Trigonometric Functions

Class 11 NCERT

NCERT

1   Find the radian measures corresponding to the following degree measures:$\\$ (i) $25^o$$\\$ (ii)$-47^o30'$$\\$ (iii)$240^o$$\\$ (iv)$520^o$

Solution :

(i)$25^o$$\\$ We know that $180^o=\pi $ radian$\\$ $\therefore 25^o=\dfrac{\pi}{180}*25$ radian=$\dfrac{5 \pi}{36}$ radian$\\$ (ii)$-47^o30'$$\\$ $-47^o30'= -47\dfrac{1}{2}\\ =\dfrac{-95}{2}$degree$\\$ Since $180^o=\pi $ radian$\\$ $\dfrac{-95}{2}$degree$=\dfrac{\pi}{180}*(\dfrac{-95}{2})$ radian$\\$ $=(\dfrac{-19}{36*2})\pi$ radian$=\dfrac{-19}{72}\pi $ radian$\\$ $\therefore -47^o30'=\dfrac{-19}{72}\pi $ radian$\\$ (iii)$240^o$$\\$ We know that $180^o=\pi$ radian$\\$ $\therefore 240^o=\dfrac{\pi}{180}*240$ radian$\\$ $=\dfrac{4}{3}\pi$ radian$\\$ (iv) $ 520^o$$\\$ We know that $180^o=\pi$ radian$\\$ $\therefore 520^o=\dfrac{\pi}{180}*520$ radian$\\$ $=\dfrac{26\pi}{9}$ radian

2   Find the degree measures corresponding to the following radian measures(Use $\pi=\dfrac{22}{7}$)$\\$ (i)$\dfrac{11}{16}$$\\$ (ii)$-4$$\\$ (iii)$\dfrac{5\pi}{3}$$\\$ (iv)$\dfrac{7\pi}{6}$

Solution :

(i)$\dfrac{11}{16}$$\\$ We know that $\pi$radian$=180^o$$\\$ $\therefore \dfrac{11}{16}$radian$=\dfrac{180}{\pi}*\dfrac{11}{16}$degree$\\$ $=\dfrac{45*11}{\pi*4}$degree$\\$ $=\dfrac{45*11*7}{22*4}$degree $\\$ $=\dfrac{315}{8}$degree$\\$ $39\dfrac{3}{8}$degree$\\$ $=39^o+\dfrac{3*60}{8}$minutes $\ \ \ \ \ \ \ \ \ \ [1^o=60']$$\\$ $=39^o+22'+\dfrac{1}{2}$minutes$\\$ $=39^o22'30'' \ \ \ \ \ \ \ \ \ \ \ \ [1'=60'']$$\\$ (ii)$-4$$\\$ We know that $\pi$ radian =$180^o$$\\$ $-4$ radian =$\dfrac{180}{\pi}*(-4)$degree$\\$ $=\dfrac{180*7(-4)}{22}$degree$\\$ $=\dfrac{-2520}{11}$degree$=-229\dfrac{1}{11}$degree$\\$ $=-229^o-\dfrac{1*60}{11}$minutes $\ \ \ \ \ \ \ \ [1^o=60'] $$\\$ $=-229^o-5'-\dfrac{5}{11}$minutes$\\$ $=-229^o5'27'' \ \ \ \ \ \ \ \ \ [1'=60'']$$\\$ (iii)$\dfrac{5\pi}{3}$$\\$ We know that $\pi $ radian=$180^o$$\\$ $\therefore \dfrac{5\pi}{3}$radian$\\$ $=\dfrac{180}{\pi}*\dfrac{5\pi}{3}$degree $\\$ $=300^o$ $\\$ (iv) $\dfrac{7\pi}{6}$$\\$ We know that $\pi$radian$=180^o$$\\$ $\therefore \dfrac{7\pi}{6}$radian$\\$ $=\dfrac{180}{\pi}*\dfrac{7\pi}{6}\\ =210^o$

3   A wheel makes $360$ revolutions in one minute. Through how many radians does it turn in one second?

Solution :

Number of revolutions made by the wheel in $1$ minute = $360$$\\$ $\therefore$ Number of revolutions made by the wheel in $1$ second =$\dfrac{360}{60}=6$$\\$ In one complete revolution, the wheel turns an angle of $2\pi $radian$\\$ Hence, in $6$ complete revolutions, it will turn an angle of $6 *2 \pi$ radian, i.e., $12 \pi $ radian$\\$ Thus, in one second, the wheel turns an angle of $12\pi $ radian

4   Find the degree measure of the angle subtended at the centre of a circle of radius $100 cm$ by an arc of length $22 cm.$ (Use$\pi =\dfrac{22}{7}$)

Solution :

We know that in a circle of radius $r$ unit, if an arc of length $l$ unit subtends an angle $\theta $ radian at the centre, then$\\$ $\theta =\dfrac{1}{r}$$\\$ Therefore, for $r =100 cm , l = 22 cm ,$ we have$\\$ $\theta =\dfrac{22}{100}$radian $=\dfrac{180}{\pi}*\dfrac{22}{100}$degree$\\$ $=\dfrac{180*7*22}{22*100}$degree$\\$ $=\dfrac{126}{10}$degree$\\$ $=12\dfrac{3}{5}$degree=$12^o36' \ \ \ \ \ \ \ [1^o=60']$$\\$ Thus, the required angle is $12^o 36'.$

5   In a circle of diameter $40 cm,$ the length of a chord is $20 cm.$ Find the length of minor arc of the chord.

Solution :

Diameter of the circle = $40 cm$ $\therefore $ Radius $(r)$ of the circle =$\dfrac{40}{2} cm =20 cm$$\\$ Let $AB $ be a chord(length=$20 cm$) of the circle.$\\$

In $\Delta OAB,OA=OB=\text{Radiusof circle}=20cm$$\\$ Also,$AB=20cm $$\\$ Thus,$\Delta OAB$ is an equilateral trigle.$\\$ $\therefore \theta =60^o=\dfrac{\pi}{3}$ radian$\\$ We know that in a circle of radius $r$ unit, if an arc of length $l$ unit subtends an angle $\theta $ radian at the centre then $\\$ $\theta =\dfrac{l}{r} $$\\$ $\dfrac{\pi}{3}=\dfrac{\widehat{AB}}{20}\implies \widehat{AB}=\dfrac{20\pi}{3} cm$$\\$ Thus, the length of the minor arc of the chord is $\dfrac{20\pi}{3}cm$

6   If in two circles, arcs of the same length subtend angles $ 60^o $ and $ 75^o $ at the centre, find the ratio of their radii.

Solution :

Let the radii of the two circles be $r_ 1$ and $ r_ 2 $ . Let an arc of length $ l $ subtend an angle of $ 60^o $ at the centre of the circle of radius $r _1$ , while let an arc of length/subtend an angle of $75^o $ at the centre of the circle of radius $ r_ 2$ .$\\$ Now, $60^o =\dfrac{\pi}{3}$ radian and $ 75^o =\dfrac{5 \pi }{12} $ radian$\\$ We know that in a circle of radius $r$ unit, if an arc of length $ l $ unit subtends an angle $\theta $. radian at the centre then$\\$ $\theta =\dfrac{1}{r} \text{or} l=r \theta\\ \therefore l=\dfrac{r_1 \pi}{3} \text{and} l=\dfrac{r_2 5 \pi}{12}\\ \Rightarrow \dfrac{r_1 \pi}{3} =\dfrac{r_2 5 \pi}{12}\\ \Rightarrow r_1=\dfrac{r_25}{4}\\ \Rightarrow \dfrac{r_1}{r_2}=\dfrac{5}{4}$$\\$ Thus, the ratio of the radii is $5:4.$

7   Find the angle in radian through which a pendulum swings if its length is $75 \ cm $ and the tip describes an arc of length.$\\$ (i) $10 \ cm$$\\$ (ii) $ 15 \ cm$$\\$ (iii) $ 21 \ cm $

Solution :

We know that in a circle of radius $ r $ unit, if an arc of length $ l $ unit subtends $\\$ An angle $\theta $ radian at the centre, then $\theta =\dfrac{l}{r}$$\\$ It is given that $r = 75 \ cm $$\\$ (i) Here, $l = 10 \ cm$$\\$ $ \theta =\dfrac{10}{75} \text{radian} =\dfrac{2}{15} \text{radian}$$\\$ (ii) Here, $ l=15 cm $$\\$ $\theta =\dfrac{15}{75}\text{radian}=\dfrac{1}{5} \text{radian}$$\\$ (iii)Here,$l=21 cm $$\\$ $\theta =\dfrac{21}{75}\text{radian}=\dfrac{7}{25}\text{radian}$$\\$

8   Find the values of other five trigonometric functions if $\cos x =-\dfrac{1}{2},x $ lies in third quadrant.

Solution :

$\cos x=-\dfrac{1}{2}\\ \therefore \sec x=\dfrac{1}{\cos x}\\ =\dfrac{1}{(-\dfrac{1}{2})} = -2 \\ \sin^2 x+ \cos^2 x=1\\ \Rightarrow \sin^2 x=1-\cos^2 x\\ \Rightarrow \sin^2 x= 1-(-\dfrac{1}{2})^2\\ \Rightarrow \sin^2 x=1-\dfrac{1}{4}=\dfrac{3}{4}\\ \Rightarrow \sin x=\pm \dfrac{\sqrt{3}}{2}$$\\$ Since $x$ lies in the $3 ^{rd}$ quadrant, the value of sin $x$ will be negative.$\\$ $\therefore \sin x=-\dfrac{\sqrt{3}}{2}\\ \csc x=\dfrac{1}{\sin x}=\dfrac{1}{-\dfrac{\sqrt{3}}{2}}=-\dfrac{2}{\sqrt{3}}\\ \tan x =\dfrac{\sin x}{\cos x}=\dfrac{(-\dfrac{\sqrt{3}}{2})}{-\dfrac{1}{2}} =\sqrt{3}\\ \cot x=\dfrac{1}{\tan x}=\dfrac{1}{\sqrt{3}}$

9   Find the values of other five trigonometric functions if $\sin x =\dfrac{3}{5}, x $ lies in second quadrant.

Solution :

$\sin x=\dfrac{3}{5}\\ \csc x=\dfrac{1}{\sin x}=\dfrac{1}{\dfrac{3}{5}} =\dfrac{5}{3}\\ \sin^2 x+ \cos^2 x =1\\ \Rightarrow \cos^2 x=1-\sin^2 x\\ \Rightarrow \cos^2 x =1-(\dfrac{3}{5})^2\\ \Rightarrow \cos^2 x=1-(\dfrac{9}{25})\\ \Rightarrow \cos^2 x=\dfrac{16}{25}\\ \Rightarrow \cos x= \pm \dfrac{4}{5}$$\\$ Since $ x $ lies in the $ 2 ^{nd} $ quadrant, the value of $\cos x $ will be negative $\\$ $\therefore \cos x=-\dfrac{4}{5}\\ \sec x=\dfrac{1}{\cos x}=\dfrac{1}{\dfrac{-4}{5}} =-\dfrac{5}{4}\\ \tan x=\dfrac{\sin x}{\cos x}=\dfrac{(\dfrac{3}{5})}{(\dfrac{-4}{5})}=-\dfrac{3}{4}\\ \cot x=\dfrac{1}{\tan x}=-\dfrac{4}{3}$

10   Find the values of other five trigonometric functions if $cot x=\dfrac{3}{4},x $ lies in third quadrant.

Solution :

$cot x=\dfrac{3}{4}\\ \tan x=\dfrac{1}{\cot x} =\dfrac{1}{(\dfrac{3}{4})} =\dfrac{4}{3}\\ 1+ \tan^2 x=\sec^2 x\\ \Rightarrow 1+(\dfrac{4}{3})^2= \sec^2 x\\ \Rightarrow 1+\dfrac{16}{9}=\sec^2 x\\ \Rightarrow \dfrac{25}{9} =\sec^2 x\\ \Rightarrow \sec x=\pm \dfrac{5}{3}$$\\$ Since $x$ lies in the $3 ^{rd}$ quadrant, the value of sec $x$ will be negative.$\\$ $\therefore \sec x=-\dfrac{5}{3}\\ \cos x=\dfrac{1}{\sec x}=\dfrac{1}{(\dfrac{-5}{3})}=-\dfrac{3}{5}\\ \tan x=\dfrac{\sin x}{\cos x}\\ \Rightarrow \dfrac{4}{3} =\dfrac{\sin x}{(-\dfrac{3}{5})}\\ \Rightarrow \sin x=(\dfrac{4}{3})*(\dfrac{-3}{5})=-\dfrac{4}{5}\\ \csc x =\dfrac{1}{\sin x}=-\dfrac{5}{4}$

11   Find the values of other five trigonometric functions if $\sec x=\dfrac{13}{5},x$ lies in fourth quadrant.

Solution :

$\sec x=\dfrac{13}{5} $ $\\$ cos $x=\dfrac{1}{\sec x}=\dfrac{1}{(\dfrac{13}{5})}=\dfrac{5}{13}$ $\\$ $\sin^2 x+$ cos$^2=1$ $\\$ $\implies$ $\sin^2 x=1-$ cos $^2x$$\\$ $\implies$ $\sin^2 x=1-(\dfrac{5}{13})^2 $ $\\$ $\implies$ $\sin^2 x=1-\dfrac{25}{169}=\dfrac{144}{169} $ $\\$ $\implies$ $\sin x=\pm \dfrac{12}{13}$ $\\$ Since x lies in the 4 th quadrant, the value of sin x will be negative. $\\$ $\therefore \sin x=-\dfrac{12}{13}$ $\\$ cosec $ x=\dfrac{1}{\sin x}=\dfrac{1}{(-\dfrac{12}{13})}=-\dfrac{13}{12}$ $\\$ $\tan x=\dfrac{\sin x}{\text{cos}\ x}=\dfrac{(-\dfrac{12}{13})}{\dfrac{5}{13}}=-\dfrac{12}{5}$ $\\$ $\cot x=\dfrac{1}{\tan x}=\dfrac{1}{(-\dfrac{12}{5})}=-\dfrac{5}{12}.$

12   Find the values of other five trigonometric functions if $\tan x=-\dfrac{5}{12},x$ lies in second quadrant.

Solution :

$\tan x =-\dfrac{5}{12}$ $\\$ $\cot x=\dfrac{1}{\tan x}=\dfrac{1}{(-\dfrac{5}{12})}=-\dfrac{12}{5}$ $\\$ $1+\tan^2 x=\sec^2 x$ $\\$ $\implies 1+(-\dfrac{5}{12})^2=\sec^2 x$ $\\$ $\implies 1+\dfrac{25}{144} =\sec^2 x$ $\\$ $\implies \dfrac{169}{144}=\sec^2 x$ $\\$ $\implies \sec x=\pm \dfrac{13}{12}$ $\\$ Since x lies in the $2 ^{nd}$ quadrant, the value of sec x will be negative. $\\$ $\therefore \sec x=-\dfrac{13}{12}$ $\\$ cos $x=\dfrac{1}{\sec x}=\dfrac{1}{(-\dfrac{13}{12})}=-\dfrac{12}{13}$ $\\$ $ \tan x=\dfrac{\sin x}{\text{cos} \ x}$ $\\$ $\implies -\dfrac{5}{12}=\dfrac{\sin x}{(-\dfrac{12}{13})}$ $\\$ $\implies \sin x=(-\dfrac{5}{12})*(-\dfrac{12}{13})=\dfrac{5}{13}$ $\\$ cosec $ x=\dfrac{1}{\sin x}=\dfrac{1}{(\dfrac{5}{13})}=\dfrac{13}{5}. $ $\\$

13   Find the value of the trigonometric function sin 765$^o$ .

Solution :

It is known that the values of $\sin x$ repeat after an interval of 2n or 360$^o$ .$\\$ $\therefore \sin 765^o= \sin ( 2* 360 ^o+ 45 ^o)=\sin 45^o=\dfrac{1}{\sqrt{2}}$

14   Find the value of the trigonometric function cosec $(- 1410^o)$

Solution :

It is known that the values of cosec x repeat after an interval of 2n or 360$^o$ . $\\$ $\therefore$ cosec$ (-1410 ^o)=$ cosec $(- 1410 ^o + 4 * 360 ^o)$ $\\$ = cosec $(- 1410 ^o+ 1440 ^o)$ $\\$ = cosec $30^o =2 .$

15   Find the value of the trigonometric function $\tan \dfrac{19\pi}{3}.$

Solution :

It is known that the values of tan x repeat after an interval of n or 180$^o$ .$\\$ $\therefore \tan \dfrac{19\pi}{3}=\tan 6 \dfrac{1}{3}\pi=\tan(6\pi+\dfrac{pi}{3})=\tan\dfrac{\pi}{3}=\tan 60^o=\sqrt{3}.$

16   Find the value of the trigonometric function $\sin (-\dfrac{11\pi}{3})$

Solution :

It is known that the values of sin x repeat after an interval of 2n or 360$^o$ .$\\$ $\therefore \sin(-\dfrac{11\pi}{3})=\sin(-\dfrac{11\pi}{3}+2*2\pi)=\sin(\dfrac{\pi}{3})=\dfrac{\sqrt{3}}{2}.$

17   Find the value of the trigonometric function $cot (-\dfrac{15\pi}{4})$

Solution :

It is known that the values of cot x repeat after an interval of n or 180$^o$ . $\\$ $\therefore $$(-\dfrac{15 p \i}{4})=cot(-\dfrac{15^\pi}{4}+4 \pi)=$cot \dfrac{\pi }{ 4}=1

18   $\sin^2 \dfrac{\pi}{6} +\cos^2 \dfrac{\pi}{3}-\tan^2 \dfrac{\pi}{4}=-\dfrac{1}{2}$

Solution :

L.H.S= $\sin^2 \dfrac{\pi}{6} +\cos^2 \dfrac{\pi}{3}-\tan^2 \dfrac{\pi}{4} \\ =(\dfrac{1}{2})^2+(\dfrac{1}{2})^2-(1)^2\\ =\dfrac{1}{4}+=\dfrac{1}{4}-1=-\dfrac{1}{2}\\ $=R.H.S

19   Prove taht $2\sin^2\dfrac{\pi}{6}+$ cosec$^2\dfrac{7 \pi}{6} \cos^2 \dfrac{\pi}{3} =\dfrac{3}{2}$

Solution :

L.H.S.=$2\sin^2\dfrac{\pi}{6}+$ cosec$^2\dfrac{7 \pi}{6} \cos^2 \dfrac{\pi}{3}\\ =2(\dfrac{1}{2})^2+$cosec$^2(\pi+\dfrac{\pi}{6})(\dfrac{1}{2})^2\\ =2*\dfrac{1}{4}+($-cosec$\dfrac{\pi}{6})^2 (\dfrac{1}{4})\\ =\dfrac{1}{2}+(-2)^2(\dfrac{1}{4})\\ =\dfrac{1}{2}+\dfrac{4}{4}=\dfrac{1}{2}+1=\dfrac{3}{2}$ = R.H.S.

20   Prove that $\cot^2 \dfrac{\pi}{6}+$cosec$\dfrac{5\pi}{6}+3\tan^2 \dfrac{\pi}{6}=6$

Solution :

L.H.S.=$\cot^2 \dfrac{\pi}{6}+$cosec$\dfrac{5\pi}{6}+3\tan^2 \dfrac{\pi}{6}$ $\\$ $=(\sqrt{3})^2+$cosec$(\pi-\dfrac{\pi}{6})+3(\dfrac{1}{\sqrt{3}})^2$ $\\$ $=3+$cosec$\dfrac{\pi}{6}+3*\dfrac{1}{\sqrt{3}}$$\\$ $=3+$cosec$ \dfrac{\pi}{6}+3*\dfrac{1}{3}$ $\\$ $=3+2+1=6$ $\\$ =R.H.S

21   Prove that $2\sin^2 \dfrac{3\pi}{4}+2 \cos^2\dfrac{\pi}{4}+2\sec^2\dfrac{\pi}{3}=10$

Solution :

L.H.S.=$2\sin^2 \dfrac{3\pi}{4}+2 \cos^2\dfrac{\pi}{4}+2\sec^2\dfrac{\pi}{3}\\ =2{\sin(\pi-\dfrac{\pi}{4})}^2+2(\dfrac{1}{\sqrt{2}})^2+2(2)^2\\ =2{\sin \dfrac{\pi}{4}}^2+2*\dfrac{1}{2}+8\\ =1+1+8\\ =10$ =R.H.S

22   Find the value of :$\\$ (i)$\sin 75^o$ $\\$ (ii)$\tan 15^o$

Solution :

(i)$\sin 75^o=\sin(45^o+30^o)\\ =\sin 45^o \cos 30^o+\cos 45^o \sin 30^o\\ [\sin(x+y)=\sin x \cos y+\cos x \sin y]\\ =(\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{3}}{2})+(\dfrac{1}{\sqrt{2}})(\dfrac{1}{2})\\ =\dfrac{\sqrt{3}}{2 \sqrt{2}}+\dfrac{1}{2 \sqrt{2}}=\dfrac{\sqrt{3}+1}{2 \sqrt{2}}$ $\\$ (iii)$\tan 15^o=\tan(45^o-30^o)\\ =\dfrac{\tan 45^o-\tan 30^o}{1+\tan 45^o \tan 30^o} \qquad [\tan(x-y)=\dfrac{\tan x-\tan y}{1+\tan x \tan y}]\\ =\dfrac{1-\dfrac{1}{\sqrt{3}}}{1+1(\dfrac{1}{\sqrt{3}})}\\ =\dfrac{\dfrac{\sqrt{3}-1}{\sqrt{3}}}{\dfrac{\sqrt{3}+1}{\sqrt{3}}}\\ =\dfrac{\sqrt{3}-1}{\sqrt{3}+1}=\dfrac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)}\\ =\dfrac{3+1-2\sqrt{3}}{(\sqrt{3})^2-(1)^2}\\ =\dfrac{4-2\sqrt{3}}{3-1}=2-\sqrt{3}$

23   Prove that $\cos(\dfrac{\pi}{4}-x) \cos(\dfrac{\pi}{4}-y)-\sin(\dfrac{\pi}{4}-x)\sin(\dfrac{\pi}{4}-y)=\sin(x+y)$

Solution :

$\cos(\dfrac{\pi}{4}-x) \cos(\dfrac{\pi}{4}-y)-\sin(\dfrac{\pi}{4}-x)\sin(\dfrac{\pi}{4}-y)\\ =\dfrac{1}{2}[ 2\cos(\dfrac{\pi}{4}-x)\cos(\dfrac{\pi}{4}-y)]+\dfrac{1}{2}[-2 \sin(\dfrac{\pi}{4}-x) \sin(\dfrac{\pi}{4}-y)]\\ =\dfrac{1}{2}[\cos{(\dfrac{\pi}{4}-x)+(\dfrac{\pi}{4}-y)}+\cos{(\dfrac{\pi}{4}-x)-(\dfrac{\pi}{4}-y)}]\\ +\dfrac{1}{2}[\cos{(\dfrac{\pi}{4}-x)+(\dfrac{\pi}{4}-y)}-\cos{\dfrac{\pi}{4}-x}-(\dfrac{\pi}{4}-y)]\quad [\because 2 \cos a \cos B=\cos(A+B)+ \cos(A-B) -2 \sin A \sin B=\cos(A+B)=\cos(A+B)-\cos(A-B) ]\\ =2*\dfrac{1}{2}[\cos{(\dfrac{\pi}{4}-x)+(\dfrac{\pi}{4}-y)}]\\ =\cos[\dfrac{\pi}{4}-(x+y)]\\ =\sin(x+y)$ =R.H.S.

24   Prove that $\dfrac{\tan(\dfrac{\pi}{4}+x)}{\tan(\dfrac{\pi}{4}-x)}=(\dfrac{1+\tan x}{1-\tan x})^2$

Solution :

It is known that $ \tan(A+B)=\dfrac{\tan a+\tan B}{1-\tan a \tan B}$ and $(A-B)=\dfrac{\tan a -\tan b}{1+ \tan A \tan B}\\ L.H.S.=\dfrac{\tan(\dfrac{\pi}{4}+x)}{\tan(\dfrac{\pi}{4}-x)}=\\ \quad \\ \dfrac{(\dfrac{\tan \frac{\pi}{4}+\tan x}{1-\tan\frac{\pi}{4}\tan x})}{\dfrac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4}\tan x}}=\\ \quad \\ \dfrac{(\dfrac{1+\tan x}{1-\tan x})}{(\dfrac{1-\tan x}{1+ \tan x})}=\\ \quad \\ (\dfrac{1+\tan x}{1-\tan x})^2=R.H.S.$

25   Prove that $\dfrac{\cos (\pi+x)\cos(-x)}{\sin(\pi-x)\cos(\frac{\pi}{2}+x)} = \cot^2 x$

Solution :

L.H.S.=$\dfrac{\cos (\pi+x)\cos(-x)}{\sin(\pi-x)\cos(\frac{\pi}{2}+x)}\\ \qquad \\ =\dfrac{[-\cos x][\cos x]}{(\sin x)(-\sin x)}\\ \qquad \\ =\dfrac{-\cos^2 x}{-\sin^2 x} \\ =\cot62 x$ = R.H.S.

26   $\cos(\dfrac{3\pi}{2}+x)\cos(2\pi+x)[\cot(\dfrac{3\pi}{2}-x)+\cot (2\pi+x)]=1$

Solution :

L.H.S.=$\cos(\dfrac{3\pi}{2}+x)\cos(2\pi+x)[\cot(\dfrac{3\pi}{2}-x)+\cot (2\pi+x)]\\ =\sin x\cos x[\tan x+ \cot x]\\ =\sin x \cos x(\dfrac{\sin x}{\cos x}+\dfrac{cos x}{\sin x})\\ =(\sin x\cos x)[\dfrac{\sin^2 x+\cos^2 x}{\sin x \cos x}]\\ =1$= R.H.S.

27   Prove that $\sin(n+1)x \sin(n+2)x+\cos(n+1)x \cos(n+2)x=\cos x$

Solution :

L.H.S.=$\sin(n+1)x \sin(n+2)x+\cos(n+1)x \cos(n+2)x\\ =\dfrac{1}{2}[2\sin(n+1)x \sin(n+2)x+ \cos(n+1)x \cos(n+2)x]\\ =\dfrac{1}{2}[\cos{(n+1)x-(n+2)x}-\cos{(n+1)x+(n+2)x}+\\ \cos{(n+1)x+(n+2)x}+\cos{(n+1)x-(n+2)x}]\\ \qquad \quad \\ \qquad [\because -2 \sin a \sin B=\cos(A+B)-\cos(A-B)\\ 2\cos a \cos B=\cos(A+B)+\cos(A-b)]\\ =\dfrac{1}{2}* 2\cos{(n+1)x-(n+2)x}\\ =\cos(-x)=\cos x$= R.H.S.

28   Prove that $ \cos(\dfrac{3\pi}{4}+x)-\cos(\dfrac{3\pi}{4}-x)=-\sqrt{2}\sin x$

Solution :

It is know that $ \cos A-\cos B=-2 \sin(\dfrac{A+B}{2}).\sin(\dfrac{A-B}{2})\\ \quad\\ \therefore L.H.S.=\cos(\dfrac{3\pi}{4}+x)-\cos(\dfrac{3 \pi}{4}-x)\\ \quad\\ =-2 \sin {\dfrac{(\dfrac{3 \pi}{4}+x)(\dfrac{3 \pi}{4}-x)}{2}}.\sin{\dfrac{(\dfrac{3 \pi}{4}+x)-(\dfrac{3 \pi}{4}-x)}{2}}\\ \quad\\ =-2 \sin(\dfrac{3 \pi}{4}) \sin x \\ \quad\\ =-2 \sin(\pi-\dfrac{\pi}{4})\sin x\\ \quad\\ =-2 \sin \dfrac{\pi}{4}\sin x\\ \quad\\ =-2 * \dfrac{1}{\sqrt{2}}* \sin x\\ \quad\\ =-\sqrt{2} \sin x$ =R.H.s.

29   Prove that $\sin ^2 6 x - \sin^ 2 4 x = \sin ^2 x \sin10 x$

Solution :

It is know that $\\$ $\sin A + \sin B=2 \sin(\dfrac{A+B}{2}) \cos(\dfrac{A-B}{2}),\sin A- \sin B=2 \cos(\dfrac{A+B}{2}) \sin(\dfrac{A-B}{2})\\ \quad \\ \because L.H.S. =\sin^2 6x-\sin^2 4x\\ \quad \\ =(\sin 6x+\sin 4x)(\sin 6x-\sin 4x)\\ \quad \\ =[2 \sin(\dfrac{6x+4x}{2})\cos(\dfrac{6x-4x}{2})][2 \sin(\dfrac{6x+4x}{2}).\sin(\dfrac{6x-4x}{2})]\\ \quad \\ =(2\sin 5x \cos x)(2 \cos 5x \sin x)=(2 \sin 5x \cos 5x)(2 \sin x \cos x)\\ \quad \\ =\sin 10 x \sin 2x$ =R.H.S.

30   Prove that $\cos^ 2 2 x - \cos ^2 6 x= \sin 4 x \sin8 x$

Solution :

It is known that $\\$ $\cos A + \cos B= 2 \cos(\dfrac{A+B}{2})\cos (\dfrac{A-B}{2}),\cos A-\cos B =-2\sin(\dfrac{A+B}{2})\sin (\dfrac{A-B}{2})\\ \because L.H.S.=\cos^2 2x-\cos^2 6x\\ =(\cos 2x+\cos 6x)(\cos 2x-6x)\\ =[2 \cos (\dfrac{2x+6x}{2})\cos(\dfrac{2x-6x}{2})][2 \sin (\dfrac{2x+6x}{2})\sin(\dfrac{2x-6x}{2})]\\ =[2 \cos 4x \cos 4x][-2 \sin 4x(-\sin 2x)]\\ =(2 \sin 4x \cos 4x)(2 \sin 2x \cos 2x)\\ =\sin 8x \sin 4x =$ R.H.s.

31   Prove that $\sin 2 x + 2\sin 4 x + \sin 6 x = 4\cos ^2 x \sin 4 x$

Solution :

L.H.S.=$\sin 2x + 2 \sin 4x +\sin 6x\\ =[\sin 2x + \sin 6x]+2 \sin 4x\\ =[2\sin(\dfrac{2x+6x}{2})\cos(\dfrac{2x-6x}{2})]+2\sin4 x\\ [\because \sin A+\sin B=2 \sin(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2})]\\ =2 \sin 4x \cos(-2x)+2 \sin 4x\\ =2 \sin 4x \cos 2x+2 \sin 4x\\ =2 \sin 4x(\cos 2x +1)\\ =2 \sin 4x(2 \cos^2 x-1+1)\\ =2 \sin 4x(2 \cos^2 x)\\ =4 \cos^2 x \sin 4x $ =R.H.S.

32   Prove that $\cot 4x(\sin 5x+ \sin 3x)=\cot x(\sin 5x-\sin 3x)$

Solution :

L.H.S.= $ \cot 4x(\sin 5x+\sin 3x)\\ =\dfrac{\cot 4x}{\sin 4x}[2 \sin(\dfrac{5x+3x}{2})\cos(\dfrac{5x-3x}{2})]\\ [\therefore \sin A + \sin B=2 \sin(\dfrac{A+B}{2})\cos (\dfrac{A-B}{2})]\\ =(\dfrac{\cos 4x}{\sin 4x})[2 \sin 4x \cos x]\\ =2 \cos 4x \cos x\\ R.H.S.=\cot x(\sin 5x -\sin 3x)\\ =\dfrac{\cos x}{\sin x}[2 \cos(\dfrac{5x+3x}{2})\sin(\dfrac{5x-3x}{2})]\\ [\because \sin A- \sin B=2 \cos(\dfrac{A+B}{2})\sin(\dfrac{A-B}{2})]\\ =\dfrac{\cos x}{\sin x}[2 \cos 4x \sin x]\\ =2 \cos 4x. \cos x$ L.H.S=R.H.S.

33   Prove that $\dfrac{\cos 9x-\cos 5x}{\sin 17x -\sin 3x}=-\dfrac{\sin 2 x}{\cos 10x}$

Solution :

It is known that $\\$ $\cos A-\cos B =-2 \sin(\dfrac{A+B}{2})\sin(\dfrac{A-B}{2}),\\ \sin A-\sin b=2\cos(\dfrac{A+B}{2})\sin(\dfrac{A-B}{2})\\ \therefore L.H.S.=\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}\\ =\dfrac{-2 \sin(\dfrac{9x+5x}{2}).\sin(\dfrac{9x-5x}{2})}{2\cos(\dfrac{17x+3x}{2}).\sin(\dfrac{17x-3x}{2})}\\ =\dfrac{-2 \sin 7x.\sin 2x}{2 \cos 10 x. \sin 7x}\\ =-\dfrac{\sin 2x}{\cos 10x}$ R.H.S.

34   Prove that: $\dfrac{\sin 5x+\sin 3x}{\cos 5x+\cos 3x}=\tan 4x$

Solution :

It is known that $\\$ $\sin A+\sin B=2 \sin(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2}),\\ \cos A + \cos B=2 \cos(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2})\\ \therefore L.H.S.=\dfrac{\sin 5x +\sin 3x}{\cos 5x+\cos 3x}\\ =\dfrac{2 \sin(\dfrac{5x+3x}{2}).\cos(\dfrac{5x-3x}{2})}{2 \cos(\dfrac{5x+3x}{2}).\cos(\dfrac{5x-3x}{2})}\\ =\dfrac{2 \sin 4x,\cos x}{2 \cos 4x.\cos x}\\ =\tan 4x $ = R.H.S.

35   Prove that $\dfrac{\sin x-\sin y}{\cos x+\cos y}=\tan\dfrac{x-y}{2}$

Solution :

It is known that $\\$ $\sin A-\sin B=2 \cos(\dfrac{A+B}{2})\sin(\dfrac{A-B}{2}),\\ \cos A+ \cos B=2 \cos(\dfrac{A+B}{2}) \cos(\dfrac{A-B}{2})\\ \therefore L.H.S.=\dfrac{\sin x-\sin y}{\cos x+\cos y}\\ =\dfrac{2 \cos(\dfrac{x+y}{2}).\sin(\dfrac{x-y}{2})}{2 \cos(\dfrac{x+y}{2}).\cos(\dfrac{x-y}{2})}\\ =\dfrac{\sin(\dfrac{x-y}{2})}{\cos(\dfrac{x-y}{2})}\\ =\tan(\dfrac{x-y}{2})$ =R.H.S.

36   Prove that $\dfrac{\sin x+ \sin3x}{\cos x+\cos 3x}=\tan 2x$

Solution :

It is known that $\\$ $\sin A +\sin B=2 \sin(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2})\\ \cos A+\cos B=2 \cos(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2})\\ \therefore L.H.S.=\dfrac{\sin x+\sin 3x}{\cos x+\cos 3x}\\ =\dfrac{2 \sin(\dfrac{x+3x}{2})\cos(\dfrac{x-3x}{2})}{2 \cos(\dfrac{x+3x}{2})\cos(\dfrac{x-3x}{2})}\\ =\dfrac{\sin 2x}{\cos 2x}\\ =\tan 2x$ =R.H.S.

37   Prove that $ \dfrac{\sin x-\sin 3x}{\sin^2 x-\cos^2 x}=2 \sin x$

Solution :

It is know that $\\$ $ \sin A-\sin b=2 \cos(\dfrac{A+B}{2})\sin(\dfrac{A-B}{2}),\\ \cos^2 A-\sin^2A=\cos 2A\\ \therefore L.H.S.=\dfrac{\sin x-\sin 3x}{\sin^2 x-\cos^2 x}\\ =\dfrac{2 \cos(\dfrac{x+3x}{2})\sin(\dfrac{x-3x}{2})}{-\cos 2x}\\ =\dfrac{2\cos2x \sin(-x)}{-\cos 2x}\\ =-2*(-\sin x)\\ = 2\sin x$ =R.H.S.

38   Prove that $ \dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin3x+\sin 2x}=\cot 3x$

Solution :

L.H.S.=$\dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin3x+\sin 2x}\\ =\dfrac{(\cos 4x+\cos 2x)+\cos 3x}{(\sin 4x +\sin 2x)+\sin 3x}\\ =\dfrac{2 \cos(\dfrac{4x+2x}{2})\cos(\dfrac{4x-2x}{2})+\cos 3x} {2 \sin(\dfrac{4x+2x}{2})\cos(\dfrac{4x-2x}{2})+\sin 3x}\\ [\therefore \cos a +\cos B=2 \cos(\dfrac{A+B}{2}) \cos(\dfrac{A-B}{2}),\sin A + \sin B=2 \sin(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2})]\\ =\dfrac{2 \cos 3x \cos+ \cos 3x}{2\sin3x \cos x+\sin 3x}\\ =\dfrac{\cos 3x(2 \cos x+1)}{2 \sin 3x(2 \cos x + 1)}\\ \cot 3x $ =R.H.S.

39   Prove that $\cot x \cot 2x-\cot 2x \cot 3x -\cot 3x \cot x=1$

Solution :

L.H.S.=$\cot x\cot 2x-\cot 2x \cot 3x-\cot x\\ =\cot x \cot 2x-\cot 3x(\cot 2x+\cot x)\\ =\cot x \cot 2x-\cot(2x+x)(\cot 2x+ \cot x)\\ =\cot x \cot 2x-[\dfrac{\cot 2x \cot x-1}{\cot x+\cot 2x}](\cot 2x +\cot x)\\ [\because cot(A+B)=\dfrac{\cot A \cot B-1}{\cot A + \cot B}]\\ =\cot x \cot 2x-(\cot 2x \cot x-1)=1 $ = R.H.S.

40   Prove that $ \tan 4x=\dfrac{4 \tan x(1-\tan^2 x)}{1-6 \tan^2 x+\tan^4 x}$

Solution :

It is known that $ \tan 2A=\dfrac{2 \tan A}{1-\tan^2 A}\\ \quad \therefore L.H.S.=\tan 4x=\tan 2(2x)\\ \quad =\dfrac{2 \tan 2x}{1-\tan^2(2x)}\\ \quad =\dfrac{2(\dfrac{2 \tan x}{1-\tan^2 x})}{1-(2 \tan x)}{1-(\dfrac{2 \tan x}{1-\tan^2 x})^2}\\ \quad =\dfrac{(\dfrac{4 \tan x}{1- \tan^2 x})}{[1-\dfrac{4 \tan^2 x}{(1-\tan^2 x)^2}]}\\ \quad =\dfrac{(\dfrac{4 \tan x}{1-\tan^2 x})}{[\dfrac{(1-\tan^2 x)^2-4 \tan^2 x}{(1-\tan^2 x)^2}]} \\ \quad =\dfrac{4 \tan x(1-\tan62 x)}{(1-\tan^2 x)^2-4 \tan^2 x}\\ \quad =\dfrac{4 \tan x(1-\tan^2 x)}{1-6 \tan^2 x +\tan^4 x}$ = R.H.S.

41   Prove that: $ \cos 4x= 1- 8 \sin^2 x \cos^2 x$

Solution :

L.H.S.=$ \cos 4x\\ =\cos 2(2x) \\ =1- 2 \sin^2 2x[\cos 2A = 1- 2 \sin^2 A]\\ =1-2(2 \sin x \cos x)^2[\sin 2A= 2\sin A \cos A]\\ 1-8 sin^2 x \cos^2 x$ =R.H.S.

42   Prove that: $ \cos 6x = 32 x \cos^6 x - 48 \cos^4 x+ 18 \cos^2 x-1$

Solution :

L.H.S.=$\cos 6x\\ =\cos 3(2x)\\ =4 \cos^3 2x-3 \cos 2x[\cos 3A = 4 \cos^3 A - 3 \cos A]\\ =4[(2 \cos^2 x-1)^3-3( 2 \cos^2 x-1)][\cos 2x= 2\cos^2 x-1]\\ =4[8 \cos^6 x-1-12 \cos^4 x + 6 \cos^2 x]-6 \cos^2 x+3\\ =32 \cos^6 x-4 -48 \cos^4 x+ 24 \cos^2 x-6 \cos^2 x+3\\ =32 \cos^6 x-48 \cos^4 x+18 \cos^2 x-1$ =R.H.S.

43   Find the principal and general solutions of the question tan $x =\sqrt{ 3}$ .

Solution :

$\tan x=\sqrt{3}$ $\\$ It is known that $\tan\dfrac{\pi}{3}=\sqrt{3}$ and $ \tan(\pi+\dfrac{\pi}{3})=\tan(\dfrac{\pi}{3})=\sqrt{3}$ $ \\$ Therefore, the principal solutions are $ x=\dfrac{\pi}{3}$ and $ \dfrac{4\pi}{3}.$ $\\$ Now, $ \tan x=\tan\dfrac{\pi}{3}$ $\\$ $\implies x= n\pi + \dfrac{\pi}{3},$ where $ n\in Z$ $\\$ Therefore, the general solution is $ x=n \pi+ \dfrac{\pi}{3},$ where $ n \in Z.$

44   Find the principal and general solutions of the equation $\sec x = 2$

Solution :

$\sec x=2 $ $\\$ It is known that $ \sec \dfrac{\pi}{3}=2 $ and $\sec\dfrac{5 \pi}{3}=\sec(2 \pi-\dfrac{\pi}{3})=\sec \dfrac{\pi}{3}=2$ $\\$ Therefore, the principal solutions are $ x=\dfrac{\pi}{3}$ and $ \dfrac{5 \pi}{3}.$ $\\$ Now,$ \sec x =\sec \dfrac{\pi}{3}\\ \implies \cos x=\cos \dfrac{\pi}{3} \qquad \quad [\sec x=\dfrac{1}{\cos x}]\\ \implies 2 n \pi \pm \dfrac{\pi}{3},$ where $ n\in Z.$$\\$ Therefore, the general solution is $ x=2 n \pi \pm \dfrac{\pi}{3},$ where $ n\in Z.$

45   Find the principal and general solutions of the equation $ \cot x=-\sqrt{3}$

Solution :

$\cot x=-\sqrt{3}$ $\\$ It is known that $ \cot\dfrac{\pi}{6}=\sqrt{3}\\ \therefore \cot(\pi-\dfrac{\pi}{6})=-\cot \dfrac{\pi}{6} =-\sqrt{3} $ and $ \cot(2\pi-\dfrac{\pi}{6})=-\cot \dfrac{\pi}{6}=-\sqrt{3}\\ i.e., \cot \dfrac{5\pi}{6}=-\sqrt{3} $ and $ \cot\dfrac{11 \pi}{6}=-\sqrt{3}$ $\\$ Therefore, the principal solutions are $ x=\dfrac{5\pi}{6}$ and $ \dfrac{11 \pi}{6}.$ $\\$ Now,$ \cot x=\cot \dfrac{5\pi}{6}\\ \implies \tan x=\tan \dfrac{5 \pi}{6} \quad[\cot x=\dfrac{1}{\tan x}]\\ \implies x= n\pi+\dfrac{5 \pi}{6},$ where $ n\in Z$$\\$ Therefore, the general solution is $ x=n\pi+\dfrac{5 \pi}{6},$ where $ n\in Z.$

46   Find the general solution of cosec $x=-2$

Solution :

cosec$ x=-2$ $\\$ It is known that $\\$ cosec $ \dfrac{\pi}{6}=2$ $\\$ $\therefore $ cosec $ \dfrac{7 \pi}{6}=-2$ and cosec $\dfrac{cosec $ (2 \pi-\dfrac{\pi}{6})}=-$ cosec $ \dfrac{\pi}{6}=-2$ $\\$ cosec$ \dfrac{7 \pi}{6}=-2$ and cosec $ \dfrac{11 \pi}{6}=-2 $ $\\$ Therefore, the principal solutions are $ x=\dfrac{7 \pi}{6} $ and $\dfrac{11 \pi}{6}.$ $\\$ Now, cosec $ c=$ cosec $ \dfrac{7 \pi}{6}$ $\\$ $\implies \sin x=\sin \dfrac{7 \pi}{6 } \quad [$cosec $x=\dfrac{1}{\sin x}]$ $\\$ $\implies x= n \pi+(-1)^n \dfrac{7 \pi}{6},$ where $ n\in Z $ $\\$ Therefore, the general solution is $ x= n\pi+(-1)^n \dfrac{7 \pi}{6}, $ where $ n\in Z.$

47   Find the general solution of the equation $ \cos 4x=\cos 2x$

Solution :

$\cos 4x =\cos 2x\\ \implies \cos 4x-\cos 2x=0\\ \implies -2 \sin(\dfrac{4x+2x}{2})\sin(\dfrac{4x-2x}{2})=0\\ [\because \cos A -\cos b=-2 \sin(\dfrac{A+B}{2}) \sin(\dfrac{A-B}{2})]\\ \implies \sin 3x \sin x=0\\ \implies \sin 3x=0 or \sin x =0\\ \therefore 3x=n \pi \quad or \quad \sin x =0\\ \therefore 3x=n \pi \quad or \quad x= n \pi, $ where $ n\in Z $ $\\$ $\implies x=\dfrac{n \pi}{3} \quad or \quad x= n \pi,$ where $n\in Z $

48   Find the general solution of the equation $\cos 3x +\cos x- \cos 2x =0.$

Solution :

$ \cos 3x+ \cos x-\cos 2 x=0\\ \implies 2 \cos(\dfrac{3x+2}{2}) \cos(\dfrac{3x-x}{2})-\cos 2x=0 \quad [\cos A +\cos B=2 \cos(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2})]\\ \implies 2 \cos 2 x \cos x-\cos 2x=0\\ \implies \cos 2x(2 \cos x-1)=0\\ \implies \cos 2x=0 \quad or \quad 2 \cos x-1=0\\ \implies \cos 2x=0 \quad or \cos x=\dfrac{1}{2}\\ \therefore 2x=(2n+1)\dfrac{\pi}{2} \quad or \cos x=\cos\dfrac{\pi}{3},$ where $ n\in Z$ $\\$ $\implies x=(2n+1)\dfrac{\pi}{4}\quad or \quad x=2n\pi\pm \dfrac{\pi}{3},$ where $ n\in Z $ $\\$

49   Find the general solution of the equation $\sin 2 x+\cos x=0.$

Solution :

$\sin 2x+\cos x=0\\ \implies 2 \sin x \cos x+ \cos x=0\\ \implies\cos x(2 \sin x+1)=0\\ \implies \cos x=0 \quad or \quad 2 \sin x+1=0$ $\\$ Now,$ \cos x=0 \implies \cos x=(2n+1)\dfrac{\pi}{2},$ where $ n\in Z$ $\\$ $ 2 \sin x+1 =0\\ \\implies \sin x=\dfrac{-1}{2}=-\sin(\pi+ \dfrac{\pi}{6})=\sin\dfrac{7 \pi}{6}\\ \implies x= n \pi+(-1)^n \dfrac{7 \pi}{6},$ where $ n\in Z $ $\\$ Therefore, the general solution is $(2n+1)\dfrac{\pi}{2} $ or $ n \pi+(-1)^n \dfrac{7\pi}{6},n \in Z.$

50   Find the general solution of the equation $ \sec^2 2 x=1-\tan 2x$

Solution :

$\sec^2 2x=1-\tan 2x\\ \implies 1+ \tan^2 2x=1-\tan 2x\\ \implies \tan^2 2x+\tan 2x=0\\ \implies \tan 2x(\tan 2x +1)=0\\ \implies \tan 2x =0 \quad or \quad \tan 2x +1=0$ Now,$ \tan2x =0\\ \implies \tan 2x=\tan 0\\ \implies 2x=n\pi+0,$ where $n\in Z$$\\$ $\implies x=\dfrac{n \pi}{2}, $ where $ n \in Z$$\\$ $\implies x=\dfrac{n \pi}{2},$ where $ n\in Z$ $\\$ $ \tan 2x+1=0\\ \implies \tan 2x=-1=-\tan \dfrac{\pi}{4}=\tan (\pi-\dfrac{\pi}{4})=\tan\dfrac{3 \pi}{4}\\ \implies 2x=n\pi+\dfrac{3 \pi}{4}$, where $ n\in Z $ $\\$ $\implies x=\dfrac{n\pi}{2}+\dfrac{3 \pi}{8},$ where $ n\in Z $ $\\$ Therefore, the general solution is $ \dfrac{n\pi}{2} $ or $\dfrac{n\pi}{2}+\dfrac{3 \pi}{8}, n\in Z$

51   Find the general solution of the equation $\sin x + \sin 3 x + \sin 5 x = 0$

Solution :

$\sin x +\sin 3x+\sin 5x=0\\ (\sin x+ \sin 5x)+\sin 3x =0\\ \implies [2\sin(\dfrac{x+5x}{2})\cos(\dfrac{x-5x}{2})]+\sin 3x =0 \quad [\sin A + \sin B=2 \sin(\dfrac{A+B}{2}) \cos (\dfrac{A-B}{2})]\\ \implies 2 \sin 3x \cos(-2 x)+\sin 3x =0\\ \implies2 \sin 3x \cos 2x+\sin 3x=0\\ \implies \sin 3x(2 \cos 2x+1)=0\\ \implies \sin 3x=0 \quad or \quad 2 \cos 2x +1=0$ $\\$ Now,$ \sin 3x=0\implies 3x =n\pi,$ where$ n\in Z$$\\$ $x=\dfrac{n\pi}{3},$ where$ n\in Z $$\\$ $ 2\cos2x+1=0\\ \implies \cos 2x =\dfrac{-1}{2}=-\cos\dfrac{\pi}{3}=\cos(\pi-\dfrac{\pi}{3})\\ \implies \cos 2x=\cos \dfrac{2\pi}{3}\\ \implies 2x=2n\pi \pm \dfrac{2 \pi}{3},$ where $ n\in Z$ $\\$ $\implies x=n\pi \pm \dfrac{\pi}{3},$ where $ n\in Z $ $\\$ Therefore, the general solution is $ \dfrac{n \pi}{3}$ or $ n \pi \pm \dfrac{\pi}{3},n\in Z.$

52   Prove that: $( \sin3 x + \sin x ) \sin x+( \cos3 x - \cos x ) \cos x = 0$

Solution :

L.H.S. =$( \sin3 x+ \sin x ) \sin x +( \cos3 x - \cos x ) \cos x\\ = \sin3 x \sin x + \sin ^2 x+ \cos3 x \cos x - \cos ^2 x\\ = \cos3 x \cos x + \sin 3 x \sin x -( \cos ^2 - \sin ^2 x )\\ = \cos ( 3 x - x )- \cos 2 x \quad [ \cos ( A - B )= \cos A \cos B + \sin A \sin B ]\\ = \cos 2 x - \cos 2 x\\ = 0$ = R.H.S.

53   Prove that: $( \cos x + \cos y )^2+( \sin x - \sin y )^2= 4\cos^2 \dfrac{x+y}{2}$

Solution :

L.H.S. $=( \cos x + \cos y )^2+( \sin x - \sin y )^2\\ = \cos ^2 x + \cos ^2 y + 2\cos x \cos y + \sin ^2 x + \sin ^2 y - 2\sin x \sin y\\ =( \cos ^2 x + \sin ^2 x )+( \cos^ 2 y + \sin ^2 y )+2( 2 \cos x \cos y - \sin x \sin y )\\ = 1 +1 +2\cos( x + y )\quad [ \cos( A + B )=( \cos A \cos B - \sin A \sin B )]\\ = 2 + 2\cos ( x + y )\\ = 2 [ 1 + cos ( x + y )]\\ =2[1+2\cos^2(\dfrac{x+y}{2})-1] \quad [\cos 2A=2\cos^2 A-1]\\ =4\cos^2(\dfrac{x+y}{2})$=R.H.S.

54   Prove that: $( \cos x - \cos y )^2+( \sin x - \sin y )^2= 4\cos^2 \dfrac{x-y}{2}$

Solution :

L.H.S. $=( \cos x - \cos y )^2+( \sin x - \sin y )^2\\ = \cos ^2 x + \cos ^2 y - 2\cos x \cos y + \sin ^2 x + \sin ^2 y - 2\sin x \sin y\\ =( \cos ^2 x + \sin ^2 x )+( \cos^ 2 y + \sin ^2 y )-2( 2 \cos x \cos y + \sin x \sin y )\\ = 1 +1 -2\cos( x - y )\quad [ \cos( A - B )=( \cos A \cos B + \sin A \sin B )]\\ = 2 [ 1 - cos ( x - y )]\\ =2[1-\{1-2\sin^2(\dfrac{x-y}{2})-1] \quad [\cos 2A=1-2\sin^2 A]\\ =4\sin^2(\dfrac{x-y}{2})$=R.H.S.

55   Prove that: $ \ \ \sin x + \sin3 x + \sin5 x + \sin7 x = 4\cos x \cos2 x \sin 4 x$

Solution :

It is known that $\ \ \sin A + \sin B = 2\sin(\dfrac{A+B}{2}).\cos(\dfrac{A-B}{2})$ $\\$ L.H.S.$ =\sin x+\sin3 x + \sin5 x + \sin7 x\\ =( \sin x + \sin5 x )+( \sin3 x + \sin 7 x )\\ =2 \sin(\dfrac{x+5x}{2}).\cos(\dfrac{x-5x}{2})+2\sin(\dfrac{3x+7x}{2})\cos(\dfrac{3x-7x}{2})\\ = 2\sin3 x \cos (- 2 x )+ 2\sin5 x \cos (- 2 x )\\ = 2\sin3 x \cos2 x + 2\sin5 x \cos2 x\\ = 2\cos 2 x [ \sin3 x + \sin5 x ]\\ =2\cos 2x[2 \sin(\dfrac{3x+5x}{2}).\cos(\dfrac{3x-5x})]\\ =2\cos 2x[2\sin4x.\cos(-x)]\\ =4\cos 2x \sin 4x \cos x$=R.H.S.

56   Prove that: $\dfrac{(\sin 7x+\sin 5x)+(\sin 9x+\sin3x)}{(\cos 7x+\cos5x)+(\cos 9x+\cos 3x)}=\tan 6x$

Solution :

It is known that $\ \ \sin A+\sin B=2\sin(\dfrac{A+B}{2}).\cos(\dfrac{A-B}{2}),\cos A+\cos B=2 \cos(\dfrac{A+B}{2}).\cos(\dfrac{A-B}{2})\\ L.H.S.=\dfrac{(\sin7 x + \sin 5 x )+( \sin 9 x +\sin3 x )}{( \cos 7 x + \cos5 x )+( \cos9 x + \cos3 x )}\\ =\dfrac{[2\sin(\dfrac{7x+5x}{2}).\cos(\dfrac{7x-5x}{2})]+[2\sin(\dfrac{9x+3x}{2}).\cos(\dfrac{9x-3x}{2})]}{[2\cos(\dfrac{7x+5x}{2}).\cos(\dfrac{7x-5x}{2})]+[2\cos(\dfrac{9x+3x}{2}).\cos(\dfrac{9x-3x}{2})]}\\ =\dfrac{[2\sin6x.\cos x]+[2 \sin 6x.\cos 3x]}{[2\cos 6x.\cos x]+[2\cos x+\cos 3x]}\\ =\tan 6x\\ =R.H.S.$

57   Prove that: $ \ \ \sin 3 x + \sin 2 x - \sin x = 4\sin x \cos\dfrac{x}{2}\cos\dfrac{3x}{2}$

Solution :

L.H.S. =$ \ \ \sin 3 x + \sin 2 x - \sin x\\ =\sin 3x+[2\cos(\dfrac{2x+x}{2})\sin(\dfrac{2x-x}{2})]\quad [\sin A- \sin B=2 \cos(\dfrac{A+B}{2})\sin(\dfrac{A-B}{2})]\\ =\sin 3x+[2 \cos(\dfrac{3x}{2})\sin(\dfrac{x}{2})]\\ =\sin3x+2\cos\dfrac{3x}{2}\sin\dfrac{3}{2}\\ =2\sin \dfrac{3x}{2}.\cos\dfrac{3x}{2}+2 \cos \dfrac{3x}{2}\sin\dfrac{x}{2}\quad [\sin2a=2\sin A.\cos B]\\ =2 \cos(\dfrac{3x}{2})[\sin(\dfrac{3x}{2})+\sin(\dfrac{x}{2})]\\ =2\cos(\dfrac{3x}{2})[2\sin{\dfrac{(\dfrac{3x}{2})+(\dfrac{x}{2})}{2}}\cos{\dfrac{(\dfrac{3x}{2})-(\dfrac{x}{2})}{2}}]\\ [\sin A+\sin B=2\sin(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2})]\\ =2\cos(\dfrac{3x}{2}).2 \sin x \cos (\dfrac{x}{2})\\ =2 \sin x \cos(\dfrac{x}{2})\cos(\dfrac{3x}{2})$=R.H.S.

58   Find $ \sin \dfrac{x}{2},\cos\dfrac{x}{2}$ and $ \tan\dfrac{x}{2},$ if $ \tan x=\dfrac{-4}{3}, x $ in quadrant II

Solution :

Here, x is in quadrant II.$\\$ i.e., $\dfrac{\pi}{2} < x < \pi $$\\$ $\implies \dfrac{\pi}{4} < \dfrac{x}{2} < \dfrac{\pi}{2}$$\\$ There, $\sin \dfrac{x}{2},\cos\dfrac{x}{2}$ and $ \tan \dfrac{x}{2}$ $\\$ are lies in first quadrant.$\\$ It is given that $ \tan x=-\dfrac{4}{3}$ $\\$ $\sec^2 x=1+\tan^2 x=1+(\dfrac{-4}{3})^2=1+\dfrac{16}{9}=\dfrac{25}{9}$$\\$ $\therefore \cos^2 x=\dfrac{9}{25}$ $\\$ $\implies \cos x=\pm \dfrac{3}{5}$ $\\$ As x is in quadrant ii, $ \cos x $ is negative. $\cos x=\dfrac{-3}{5}$ $\\$ Now, $ \cos x=2 \cos^2 \dfrac{x}{2}-1$$\\$ $\implies \dfrac{-3}{5} = 2\cos^2 \dfrac{x}{2}-1$ $\\$ $\implies 2 \cos^2 \dfrac{x}{2}=1-\dfrac{3}{5}$ $\\$ $\implies 2 \cos^2 \dfrac{x}{2}=\dfrac{2}{5}$ $\\$ $\implies \cos^2 \dfrac{x}{2}=\dfrac{1}{5} $$\\$ $\implies \cos \dfrac{x}{2}=\dfrac{1}{\sqrt{5}} \quad [\because \cos \dfrac{x}{2} \text{is positive}]$ $\\ $

$ \therefore \cos \dfrac{x}{2}=\dfrac{\sqrt{5}}{5}\\ \sin^2 \dfrac{x}{2}+\cos^2\dfrac{x}{2}=1\\ \implies \sin^2\dfrac{x}{2}+(\dfrac{1}{\sqrt{5}})^2=1\\ \implies \sin^2\dfrac{x}{2}=1-\dfrac{1}{5}=\dfrac{4}{5}\\ \implies \sin^2 \dfrac{x}{2}=\dfrac{2}{\sqrt{5}}[\because \sin \dfrac{x}{2} \text{is positive}]\\ \tan \dfrac{x}{2}=\dfrac{\sin\dfrac{x}{2}}{\cos\dfrac{x}{2}}=\dfrac{(\dfrac{2}{\sqrt{5}})}{\dfrac{1}{\sqrt{5}}}=2$ Thus, the respective values of $ \sin \dfrac{x}{2},\cos \dfrac{x}{2}$ and $\tan \dfrac{x}{2}$ are $ \dfrac{2\sqrt{5}}{5},\dfrac{\sqrt{5}}{5},$ and $2$.

59   Find $ \sin \dfrac{x}{2},\cos\dfrac{x}{2} $ and $ \tan \dfrac{x}{2}$ for $ \cos x=-\dfrac{1}{3},x$ in quadrant III

Solution :

Here, x is in quadrant III.$\\$ i.e., $ \pi < x < \dfrac{3\pi}{2}\\ \implies \dfrac{\pi}{2} < \dfrac{x}{2} < \dfrac{3 \pi}{4}$$\\$ Therefore, $\cos \dfrac{x}{2}$ and $\tan \dfrac{x}{2}$ are negative, where $ \sin \dfrac{x}{2}$ as is positive.$\\$ It is given that $ \cos x=-\dfrac{1}{3}.$ $\\$ $ \cos x =1-2 \sin^2 \dfrac{x}{2}$ $\\$ $\implies \sin^2 \dfrac{x}{2}=\dfrac{1-\cos x}{2}\\ \implies \sin^2 \dfrac{x}{2}=\dfrac{1-(-\dfrac{1}{3})}{2}=\dfrac{(1+\dfrac{1}{3})}{2}=\dfrac{4/3}{2}=\dfrac{2}{3}\\ \implies \sin \dfrac{x}{2}=\dfrac{\sqrt{2}}{\sqrt{3}}\quad [\because \sin \dfrac{x}{2}\text{is positive}]\\ \therefore \sin \dfrac{x}{2}=\dfrac{\sqrt{2}}{\sqrt{3}}*\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{\sqrt{6}}{3}$ $\\$ Now $ \cos x=2 \cos^2 \dfrac{x}{2}-1\\ \implies \cos^2 \dfrac{x}{2}=\dfrac{1+\cos x}{2}=\dfrac{1+(\dfrac{-1}{3})}{2}=\dfrac{(\dfrac{3-1}{3})}{2}=\dfrac{(\dfrac{2}{3})}{2}=\dfrac{1}{3}\\ \implies \cos \dfrac{x}{2}=-\dfrac{1}{\sqrt{3}} \quad [\because \cos \dfrac{x}{2} \text{is negative}]\\ \therefore \cos \dfrac{x}{2}=-\dfrac{1}{\sqrt{3}}*\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{-\sqrt{3}}{3}\\ \tan \dfrac{x}{2}=\dfrac{\sin\dfrac{x}{2}}{\cos\dfrac{x}{2}}=\dfrac{(\dfrac{\sqrt{2}}{\sqrt{3}})}{\dfrac{-1}{\sqrt{3}}}=-\sqrt{2}$ $\\$ Thus, the respective values of $ \sin \dfrac{x}{2},\cos \dfrac{x}{2}$ and $ \tan \dfrac{x}{2}$ are $ \dfrac{\sqrt{6}}{3},\dfrac{-\sqrt{3}}{3},$ and $-\sqrt{2}.$