[cos (180\degree-80\degree) + cos (60\degree)]\\ = \frac{1}{2} cos 40\degree [- cos (80\degree) + \frac{1}{2}]\\ = \frac{1}{2} cos 40\degree [\frac{1}{2}- cos (80\degree)]\\ = \frac{1}{2} [\frac{1}{2} cos 40\degree - cos (80\degree) cos 40\degree]\\

Again using the identity, we get

= \frac{1}{2} [\frac{1}{2} cos 40\degree - \frac{1}{2}(cos (80\degree+40\degree) + cos (80\degree-40\degree))]\\ = \frac{1}{2} [\frac{1}{2} cos 40\degree - \frac{1}{2}(cos (120\degree) + cos (40\degree))]\\ = \frac{1}{2} [\frac{1}{2} cos 40\degree - \frac{1}{2}(cos (180\degree-60\degree) + cos (40\degree))]\\ = \frac{1}{2} [\frac{1}{2} cos 40\degree - \frac{1}{2}(- cos (60\degree) + cos (40\degree))]\\ = \frac{1}{2} [\frac{1}{2} cos 40\degree + \frac{1}{2} cos (60\degree) - \frac{1}{2} cos (40\degree)]\\ = \frac{1}{2} [\frac{1}{2} \frac{1}{2}]\\ = \frac{1}{8}

Hence, LHS = RHS

(v) tan 20° tan 40° tan 60° tan 80° = 3

Solution:

tan 20° tan 40° tan 60° tan 80° = tan 60° \frac{sin 20\degree sin 40\degree sin 80\degree}{cos 20\degree cos 40\degree cos 80\degree}

= \sqrt{3} \frac{sin 20\degree sin 40\degree sin 80\degree}{cos 20\degree cos 40\degree cos 80\degree}

= \sqrt{3} \frac{(sin 20\degree sin 40\degree) sin 80\degree}{(cos 20\degree cos 40\degree) cos 80\degree}

By using the trigonometric identity,

2 cos A cos B = cos (A+B) + cos (A-B)

cos A cos B = \frac{1}{2} [cos (A+B) + cos (A-B)]

and, 2 sin A sin B = cos (A-B) – cos (A+B)

sin A sin B = \frac{1}{2} [cos (A-B) – cos (A+B)]

Taking A = 40° and B = 20°

= \sqrt{3} (\frac{(\frac{1}{2}[cos ( 40\degree- 20\degree) - cos ( 40\degree+ 20\degree)]) sin 80\degree}{(\frac{1}{2}[cos ( 40\degree+ 20\degree) + cos ( 40\degree- 20\degree)]) cos 80\degree})\\ = \sqrt{3} (\frac{[cos ( 20\degree) - cos (60\degree)] sin 80\degree}{([cos ( 60\degree) + cos (20\degree)) cos 80\degree})\\ = \sqrt{3} (\frac{[cos ( 20\degree) - \frac{1}{2}] sin 80\degree}{(\frac{1}{2} + cos (20\degree)) cos 80\degree})\\ = \sqrt{3} (\frac{(cos ( 20\degree) sin 80\degree - \frac{1}{2} sin 80\degree)}{\frac{1}{2} cos 80\degree + cos (20\degree) cos 80\degree})\\

Again using the identity, we get

2 sin A cos B = sin (A+B) + sin (A-B)

sin A cos B = \frac{1}{2} [sin (A+B) + sin (A-B)]

Taking A = 80° and B = 20°

= \sqrt{3} (\frac{(\frac{1}{2}[sin (80\degree+20\degree) + sin (80\degree-20\degree)] - \frac{1}{2} sin 80\degree)}{\frac{1}{2} cos 80\degree + \frac{1}{2}[cos (80\degree+20\degree) + cos (80\degree-20\degree)]})\\ = \sqrt{3} (\frac{(\frac{1}{2}[sin (100\degree) + sin (60\degree)] - \frac{1}{2} sin 80\degree)}{\frac{1}{2} cos 80\degree + \frac{1}{2}[cos (100\degree) + cos (60\degree)]})\\ = \sqrt{3} (\frac{(\frac{1}{2}[sin (180\degree-80\degree) + sin (60\degree)] - \frac{1}{2} sin 80\degree)}{\frac{1}{2} cos 80\degree + \frac{1}{2}[cos (180\degree-80\degree) + cos (60\degree)]})\\ = \sqrt{3} (\frac{(\frac{1}{2} sin (80\degree) + \frac{1}{2} sin (60\degree) - \frac{1}{2} sin 80\degree)}{\frac{1}{2} (- cos 80\degree) + \frac{1}{2} cos (80\degree) + \frac{1}{2} cos (60\degree)})\\ = \sqrt{3} (\frac{(\frac{1}{2} sin (60\degree)}{\frac{1}{2} cos (60\degree)})\\ = \sqrt{3} (tan (60\degree))\\ = \sqrt{3} (\sqrt{3})\\ = 3

Hence, LHS = RHS

(vi) tan 20° tan 30° tan 40° tan 80° = 1

Solution:

tan 20° tan 30° tan 40° tan 80° = tan 30° \frac{sin 20\degree sin 40\degree sin 80\degree}{cos 20\degree cos 40\degree cos 80\degree}

= \frac{1}{\sqrt{3}} \frac{sin 20\degree sin 40\degree sin 80\degree}{cos 20\degree cos 40\degree cos 80\degree}

= \frac{1}{\sqrt{3}}\frac{(sin 20\degree sin 40\degree) sin 80\degree}{(cos 20\degree cos 40\degree) cos 80\degree}

By using the trigonometric identity,

2 cos A cos B = cos (A+B) + cos (A-B)

cos A cos B = \frac{1}{2} [cos (A+B) + cos (A-B)]

and, 2 sin A sin B = cos (A-B) – cos (A+B)

sin A sin B = \frac{1}{2} [cos (A-B) – cos (A+B)]

Taking A = 40° and B = 20°

= \frac{1}{\sqrt{3}}\frac{(\frac{1}{2}[cos ( 40\degree- 20\degree) - cos ( 40\degree+ 20\degree)]) sin 80\degree}{(\frac{1}{2}[cos ( 40\degree+ 20\degree) + cos ( 40\degree- 20\degree)]) cos 80\degree}\\ = \frac{1}{\sqrt{3}}\frac{[cos ( 20\degree) - cos (60\degree)] sin 80\degree}{([cos ( 60\degree) + cos (20\degree)) cos 80\degree}\\ = \frac{1}{\sqrt{3}} \frac{[cos ( 20\degree) - \frac{1}{2}] sin 80\degree}{(\frac{1}{2}+ cos (20\degree)) cos 80\degree}\\ = \frac{1}{\sqrt{3}} \frac{(cos ( 20\degree) sin 80\degree - \frac{1}{2}sin 80\degree)}{\frac{1}{2} cos 80\degree + cos (20\degree) cos 80\degree}

Again using the identity, we get

2 sin A cos B = sin (A+B) + sin (A-B)

sin A cos B = \frac{1}{2}[sin (A+B) + sin (A-B)]

Taking A = 80° and B = 20°

= \frac{1}{\sqrt{3}} (\frac{1}{2}[sin (80\degree+20\degree) + sin (80\degree-20\degree)] - \frac{1}{2} sin 80\degree){\frac{1}{2} cos 80\degree + \frac{1}{2}[cos (80\degree+20\degree) + cos (80\degree-20\degree)]}\\ = \frac{1}{\sqrt{3}} (\frac{1}{2}[sin (100\degree) + sin (60\degree)] - \frac{1}{2} sin 80\degree){\frac{1}{2} cos 80\degree + \frac{1}{2}[cos (100\degree) + cos (60\degree)]}\\ = \frac{1}{\sqrt{3}} (\frac{1}{2}[sin (180\degree-80\degree) + sin (60\degree)] - \frac{1}{2} sin 80\degree){\frac{1}{2} cos 80\degree + \frac{1}{2}[cos (180\degree-80\degree) + cos (60\degree)]}\\ = \frac{1}{\sqrt{3}} (\frac{1}{2} sin (80\degree) + \frac{1}{2} sin (60\degree) - \frac{1}{2} sin 80\degree){\frac{1}{2} (- cos 80\degree) + \frac{1}{2} cos (80\degree) + \frac{1}{2} cos (60\degree)}\\ = \frac{1}{\sqrt{3}} (\frac{1}{2} sin (60\degree){\frac{1}{2} cos (60\degree)}\\ = \frac{1}{\sqrt{3}} (tan (60\degree))\\ = \frac{1}{\sqrt{3}} (\sqrt{3})\\ = 1

Hence, LHS = RHS

(vii) sin 10° sin 50° sin 60° sin 70° = \frac{\sqrt{3}}{16}

Solution:

sin 10° sin 50° sin 60° sin 70° = sin 60° (sin 10° sin 50° sin 70°)

= \frac{\sqrt{3}}{2} (sin (90-80°) sin (90-40°) sin (90-20°))

= \frac{\sqrt{3}}{2}  (cos (80°) cos (40°) cos (20°))

= \frac{\sqrt{3}}{2}  cos 40° (cos 80° cos 20°)

By using the trigonometric identity,

2 cos A cos B = cos (A+B) + cos (A-B)

cos A cos B = \frac{1}{2} [cos (A+B) + cos (A-B)]

Taking A = 80° and B = 20°

= \frac{\sqrt{3}}{2} cos 40\degree (\frac{1}{2}[cos (80\degree+20\degree) + cos (80\degree-20\degree)])\\ = \frac{\sqrt{3}}{2} \frac{1}{2} cos 40\degree [cos (100\degree) + cos (60°)]\\ = \frac{\sqrt{3}}{4} cos 40\degree[cos (180\degree-80\degree) + cos (60\degree)]\\ = \frac{\sqrt{3}}{4} cos 40\degree [- cos (80\degree) + \frac{1}{2}]\\ = \frac{\sqrt{3}}{4} cos 40\degree [\frac{1}{2}- cos (80\degree)]\\ = \frac{\sqrt{3}}{4} [\frac{1}{2} cos 40\degree - cos (80\degree) cos 40\degree]

Again using the identity, we get

= \frac{\sqrt{3}}{4} [\frac{1}{2} cos 40\degree - \frac{1}{2}(cos (80\degree+40\degree) + cos (80\degree-40\degree))]\\ = \frac{\sqrt{3}}{4} [\frac{1}{2} cos 40\degree - \frac{1}{2}(cos (120\degree) + cos (40\degree))]\\ = \frac{\sqrt{3}}{4} [\frac{1}{2} cos 40\degree - \frac{1}{2}(cos (180\degree-60\degree) + cos (40\degree))]\\ = \frac{\sqrt{3}}{4} [\frac{1}{2} cos 40\degree - \frac{1}{2}(- cos (60\degree) + cos (40\degree))]\\ = \frac{\sqrt{3}}{4} [\frac{1}{2} cos 40\degree + \frac{1}{2} cos (60\degree) - \frac{1}{2} cos (40\degree)]\\ = \frac{\sqrt{3}}{4} [\frac{1}{2} \frac{1}{2}]\\ = \frac{\sqrt{3}}{16}

Hence, LHS = RHS

(viii) sin 20° sin 40° sin 60° sin 80° = \frac{3}{16}

Solution:

sin 20° sin 40° sin 60° sin 80° = sin 60° (sin 20° sin 40° sin 80°)

= \frac{\sqrt{3}}{2} (sin 20° sin 40°) sin 80°

By using the trigonometric identity,

2 sin A sin B = cos (A-B) – cos (A+B)

sin A sin B = \frac{1}{2} [cos (A-B) – cos (A+B)]

Taking A = 40° and B = 20°

= \frac{\sqrt{3}}{2} (\frac{1}{2}[cos (40\degree-20\degree) - cos (40\degree+20\degree)]) sin 80\degree\\ = \frac{\sqrt{3}}{2} \frac{1}{2} sin 80\degree [cos (20\degree) - cos (60\degree)]\\ = \frac{\sqrt{3}}{4} sin 80\degree [cos (20\degree) - \frac{1}{2}]\\ = \frac{\sqrt{3}}{4} [sin 80\degree cos (20\degree) - \frac{1}{2} sin 80\degree]\\

By using the trigonometric identity,

2 sin A cos B = sin (A+B) + sin (A-B)

sin A cos B = \frac{1}{2} [sin (A+B) + sin (A-B)]

Taking A = 80° and B = 20°

= \frac{\sqrt{3}}{4} [(\frac{1}{2}[sin (80\degree+20\degree) + sin (80\degree-20\degree)]) - \frac{1}{2} sin 80\degree]\\ = \frac{\sqrt{3}}{8} [(sin (80\degree+20\degree) + sin (80\degree-20\degree)) - sin 80\degree]\\ = \frac{\sqrt{3}}{8} [sin (100\degree) + sin (60\degree) - sin 80\degree]\\ = \frac{\sqrt{3}}{8} [sin (180\degree-80\degree) + \frac{\sqrt{3}}{2} - sin 80\degree]\\ = \frac{\sqrt{3}}{8} [sin (80\degree) + \frac{\sqrt{3}}{2} - sin 80\degree]\\ = \frac{\sqrt{3}}{8} [\frac{\sqrt{3}}{2}]\\ = \frac{3}{16}

Hence, LHS = RHS

Question 6. Show that

(i) sin A sin (B-C) + sin B sin (C-A) + sin C sin (A-B) = 0

Solution:

By using the trigonometric identity,

2 sin θ sin Φ = cos (θ-Φ) – cos (θ+Φ)

sin θ sin Φ = \frac{1}{2} [cos (θ-Φ) – cos (θ+Φ)]

sin A sin (B-C) + sin B sin (C-A) + sin C sin (A-B) = (\frac{1}{2} [cos (A-(B-C)) – cos (A+(B-C))]) + (\frac{1}{2} [cos (B-(C-A)) – cos (B+(C-A))]) + (\frac{1}{2} [cos (C-(A-B)) – cos (C+(A-B))])

= \frac{1}{2} (cos (A-B+C)) – cos (A+B-C) + cos (B-C+A) – cos (B+C-A) + cos (C-A+B) – cos (C+A-B))

= \frac{1}{2} (cos (A-B+C)) – cos (C+A-B) – cos (A+B-C) + cos (B-C+A) – cos (B+C-A) + cos (C-A+B))

= \frac{1}{2}(0)

 = 0

Hence, LHS = RHS

(ii) sin (B-C) cos (A-D) + sin (C-A) cos (B-D) + sin (A-B) cos (C-D) = 0

Solution:

By using the trigonometric identity,

2 sin θ cos Φ = sin (θ+Φ) + sin (θ-Φ)

sin θ cos Φ = \frac{1}{2} [sin (θ+Φ) + sin (θ-Φ)]

sin (B-C) cos (A-D) + sin (C-A) cos (B-D) + sin (A-B) cos (C-D) = (\frac{1}{2} [sin (B-C+(A-D)) + sin (B-C-(A-D))]) + (\frac{1}{2} [sin (C-A+(B-D)) + sin (C-A-(B-D))]) +(\frac{1}{2} [sin (A-B+(C-D)) + sin (A-B-(C-D))])

= (\frac{1}{2} [sin (A+B-C-D) + sin (-A+B-C+D)]) + (\frac{1}{2} [sin (-A+B+C-D) + sin (-A-B+C+D)]) +(\frac{1}{2} [sin (A-B+C-D) + sin (A-B-C+-D)])

= \frac{1}{2} (sin (A+B-C-D) + sin (-(A-B+C-D)) + sin (-(A-B-C+D)) + sin (-(A+B-C-D)) +sin (A-B+C-D) + sin (A-B-C+-D))

= \frac{1}{2} (sin (A+B-C-D) – sin(A-B+C-D) – sin (A-B-C+D) – sin (A+B-C-D) +sin (A-B+C-D) + sin (A-B-C+-D))

= \frac{1}{2}(0)

= 0

Hence, LHS = RHS

Question 7. Prove that : tan θ tan (60°-θ) tan (60°+θ) = tan 3θ

Solution:

tan θ tan (60°-θ) tan (60°+θ) = tan θ (tan (60°-θ)) (tan (60°+θ))

By using the trigonometric identity,

tan (a+b) = \frac{tan\hspace{0.1cm} a + tan\hspace{0.1cm} b}{1 - tan\hspace{0.1cm} a \hspace{0.1cm}tan\hspace{0.1cm} b}

tan (a+b) = \frac{tan\hspace{0.1cm} a - tan\hspace{0.1cm} b}{1 + tan\hspace{0.1cm} a \hspace{0.1cm}tan\hspace{0.1cm} b}

= tan\hspace{0.1cm} θ (\frac{tan 60\degree - tan θ}{1 + tan 60\degree tan θ}) ( \frac{tan 60\degree + tan θ}{1 - tan 60\degree tan θ})\\ = tan\hspace{0.1cm} θ (\frac{(tan 60\degree)^2 - (tan θ)^2}{1^2 - (tan 60\degree° tan θ)^2})\hspace{0.1cm}\hspace{0.1cm}\hspace{0.1cm}((a+b)(a-b)=a^2-b^2)\\ = tan\hspace{0.1cm} θ (\frac{(tan 60\degree)^2 - tan^2 θ}{1 - (tan 60\degree)^2 tan^2 θ})\\ = tan \hspace{0.1cm}θ (\frac{(\sqrt{3})^2 - tan^2 θ}{1 - (\sqrt{3})^2 tan^2 θ})\\ = tan\hspace{0.1cm} θ (\frac{3 - tan^2 θ}{1 - 3 tan^2 θ})\\ = \frac{3 \hspace{0.1cm}tan\hspace{0.1cm} \theta - tan^3 \theta}{1 - 3 \hspace{0.1cm}tan^2 \theta}

= tan 3θ

Hence, LHS = RHS

Question 8. If α + β = 90°, show that the maximum value of cos(α) cos(β) is \frac{1}{2}.

Solution:

cos(α) cos(β) = y

By using the trigonometric identity,

2 cos A cos B = cos (A+B) + cos (A-B)

cos A cos B = \frac{1}{2} [cos (A+B) + cos (A-B)]

Taking A = α and B = β

cos(α) cos(β) = \frac{1}{2} [cos (α+β) + cos (α-β)]

As, α + β = 90°

y = \frac{1}{2} [cos (90°) + cos (α-β)]

y = \frac{1}{2} [0 + cos (α-β)]

y = \frac{1}{2} (cos (α-β))

AS, we know that range of cos function is [-1,1]

-1\leq (cos (\alpha-\beta)) \leq 1

\frac{-1}{2} \leq \frac{1}{2}(cos (\alpha-\beta)) \leq \frac{1}{2}

\frac{-1}{2} \leq y \leq \frac{1}{2}

Hence, the maximum value of cos(α) cos(β) is \frac{1}{2}.



Like Article
Suggest improvement
Previous
Class 11 RD Sharma Solutions - Chapter 7 Trigonometric Ratios of Compound Angles - Exercise 7.2
Next
Class 11 RD Sharma Solutions - Chapter 8 Transformation Formulae - Exercise 8.2 | Set 1
Share your thoughts in the comments

Similar Reads

Class 11 RD Sharma Solutions - Chapter 8 Transformation Formulae - Exercise 8.2 | Set 1
Class 11 RD Sharma Solutions - Chapter 8 Transformation Formulae - Exercise 8.2 | Set 2
Class 11 RD Sharma Solutions - Chapter 10 Sine and Cosine Formulae and Their Applications - Exercise 10.2 | Set 1
Class 11 RD Sharma Solutions - Chapter 10 Sine and Cosine Formulae and Their Applications - Exercise 10.2 | Set 2
Class 11 RD Sharma Solutions - Chapter 10 Sine and Cosine Formulae and Their Applications - Exercise 10.1 | Set 1
Class 11 RD Sharma Solutions - Chapter 10 Sine and Cosine Formulae and Their Applications - Exercise 10.1 | Set 2
Class 11 RD Sharma Solutions - Chapter 31 Derivatives - Exercise 31.1
Class 10 RD Sharma Solutions - Chapter 11 Constructions - Exercise 11.2 | Set 1
Class 12 RD Sharma Solutions - Chapter 9 Continuity - Exercise 9.1 | Set 2
Class 11 RD Sharma Solutions - Chapter 12 Mathematical Induction - Exercise 12.1
author
_shinchancode
Article Tags :
We use cookies to ensure you have the best browsing experience on our website. By using our site, you acknowledge that you have read and understood our Cookie Policy & Privacy Policy
Lightbox