Class 11: Trigonometric Functions – Exercise 5.3 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the values of the following trigonometric ratios:

i) $\sin \Big($ $\frac{5 \pi}{3}$ $\Big)$ ii) $\sin 17 \pi$ iii) $\tan \Big($ $\frac{11 \pi}{6}$ $\Big)$ iv) $\cos \Big( -$ $\frac{25\pi}{4}$ $\Big)$ v) $\tan \Big($ $\frac{7\pi}{4}$ $\Big)$

vi) $\sin \Big($ $\frac{17\pi}{6}$ $\Big)$ vii) $\cos \Big($ $\frac{19\pi}{6}$ $\Big)$ viii) $\sin \Big( -$ $\frac{11\pi}{6}$ $\Big)$ ix) $\mathrm{cosec} \Big( -$ $\frac{20\pi}{3}$ $\Big)$

x) $\tan \Big( -$ $\frac{13\pi}{4}$ $\Big)$ xi) $\cos \Big($ $\frac{19\pi}{4}$ $\Big)$ xii) $\sin \Big($ $\frac{41\pi}{4}$ $\Big)$ xiii) $\cos \Big($ $\frac{39\pi}{4}$ $\Big)$

xiv) $\sin \Big($ $\frac{151\pi}{6}$ $\Big)$

Answer:

i) $\sin \Big($ $\frac{5 \pi}{3}$ $\Big)$ $= \sin \Big($ $2\pi -$ $\frac{\pi}{3}$ $\Big)$ $= - \sin$ $\frac{\pi}{3}$ $= -$ $\frac{\sqrt{3}}{2}$

ii) $\sin 17 \pi = 0$ $(\because n\pi = 0$ for all $n \in Z)$

iii) $\tan \Big($ $\frac{11 \pi}{6}$ $\Big)$ $= \tan \Big($ $2\pi -$ $\frac{\pi}{6}$ $\Big)$ $= - \tan$ $\frac{\pi}{6}$ $= -$ $\frac{1}{\sqrt{3}}$

iv) $\cos \Big( -$ $\frac{25\pi}{4}$ $\Big)$ $= \cos \Big($ $6\pi +$ $\frac{\pi}{4}$ $\Big)$ $= \cos$ $\frac{\pi}{4}$ $=$ $\frac{1}{\sqrt{2}}$ [Note: $\cos(-x) = \cos x$ ]

v) $\tan \Big($ $\frac{7\pi}{4}$ $\Big)$ $= \tan \Big($ $2\pi -$ $\frac{\pi}{4}$ $\Big)$ $= - \tan$ $\frac{\pi}{4}$ $= -1$

vi) $\sin \Big($ $\frac{17\pi}{6}$ $\Big)$ $= \sin \Big($ $3\pi -$ $\frac{\pi}{6}$ $\Big)$ $= \sin$ $\frac{\pi}{6}$ $=$ $\frac{1}{2}$

vii) $\cos \Big($ $\frac{19\pi}{6}$ $\Big)$ $= \cos \Big($ $3\pi +$ $\frac{\pi}{6}$ $\Big)$ $= -\cos$ $\frac{\pi}{6}$ $= -$ $\frac{\sqrt{3}}{2}$

viii) $\sin \Big( -$ $\frac{11\pi}{6}$ $\Big)$ $= \sin \Big(- \Big($ $2\pi -$ $\frac{\pi}{6}$ $\Big) \Big)$ $= \sin \Big($ $2\pi -$ $\frac{\pi}{6}$ $\Big)$ $= \sin$ $\frac{\pi}{6}$ $=$ $\frac{1}{2}$

ix) $\mathrm{cosec} \Big( -$ $\frac{20\pi}{3}$ $\Big)$ $= \mathrm{cosec} \Big(- \Big($ $7\pi -$ $\frac{\pi}{3}$ $\Big) \Big)$ $= - \mathrm{cosec} \Big($ $7\pi -$ $\frac{\pi}{3}$ $\Big)$ $= -\mathrm{cosec}$ $\frac{\pi}{3}$ $= -$ $\frac{2}{\sqrt{3}}$

x) $\tan \Big( -$ $\frac{13\pi}{4}$ $\Big)$ $= - \tan \Big($ $3\pi +$ $\frac{\pi}{4}$ $\Big)$ $= - \tan \Big($ $2\pi +\pi +$ $\frac{\pi}{4}$ $\Big)$ $= - \tan$ $\frac{\pi}{4}$ $= -1$

xi) $\cos \Big($ $\frac{19\pi}{4}$ $\Big)$ $= \cos \Big($ $5\pi -$ $\frac{\pi}{4}$ $\Big)$ $= \cos \Big($ $4\pi +\pi -$ $\frac{\pi}{4}$ $\Big)$ $= \cos \Big($ $\pi -$ $\frac{\pi}{4}$ $\Big)$ $= - \cos$ $\frac{\pi}{4}$ $= -$ $\frac{1}{2}$

xii) $\sin \Big($ $\frac{41\pi}{4}$ $\Big)$ $= \sin \Big($ $10\pi +$ $\frac{\pi}{4}$ $\Big)$ $= \sin$ $\frac{\pi}{4}$ $=$ $\frac{1}{\sqrt{2}}$

xiii) $\cos \Big($ $\frac{39\pi}{4}$ $\Big)$ $= \cos \Big($ $10\pi -$ $\frac{\pi}{4}$ $\Big)$ $= \cos$ $\frac{\pi}{4}$ $=$ $\frac{1}{\sqrt{2}}$

xiv) $\sin \Big($ $\frac{151\pi}{6}$ $\Big)$ $= \sin \Big($ $25\pi +$ $\frac{\pi}{6}$ $\Big)$ $= \sin \Big($ $24\pi +\pi +$ $\frac{\pi}{6}$ $\Big)$ $= \sin \Big($ $\pi +$ $\frac{\pi}{6}$ $\Big)$ $= - \sin$ $\frac{\pi}{6}$ $= -$ $\frac{1}{2}$

$\\$

Question 2: Prove that:

i) $\tan 225^o \cot 405^o + \tan 765^o \cot 675^o = 0$

ii) $\sin$ $\frac{8 \pi}{3}$ $\cos$ $\frac{23 \pi}{6}$ $+ \cos$ $\frac{13 \pi}{3}$ $\sin$ $\frac{35 \pi}{6}$ $=$ $\frac{1}{2}$

iii) $\cos 24^o + \cos 55^o + \cos 125^o + \cos 204^o + \cos 300^o =$ $\frac{1}{2}$

iv) $\tan (-225^o) \cot(-405^o) - \tan (-765^o) \cot (675^o) = 0$

v) $\cos 570^o \sin 510^o + \sin (-300^o) \cos (-390^o) = 0$

vi) $\tan$ $\frac{11 \pi}{3}$ $- 2 \sin$ $\frac{4 \pi}{6}$ $-$ $\frac{3}{4}$ $\mathrm{cosec}^2$ $\frac{\pi}{4}$ $+ 4 \cos^2$ $\frac{17\pi}{6}$ $=$ $\frac{3-4\sqrt{3}}{2}$

vii) $3\sin$ $\frac{\pi}{6}$ $\sec$ $\frac{ \pi}{3}$ $-4 \sin$ $\frac{5 \pi}{6}$ $\cot$ $\frac{\pi}{4}$ $= 1$

Answer:

i) LHS $= \tan 225^o \cot 405^o + \tan 765^o \cot 675^o$

$= \tan \Big( \pi +$ $\frac{\pi}{4}$ $\Big) \cot \Big( 2\pi +$ $\frac{\pi}{4}$ $\Big) + \tan \Big( 4\pi +$ $\frac{\pi}{4}$ $\Big) \cot \Big( 4\pi -$ $\frac{\pi}{4}$ $\Big)$

$= \tan$ $\frac{\pi}{4}$ $\cot$ $\frac{\pi}{4}$ $+ \tan$ $\frac{\pi}{4}$ $\Big( - \cot$ $\frac{\pi}{4}$ $\Big)$

$= 1 - 1 = 0 =$ RHS. Hence proved.

ii) LHS $= \sin$ $\frac{8 \pi}{3}$ $\cos$ $\frac{23 \pi}{6}$ $+ \cos$ $\frac{13 \pi}{3}$ $\sin$ $\frac{35 \pi}{6}$

$= \sin \Big( 3\pi -$ $\frac{\pi}{3}$ $\Big) \cos \Big( 4\pi -$ $\frac{\pi}{6}$ $\Big) + \cos \Big( 4\pi +$ $\frac{\pi}{3}$ $\Big) \sin \Big( 6\pi -$ $\frac{\pi}{6}$ $\Big)$

$= \sin$ $\frac{\pi}{3}$ $\cos$ $\frac{\pi}{6}$ $+ \cos$ $\frac{\pi}{3}$ $\Big( - \sin$ $\frac{\pi}{6}$ $\Big)$

$=$ $\frac{\sqrt{3}}{2}$ $\times$ $\frac{\sqrt{3}}{2}$ $+$ $\frac{1}{2}$ $\Big( -$ $\frac{1}{2}$ $\Big)$

$=$ $\frac{3}{4}$ $-$ $\frac{1}{4}$ $=$ $\frac{2}{4}$ $=$ $\frac{1}{2}$

iii) LHS $= \cos 24^o + \cos 55^o + \cos 125^o + \cos 204^o + \cos 300^o$

$= ( \cos 24^o + \cos 204^o) + ( \cos 55^o + \cos 125^o) + \cos 300^o$

$= ( \cos 24^o + \cos (\pi + 24^o) ) + ( \cos 55^o + \cos (\pi - 55^o)) + \cos (2\pi - 60^o)$

$= \cos 24^o - \cos 24^o + \cos 55^o - \cos 55^o + \cos 60^o$

$= \cos 60^o =$ $\frac{1}{2}$ $=$ RHS. Hence proved.

iv) LHS $= \tan (-225^o) \cot(-405^o) - \tan (-765^o) \cot (675^o)$

$= -\tan 225^o (-\cot 405^o ) + \tan 765^o \cot 675^o$

$= \tan \Big( \pi +$ $\frac{\pi}{4}$ $\Big) \cot \Big( 2\pi +$ $\frac{\pi}{4}$ $\Big) + \tan \Big( 4\pi +$ $\frac{\pi}{4}$ $\Big) \cot \Big( 4\pi -$ $\frac{\pi}{4}$ $\Big)$

$= \tan$ $\frac{\pi}{4}$ $\cot$ $\frac{\pi}{4}$ $+ \tan$ $\frac{\pi}{4}$ $\Big( - \cot$ $\frac{\pi}{4}$ $\Big)$

$= 1 - 1 = 0 =$ RHS. Hence proved.

v) LHS $= \cos 570^o \sin 510^o + \sin (-300^o) \cos (-390^o)$

$= \cos \Big( 3\pi +$ $\frac{\pi}{6}$ $\Big) \sin \Big( 3\pi -$ $\frac{\pi}{6}$ $\Big) - \sin \Big( 2\pi -$ $\frac{\pi}{6}$ $\Big) \cos \Big( 2\pi +$ $\frac{\pi}{6}$ $\Big)$

$= - \cos$ $\frac{\pi}{6}$ $\sin$ $\frac{\pi}{6}$ $+ \sin$ $\frac{\pi}{6}$ $\cos$ $\frac{\pi}{6}$

$= 0 =$ RHS. Hence proved.

vi) LHS $= \tan$ $\frac{11 \pi}{3}$ $- 2 \sin$ $\frac{4 \pi}{6}$ $-$ $\frac{3}{4}$ $\mathrm{cosec}^2$ $\frac{\pi}{4}$ $+ 4 \cos^2$ $\frac{17\pi}{6}$

$= \tan \Big( 4\pi -$ $\frac{\pi}{3}$ $\Big) - 2 \sin$ $\frac{2\pi}{3}$ $-$ $\frac{3}{4}$ $\mathrm{cosec}^2$ $\frac{\pi}{4}$ $+ 4 \cos^2 \Big( 3\pi -$ $\frac{\pi}{6}$ $\Big)$

$= - \tan$ $\frac{\pi}{3}$ $- 2 \sin \Big( \pi -$ $\frac{\pi}{3}$ $\Big) -$ $\frac{3}{4}$ $(\sqrt{2})^2 + 4 \cos^2$ $\frac{\pi}{3}$

$= - \sqrt{3} - 2$ $\frac{\sqrt{3}}{2}$ $-$ $\frac{3}{2}$ $+ 4.$ $\frac{3}{4}$

$= - \sqrt{3} - \sqrt{3} +$ $\frac{3}{2}$ $=$ $\frac{3-4\sqrt{3}}{2}$ $=$ RHS.

vi) LHS $= 3\sin$ $\frac{\pi}{6}$ $\sec$ $\frac{ \pi}{3}$ $-4 \sin$ $\frac{5 \pi}{6}$ $\cot$ $\frac{\pi}{4}$

$= 3 \times$ $\frac{1}{2}$ $\times 2 - 3 \times \sin \Big( \pi -$ $\frac{\pi}{6}$ $\Big) . 1$

$= 3 - 4 \sin$ $\frac{\pi}{6}$

$= 3 - 4 \times$ $\frac{1}{2}$ $= 3 - 2 = 1 =$ RHS. Hence proved.

$\\$

Question 3: Prove that:

i) $\frac{\cos (2 \pi +x) \mathrm{cosec} (2 \pi +x) \tan (\frac{\pi}{2}+x)}{\sec (\frac{\pi}{2}+x) \cos x \cot (\pi + x) }$ $= 1$

ii) $\frac{ \mathrm{cosec} (90^o+x) + \cot(450^o + x)}{ \mathrm{cosec} (90^o-x) + \tan (180^o - x)}$ $+$ $\frac{\tan (180^o+x)+ \sec (180^o-x)}{\tan (360^o+x)- \sec(-x)}$ $= 2$

iii) $\frac{\sin (\pi + x) \cos (\frac{\pi}{2} +x ) \tan (\frac{3\pi}{2} -x ) \cot (2\pi-x) }{\sin (2\pi -x) \cos (2\pi +x) \mathrm{cosec} (-x) \sin (\frac{3\pi}{2} -x ) }$ $= 1$

iv) $\Big\{ 1 + \cot x - \sec \Big($ $\frac{\pi}{2}$ $+x \Big) \Big\} \Big\{ 1 + \cot x + \sec \Big($ $\frac{\pi}{2}$ $+x \Big) \Big\} = 2 \cot x$

v) $\frac{\tan ( \frac{\pi}{2}-x ) \sec (\pi - x) \sin (-x) }{ \sin (\pi + x) \cot (2 \pi - x) \mathrm{cosec} ( \frac{\pi}{2}-x )}$ $= 1$

Answer:

i) LHS $=$ $\frac{\cos (2 \pi +x) \mathrm{cosec} (2 \pi +x) \tan (\frac{\pi}{2}+x)}{\sec (\frac{\pi}{2}+x) \cos x \cot (\pi + x) }$

$=$ $\frac{\cos x \ \mathrm{cosec} x \ (-\cot x)}{- \mathrm{cosec} x \ \cos x \ \cot x }$

$= 1 =$ RHS

ii) LHS $=$ $\frac{ \mathrm{cosec} (90^o+x) + \cot(450^o + x)}{ \mathrm{cosec} (90^o-x) + \tan (180^o - x)}$ $+$ $\frac{\tan (180^o+x)+ \sec (180^o-x)}{\tan (360^o+x)- \sec(-x)}$

$=$ $\frac{ \sec x + \cot(2\pi + \frac{\pi}{2} + x)}{ \sec x - \tan x}$ $+$ $\frac{\tan x - \sec x}{\tan x- \sec x}$

$=$ $\frac{ \sec x + \cot( \frac{\pi}{2} + x)}{ \sec x - \tan x}$ $+ 1$

$=$ $\frac{ \sec x - \tan x}{ \sec x - \tan x}$ $+ 1$

$= 1 + 1 = 2 =$ RHS. Hence proved.

iii) LHS $=$ $\frac{\sin (\pi + x) \cos (\frac{\pi}{2} +x ) \tan (\frac{3\pi}{2} -x ) \cot (2\pi-x) }{\sin (2\pi -x) \cos (2\pi +x) \mathrm{cosec} (-x) \sin (\frac{3\pi}{2} -x ) }$

$=$ $\frac{\sin x (- \sin x ) \cot x (- \cot x ) }{- \sin x \cos x ( - \mathrm{cosec} x ) (- \cos x) }$

$=$ $\frac{\sin^2 x \cot^2 x}{\cos^2 x}$

$= \tan^2 x \cot^2 x$

$= 1=$ RHS. Hence proved.

iv) LHS $= \Big\{ 1 + \cot x - \sec \Big($ $\frac{\pi}{2}$ $+x \Big) \Big\} \Big\{ 1 + \cot x + \sec \Big($ $\frac{\pi}{2}$ $+x \Big) \Big\}$

$= ( 1 + \cot x - (\mathrm{cosec} x) )( 1 + \cot x - \mathrm{cosec} x )$

$= ( 1 + \cot x + \mathrm{cosec} x )( 1 + \cot x - \mathrm{cosec} x )$

$= ( 1 + \cot^2 x )^2 - \mathrm{cosec}^2 x$

$= 1 + \cot^2 x + 2 \cot x - \mathrm{cosec}^2 x$

$= \mathrm{cosec}^2 x + 2 \cot x - \mathrm{cosec}^2 x$

$= 2 \cot x =$ RHS. Hence proved.

v) LHS $=$ $\frac{\tan ( \frac{\pi}{2}-x ) \sec (\pi - x) \sin (-x) }{ \sin (\pi + x) \cot (2 \pi - x) \mathrm{cosec} ( \frac{\pi}{2}-x )}$

$=$ $\frac{\cot x (- \sec x) (- \sin x) }{ (- \sin x) (- \cot x) \sec x }$

$= 1 =$ RHS. Hence proved.

$\\$

Question 4: Prove that: $\sin^2$ $\frac{\pi}{18}$ $+ \sin^2$ $\frac{\pi}{9}$ $+ \sin^2$ $\frac{7\pi}{18}$ $+ \sin^2$ $\frac{4\pi}{9}$ $= 2$

Answer:

LHS $= \sin^2$ $\frac{\pi}{18}$ $+ \sin^2$ $\frac{\pi}{9}$ $+ \sin^2$ $\frac{7\pi}{18}$ $+ \sin^2$ $\frac{4\pi}{9}$

$= \sin^2$ $\Big($ $\frac{\pi}{2}$ $-$ $\frac{4\pi}{9}$ $\Big)$ $+ \sin^2$ $\frac{\pi}{9}$ $+ \sin^2$ $\Big($ $\frac{\pi}{2}$ $-$ $\frac{\pi}{9}$ $\Big)$ $+ \sin^2$ $\frac{4\pi}{9}$

$= \cos^2$ $\frac{4\pi}{9}$ $+ \sin^2$ $\frac{4\pi}{9}$ $+ \sin^2$ $\frac{\pi}{9}$ $+ \cos^2$ $\frac{\pi}{9}$

$= 1 + 1 = 2 =$ RHS. Hence proved.

$\\$

Question 5: Prove that:

$\sec \Big($ $\frac{3\pi}{2}$ $-x \Big) \sec \Big( x -$ $\frac{5\pi}{2}$ $\Big) + \tan \Big($ $\frac{5\pi}{2}$ $+x \Big) \tan \Big( x -$ $\frac{3\pi}{2}$ $\Big) = -1$

Answer:

LHS $= \sec \Big($ $\frac{3\pi}{2}$ $-x \Big) \sec \Big( x -$ $\frac{5\pi}{2}$ $\Big) + \tan \Big($ $\frac{5\pi}{2}$ $+x \Big) \tan \Big( x -$ $\frac{3\pi}{2}$ $\Big)$

$= \sec \Big($ $\frac{3\pi}{2}$ $-x \Big) \sec \Big( - ($ $\frac{5\pi}{2}$ $- x \Big) + \tan \Big($ $\frac{5\pi}{2}$ $+x \Big) \tan \Big( -($ $\frac{3\pi}{2}$ $- x) \Big)$

$= -\mathrm{cosec} x \sec \Big($ $\frac{5\pi}{2}$ $- x \Big) - \cot x \Big[ - \tan \Big($ $\frac{3\pi}{2}$ $- x \Big) \Big]$

$= -\mathrm{cosec} x . \mathrm{cosec} x + \cot x . \cot x$

$= -\mathrm{cosec}^x + \cot^2 x$

$= -\mathrm{cosec}^x + \mathrm{cosec}^x - 1$

$= - 1 =$ RHS. Hence proved.

$\\$

Question 6: In a $\triangle ABC$ , prove that: i) $\cos (A+B) + \cos C = 0$

ii) $\cos \Big($ $\frac{A+B}{2}$ $\Big) = \sin \Big($ $\frac{C}{2}$ $\Big)$ iii) $\tan \Big($ $\frac{A+B}{2}$ $\Big) = \cot \Big($ $\frac{C}{2}$ $\Big)$

Answer:

i) LHS $= \cos (A+B) + \cos C$

$= \cos (A+B) + \cos (\pi - (A + B))$

$= \cos (A+B) - \cos (A+B)$

$= 0 =$ RHS. Hence proved.

ii) LHS $= \cos \Big($ $\frac{A+B}{2}$ $\Big)$ $= \cos \Big( 90^o -$ $\frac{C}{2}$ $\Big)$ $= \sin$ $\frac{C}{2}$ Hence proved.

iii) LHS $= \tan \Big($ $\frac{A+B}{2}$ $\Big)$ $= \tan \Big( 90^o -$ $\frac{C}{2}$ $\Big)$ $= \cot$ $\frac{C}{2}$ Hence proved.

$\\$

Question 7: If A, B, C, D be the angles of a cyclic quadrilateral, taken in order, prove that: $\cos (180^o - A) + \cos (180^o + B) + \cos (180^o + C ) - \sin (90^o + D) = 0$

Answer:

Given $A + C = 180^o, B+D = 180^o$

LHS $= \cos (180^o - A) + \cos (180^o + B) + \cos (180^o + C ) - \sin (90^o + D)$

$= \cos C + \cos D - \cos C - \cos D = 0 =$ RHS. Hence proved.

$\\$

Question 8: Find $x$ from the following equations:

i) $\mathrm{cosec} \Big($ $\frac{\pi}{2}$ $+ \theta \Big) + x \cos \theta \cot \Big($ $\frac{\pi}{2}$ $+ \theta \Big) = \sin \Big($ $\frac{\pi}{2}$ $+ \theta \Big)$

ii) $x \cot \Big($ $\frac{\pi}{2}$ $+ \theta \Big) + \tan \Big($ $\frac{\pi}{2}$ $+ \theta \Big) \sin \theta + \mathrm{cosec} \Big($ $\frac{\pi}{2}$ $+ \theta \Big) = 0$

Answer:

i) $\mathrm{cosec} \Big($ $\frac{\pi}{2}$ $+ \theta \Big) + x \cos \theta \cot \Big($ $\frac{\pi}{2}$ $+ \theta \Big) = \sin \Big($ $\frac{\pi}{2}$ $+ \theta \Big)$

$\Rightarrow \sec \theta + x \cos \theta ( - \tan \theta) = \cos \theta$

$\Rightarrow$ $\frac{1}{\cos \theta}$ $+ x ( - \sin \theta) = \cos \theta$

$\Rightarrow$ $\frac{1 - \cos^2 \theta}{\cos \theta}$ $= x \sin \theta$

$\Rightarrow x =$ $\frac{\sin^2 \theta}{\cos \theta . \sin \theta}$ $=$ $\frac{\sin \theta}{\cos \theta}$ $= \tan \theta$

ii) $x \cot \Big($ $\frac{\pi}{2}$ $+ \theta \Big) + \tan \Big($ $\frac{\pi}{2}$ $+ \theta \Big) \sin \theta + \mathrm{cosec} \Big($ $\frac{\pi}{2}$ $+ \theta \Big) = 0$

$\Rightarrow x ( - \tan \theta) - \cot \theta \sin \theta + \sec \theta = 0$

$\Rightarrow x \tan \theta =$ $\frac{1}{\cos \theta}$ $- \cos \theta$

$\Rightarrow x \tan \theta =$ $\frac{1 - \cos^2 \theta}{\cos \theta}$

$\Rightarrow x =$ $\frac{\sin^2 \theta . \cos \theta}{\sin \theta . \cos \theta}$ $= \sin \theta$

$\\$

Question 9: Prove that:

i) $\tan 4\pi - \cos$ $\frac{3\pi}{2}$ $- \sin$ $\frac{5\pi}{6}$ $\cos$ $\frac{2\pi}{3}$ $= \frac{1}{4}$

ii) $\sin$ $\frac{13\pi}{3}$ $\sin$ $\frac{8\pi}{3}$ $+ \cos$ $\frac{2\pi}{3}$ $\sin$ $\frac{5\pi}{6}$ $=$ $\frac{1}{2}$

iii) $\sin$ $\frac{13\pi}{3}$ $\sin$ $\frac{2\pi}{3}$ $+ \cos$ $\frac{4\pi}{3}$ $\sin$ $\frac{13\pi}{6}$ $=$ $\frac{1}{2}$

iv) $\sin$ $\frac{10\pi}{3}$ $\cos$ $\frac{13\pi}{6}$ $+ \cos$ $\frac{8\pi}{3}$ $\sin$ $\frac{5\pi}{6}$ $= -1$

v) $\tan$ $\frac{5\pi}{4}$ $\cot$ $\frac{9\pi}{4}$ $+ \tan$ $\frac{17\pi}{4}$ $\cot$ $\frac{15\pi}{4}$ $= 0$

Answer:

i) LHS $= \tan 4\pi - \cos$ $\frac{3\pi}{2}$ $- \sin$ $\frac{5\pi}{6}$ $\cos$ $\frac{2\pi}{3}$

$= 0 - 0 - \sin \Big( \pi -$ $\frac{\pi}{6}$ $\Big) \cos \Big($ $\frac{\pi}{2}$ $+$ $\frac{\pi}{6}$ $\Big)$

$= - \sin$ $\frac{\pi}{6}$ $\Big( - \sin$ $\frac{\pi}{6}$ $\Big)$

$= \sin^2$ $\frac{\pi}{6}$ $= \Big($ $\frac{1}{2}$ $\Big)^2 =$ $\frac{1}{4}$ $=$ RHS. Hence proved.

ii) LHS $= \sin$ $\frac{13\pi}{3}$ $\sin$ $\frac{2\pi}{3}$ $+ \cos$ $\frac{4\pi}{3}$ $\sin$ $\frac{13\pi}{6}$

$= \sin \Big( 4\pi +$ $\frac{\pi}{3}$ $\Big) \sin \Big( 3\pi -$ $\frac{\pi}{3}$ $\Big) + \cos \Big($ $\frac{\pi}{2}$ $+$ $\frac{\pi}{6}$ $\Big) \sin \Big( \pi -$ $\frac{\pi}{6}$ $\Big)$

$= \sin$ $\frac{\pi}{3}$ $\sin$ $\frac{\pi}{3}$ $- \sin$ $\frac{\pi}{6}$ $\sin$ $\frac{\pi}{6}$

$= \sin^2$ $\frac{\pi}{3}$ $- \sin^2$ $\frac{\pi}{6}$

$= \Big($ $\frac{\sqrt{3}}{2}$ $\Big)^2 - \Big($ $\frac{1}{2}$ $\Big)^2 =$ $\frac{3}{4}$ $-$ $\frac{1}{4}$ $=$ $\frac{1}{2}$ $=$ RHS. Hence proved.

iii) LHS $= \sin$ $\frac{13\pi}{3}$ $\sin$ $\frac{2\pi}{3}$ $+ \cos$ $\frac{4\pi}{3}$ $\sin$ $\frac{13\pi}{6}$

$= \sin \Big( 4\pi +$ $\frac{\pi}{3}$ $\Big) \sin \Big( \pi -$ $\frac{\pi}{3}$ $\Big) + \cos \Big( \pi +$ $\frac{\pi}{3}$ $\Big) \sin \Big( 2\pi -$ $\frac{\pi}{6}$ $\Big)$

$= \sin$ $\frac{\pi}{3}$ $\sin$ $\frac{\pi}{3}$ $- \cos$ $\frac{\pi}{3}$ $\sin$ $\frac{\pi}{6}$

$= \sin^2$ $\frac{\pi}{3}$ $- \cos$ $\frac{\pi}{3}$ $\sin$ $\frac{\pi}{6}$

$= \Big($ $\frac{\sqrt{3}}{2}$ $\Big)^2 - \Big($ $\frac{1}{2}$ $\Big)^2 =$ $\frac{3}{4}$ $-$ $\frac{1}{4}$ $=$ $\frac{1}{2}$ $=$ RHS. Hence proved.

iv) LHS $= \sin$ $\frac{10\pi}{3}$ $\cos$ $\frac{13\pi}{6}$ $+ \cos$ $\frac{8\pi}{3}$ $\sin$ $\frac{5\pi}{6}$

$= \sin \Big( 3\pi +$ $\frac{\pi}{3}$ $\Big) \cos \Big( 2\pi +$ $\frac{\pi}{6}$ $\Big) + \cos \Big( 3\pi -$ $\frac{\pi}{3}$ $\Big) \sin \Big( \pi -$ $\frac{\pi}{6}$ $\Big)$

$= \sin \Big( \pi +$ $\frac{\pi}{3}$ $\Big) \cos \Big($ $\frac{\pi}{6}$ $\Big) + \cos \Big( \pi -$ $\frac{\pi}{3}$ $\Big) \sin \Big($ $\frac{\pi}{6}$ $\Big)$

$= - \sin$ $\frac{\pi}{3}$ $\cos$ $\frac{\pi}{6}$ $- \cos$ $\frac{\pi}{3}$ $\sin$ $\frac{\pi}{6}$

$= - \Big($ $\frac{\sqrt{3}}{2}$ $\Big)^2 - \Big($ $\frac{1}{2}$ $\Big)^2 = -$ $\frac{3}{4}$ $-$ $\frac{1}{4}$ $= -1$ $=$ RHS. Hence proved.

v) LHS $= \tan$ $\frac{5\pi}{4}$ $\cot$ $\frac{9\pi}{4}$ $+ \tan$ $\frac{17\pi}{4}$ $\cot$ $\frac{15\pi}{4}$

$= \tan \Big( \pi +$ $\frac{\pi}{4}$ $\Big) \cot \Big( 2\pi +$ $\frac{\pi}{4}$ $\Big) + \tan \Big( 4\pi +$ $\frac{\pi}{4}$ $\Big) \cot \Big( 4\pi -$ $\frac{\pi}{4}$ $\Big)$

$= \tan$ $\frac{\pi}{4}$ $\cot$ $\frac{\pi}{4}$ $+ \tan$ $\frac{\pi}{4}$ $\Big( - \cot$ $\frac{\pi}{4}$ $\Big)$

$= 1 -1 = 0 =$ RHS. Hence proved.

$\\$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

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Class 11: Trigonometric Functions – Exercise 5.3

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