Chapter 3:Miscellaneous Exercise

Prove that(Class 11 Chapter 3 Miscellaneous Exercise trigonometry )

Question 1: $2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+\cos\frac{3\pi}{13}+\cos \frac{5\pi}{13}=0$

Solution: Given, $2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+\cos\frac{3\pi}{13}+\cos \frac{5\pi}{13}$

Using formula $\cos x+\cos y = 2\cos\frac{x+y}{2}\cos \frac{x-y}{2}$

$= 2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+2\cos(\frac{\frac{3\pi}{13}+\frac{5\pi}{13}}{2})\cos(\frac{\frac{3\pi}{13}-\frac{5\pi}{13}}{2})$

$=2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+2\cos\frac{4\pi}{13}\cos (-\frac{\pi}{13})$

$=2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+2\cos\frac{4\pi}{13}\cos (\frac{\pi}{13})$

$= 2\cos \frac{\pi}{13}(\cos\frac{9\pi}{13}+\cos \frac{4\pi}{13})$

$= 2\cos \frac{\pi}{13}\left[2\cos(\frac{\frac{9\pi}{13}+\frac{4\pi}{13}}{2})\cos(\frac{\frac{9\pi}{13}-\frac{4\pi}{13}}{2})\right]$

$= 2\cos \frac{\pi}{13}\left[2\cos \frac{\pi}{2}.\cos \frac{5\pi}{26}\right]$

$=4\cos\frac{\pi}{13}\times 0\times \cos \frac{5\pi}{26}$

$=0=RHS$

Question 2: $(\sin 3x + \sin x)\sin x + (\cos 3x - \cos x) \cos x = 0$

Solution: Taking $LHS = (\sin 3x + \sin x)\sin x + (\cos 3x - \cos x) \cos x$

Using formula: $\sin x+\sin y=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}$ and

$\cos x-\cos y=-2\sin\frac{x+y}{2}\sin\frac{x-y}{2}$

$=(2\sin \frac{3x+x}{2}\cos\frac{3x-x}{2})\sin x+(-2\sin\frac{3x+x}{2}\sin\frac{3x-x}{2})\cos x$

$=2\sin 2x \cos x\sin x-2\sin 2x \sin x \cos x$

$=0=RHS$

Question 3: $(\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4 \cos^2\frac{x+y}{2}$

Solution: Taking $LHS= (\cos x + \cos y)^2 + (\sin x - \sin y)^2$

Using formula: $\sin x-\sin y=2\cos\frac{x+y}{2}\sin\frac{x-y}{2}$ and

$\cos x+\cos y=2\cos \frac{x+y}{2}\cos\frac{x-y}{2}$

$= (2\cos \frac{x+y}{2}\cos\frac{x-y}{2})^2+(2\cos\frac{x+y}{2}\sin\frac{x-y}{2})^2$

$= 4\cos^2 \frac{x+y}{2}\cos^2\frac{x-y}{2}+4\cos^2 \frac{x+y}{2}\sin^2\frac{x-y}{2}$

$=4\cos^2\frac{x+y}{2}(\cos^2\frac{x-y}{2}+\sin^2\frac{x-y}{2}$ )

$=4\cos^2\frac{x+y}{2}=RHS$

Question 4: $(\cos x - \cos y)^2 + (\sin x - \sin y)^2 = 4 \sin^2\frac{x+y}{2}$

Solution: Taking $LHS= (\cos x - \cos y)^2 + (\sin x - \sin y)^2$

Using formulae: $\sin x-\sin y=2\cos\frac{x+y}{2}\sin\frac{x-y}{2}$ and

$\cos x-\cos y=-2\sin \frac{x+y}{2}\sin\frac{x-y}{2}$

$= (-2\sin \frac{x+y}{2}\sin\frac{x-y}{2})^2+(2\cos\frac{x+y}{2}\sin\frac{x-y}{2})^2$

$= 4\sin^2 \frac{x+y}{2}\sin^2\frac{x-y}{2}+4\cos^2 \frac{x+y}{2}\sin^2\frac{x-y}{2}$

$=4\sin^2\frac{x-y}{2}(\cos^2\frac{x+y}{2}+\sin^2\frac{x+y}{2}$ )

$=4\cos^2\frac{x-y}{2}=RHS$

Question 5: $\sin x + \sin 3x + \sin 5x + \sin 7x = 4 \cos x \cos 2x \sin 4x$

Solution: Taking $LHS=\sin x + \sin 3x + \sin 5x + \sin 7x$

$=\sin 7x+\sin x+\sin 5x+\sin x$

Using formulae: $\sin x+\sin y=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}$

$=2\sin\frac{7x+x}{2}\cos \frac{7x-x}{2}+2\sin\frac{5x+3x}{2}\cos\frac{5x-3x}{2}$

$= 2\sin 4x\cos 3x+2\sin 4x\cos x$

$=2\sin 4x(\cos 3x +\cos x)$

Again using formula: $\cos x+\cos y=2\cos \frac{x+y}{2}\cos\frac{x-y}{2}$

$=2\sin 4x(2\cos\frac{3x+x}{2}\cos\frac{3x-x}{2})$

$=4\sin 4x\cos 2x\cos x=RHS$

Question 6: $\frac{(\sin 7x+\sin 5x)+(\sin 9x+\sin 3x)}{(\cos 7x+\cos5x)+(\cos 9x+\cos 3x)}=\tan 6x$

Solution: Taking $LHS= \frac{(\sin 7x+\sin 5x)+(\sin 9x+\sin 3x)}{(\cos 7x+\cos5x)+(\cos 9x+\cos 3x)}$

Using formula: $\sin x+\sin y=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}$ and

$\cos x+\cos y=2\cos \frac{x+y}{2}\cos\frac{x-y}{2}$

$=\frac{(2\sin\frac{7x+5x}{2}\cos\frac{7x-5x}{2})+(2\sin\frac{9x+3x}{2}\cos\frac{9x-3x}{2})}{(2\cos \frac{7x+5x}{2}\cos\frac{7x-5x}{2})+(2\cos \frac{9x+3x}{2}\cos\frac{9x-3x}{2})}$

$=\frac{2\sin 6x\cos x+2\sin 6x\cos 3x}{2\cos 6x\cos x+2\cos 6x\cos 3x}$

$=\frac{2\sin 6x(\cos x+\cos 3x)}{2\cos 6x(\cos x+\cos 3x)}$

$=\tan 6x=RHS$

Question 7: $\sin 3x + \sin 2x - \sin x = 4\sin x \cos\frac{x}{2}\cos\frac{3x}{2}$

Solution: Taking $LHS = \sin 3x + \sin 2x - \sin x$

$= \sin 3x - \sin x +\sin 2x$

Using formula: $\sin x-\sin y=2\cos\frac{x+y}{2}\sin\frac{x-y}{2}$ and

$\sin 2x = 2\sin x\cos x$

$=2\cos\frac{3x+x}{2}\sin\frac{3x-x}{2}+2\sin x\cos x$

$=2\cos 2x\sin x+2\sin x\cos x$

$=2\sin x(\cos 2x+\cos x)$

Again using formula: $\cos x+\cos y=2\cos \frac{x+y}{2}\cos\frac{x-y}{2}$

$= 2\sin x(2\cos\frac{2x+x}{2}\cos\frac{2x-x}{2})$

$=4\sin x \cos\frac{x}{2}\cos \frac{3x}{2}=RHS$

Find $\sin\frac{x}{2}, \cos\frac{x}{2}$ and $\tan\frac{x}{2}$ in each of the following :

Question 8: if $\tan x=-\dfrac{\text{4}}{\text{3}} , \text{x}$ lies in 2nd quadrant.

Solution: Here, x is in 2nd quadrant.

Therefore , $\frac{\pi}{2}<x<\pi$

$\Rightarrow \frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$

hence $\frac{x}{2}$ lies in 1st quadrant.

Therefore, $\text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}\,\,$ and $\text{tan}\dfrac{\text{x}}{\text{2}}$ are positive.

Given that $\text{tan x=-}\dfrac{\text{4}}{\text{3}}$

We know that $\sec^2x=1+\tan^2x$

$\Rightarrow \sec^2x =1+(-\frac{4}{3})^2$

$\Rightarrow \sec^2 x =1+\frac{16}{9}=\frac{25}{9}$

$\Rightarrow \sec x= -\frac{5}{3}$ x lies in second quadrant

Since, $\cos x = \frac{1}{\sec x} =-\frac{3}{5}$

From the formula: $\cos x = 2\cos^2\frac{x}{2}-1$

$-\frac{3}{5} = 2\cos^2\frac{x}{2}-1$

$\Rightarrow 1-\frac{3}{5}=2\cos^2\frac{x}{2}$

$\Rightarrow 2\cos^2\frac{x}{2}=\frac{2}{5}$

$\Rightarrow \cos^2\frac{x}{2}=\frac{1}{5}$

$\Rightarrow \cos\frac{x}{2}=\frac{1}{\sqrt{5}} =\frac{\sqrt{5}}{5},\frac{x}{2}$ lies in Ist quadrant

From formula: $\sin^2\frac{x}{2}=1-\cos^2\frac{x}{2}$

$\Rightarrow \sin^2\frac{x}{2}=1- \frac{1}{5}$

$\Rightarrow \sin^2\frac{x}{2}=\frac{4}{5}$

$\Rightarrow \sin\frac{x}{2}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5},\frac{x}{2}$ lies in Ist quadrant

Since, $\tan \frac{x}{2} = \frac{ \sin\frac{x}{2}}{ \cos\frac{x}{2}}$

$\Rightarrow \tan \frac{x}{2}= \frac{\frac{2}{5}}{\frac{1}{\sqrt{5}}}$

$\Rightarrow \tan \frac{x}{2}= 2,\frac{x}{2}$ lies in Ist quadrant

Hence the value of $\sin\frac{x}{2}, \cos\frac{x}{2}$ and $\tan\frac{x}{2}$ are $\frac{2\sqrt{5}}{5},\frac{\sqrt{5}}{5},2$

Question 9: $\cos x = −\frac{1}{3}$ , x in quadrant III

Solution: –Here, x is in IIIrd quadrant.

Therefore , $\pi<x<\frac{\pi}{2}$

$\Rightarrow \frac{\pi}{2}<\frac{x}{2}<\frac{3\pi}{4}$

hence $\frac{x}{2}$ lies in IInd quadrant.

Since, $\cos x = -\frac{1}{3}$

From the formula: $\cos x = 2\cos^2\frac{x}{2}-1$

$-\frac{1}{3} = 2\cos^2\frac{x}{2}-1$

$\Rightarrow 1-\frac{1}{3}=2\cos^2\frac{x}{2}$

$\Rightarrow 2\cos^2\frac{x}{2}=\frac{2}{3}$

$\Rightarrow \cos^2\frac{x}{2}=\frac{1}{3}$

$\Rightarrow \cos\frac{x}{2}=-\frac{1}{\sqrt{3}} ,\frac{x}{2}$ lies in IInd quadrant

From formula: $\sin^2\frac{x}{2}=1-\cos^2\frac{x}{2}$

$\Rightarrow \sin^2\frac{x}{2}=1- \frac{1}{3}$

$\Rightarrow \sin^2\frac{x}{2}=\frac{2}{3}$

$\Rightarrow \sin\frac{x}{2}=\frac{\sqrt{2}}{\sqrt{3}},\frac{x}{2}$ lies in IInd quadrant

Since, $\tan \frac{x}{2} = \frac{ \sin\frac{x}{2}}{ \cos\frac{x}{2}}$

$\Rightarrow \tan \frac{x}{2}= \frac{\frac{\sqrt{2}}{3}}{-\frac{1}{\sqrt{3}}}$

$\Rightarrow \tan \frac{x}{2}= -\sqrt{2},\frac{x}{2}$ lies in Ist quadrant

Hence the value of $\sin\frac{x}{2}, \cos\frac{x}{2}$ and $\tan\frac{x}{2}$ are $\frac{\sqrt{2}}{\sqrt{3}},\frac{1}{\sqrt{3}},-\sqrt{2}$

Question 10: $\sin x =\frac{1}{4}$ , x in quadrant II

Solution: Here, x is in IInd quadrant.

Therefore , $\frac{\pi}{2}<x<\pi$

$\Rightarrow \frac{\pi}{4}<\frac{x}{2}<\frac{3\pi}{2}$

hence $\frac{x}{2}$ lies in Ist quadrant.

Formula : $\cos^2x=1-\sin^2x$

$\Rightarrow \sin x = 1-(\frac{1}{4})^2=1-\frac{1}{16}$

$\Rightarrow \cos^2 x = \frac{15}{16}$

$\Rightarrow \cos x = -\frac{\sqrt{15}}{4}$ x lies in IInd quadrant

From the formula: $\cos x = 2\cos^2\frac{x}{2}-1$

$-\frac{\sqrt{15}}{4} = 2\cos^2\frac{x}{2}-1$

$\Rightarrow 1-\frac{\sqrt{15}}{4}=2\cos^2\frac{x}{2}$

$\Rightarrow 2\cos^2\frac{x}{2}=\frac{4-\sqrt{15}}{4}$

$\Rightarrow \cos^2\frac{x}{2}=\frac{4-\sqrt{15}}{8}$

$\Rightarrow \cos\frac{x}{2}=\sqrt{\frac{4-\sqrt{15}}{8}} ,$

$\Rightarrow \cos \frac{x}{2}=\sqrt{(\frac{4-\sqrt{15}}{8})\frac{2}{2}}$

$\Rightarrow \cos \frac{x}{2}=\sqrt{\frac{8-2\sqrt{15}}{16}}=\frac{\sqrt{8-2\sqrt{15}}}{4}$ , $\frac{x}{2}$ lies in Ist quadrant

From formula: $\sin^2\frac{x}{2}=1-\cos^2\frac{x}{2}$

$\Rightarrow \sin^2\frac{x}{2}=1- (\frac{\sqrt{8-2\sqrt{15}}}{4})^2$

$\Rightarrow \sin^2\frac{x}{2}=1-\frac{8-2\sqrt{15}}{16}$

$\Rightarrow \sin^2\frac{x}{2} =\frac{16-8+2\sqrt{15}}{16}=\frac{8+2\sqrt{15}}{16}$

$\Rightarrow \sin\frac{x}{2} = \frac{\sqrt{8+2\sqrt{15}}}{4},\frac{x}{2}$ lies in IInd quadrant

Since, $\tan \frac{x}{2} = \frac{ \sin\frac{x}{2}}{ \cos\frac{x}{2}}$

$\Rightarrow \tan \frac{x}{2}= \frac{\frac{\sqrt{8+2\sqrt{15}}}{4}}{\frac{\sqrt{8-2\sqrt{15}}}{4}}$

$\Rightarrow \tan \frac{x}{2}= \frac{\sqrt{8+2\sqrt{15}}}{\sqrt{8-2\sqrt{15}}}.\frac{\sqrt{8+2\sqrt{15}}}{\sqrt{8+2\sqrt{15}}}$

$= \frac{8+2\sqrt{15}}{\sqrt{64-60}}$

$=\frac{2(4+\sqrt{15})}{2}=4+\sqrt{15}$

Hence the value of $\sin\frac{x}{2}, \cos\frac{x}{2}$ and $\tan\frac{x}{2}$ are $\frac{\sqrt{8+2\sqrt{15}}}{4}, \frac{\sqrt{8-2\sqrt{15}}}{4},4+\sqrt{15}$

Ex 3.4 Trigonomety ncert maths solution class 11

Chapter 3:Miscellaneous Exercise

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