NCERT Solutions for Miscellaneous Exercise Chapter 3 Class 11 - trigonometric Functions

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

Answer:

We know that

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

we use this in our problem

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

$\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+2\cos\frac{4\pi }{13}\cos\frac{-\pi}{13}$ ( we know that cos(-x) = cos x )

$\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+2\cos\frac{4\pi }{13}\cos\frac{\pi}{13}$
$\small 2\cos\frac{\pi }{13}(\cos\frac{9\pi }{13}+\cos\frac{4\pi }{13})$
again use the above identity

$\small 2\cos\frac{\pi }{13}(2\cos(\frac{\frac{9\pi }{13}+\frac{4\pi }{13}}{2})\cos(\frac{\frac{9\pi }{13}-\frac{4\pi }{13}}{2})$
$\small 2\cos\frac{\pi }{13}2\cos\frac{\pi }{2}\cos\frac{5\pi }{26}$
we know that

$\small \cos\frac{\pi }{2}$ = 0
So,
$\small 2\cos\frac{\pi }{13}2\cos\frac{\pi }{2}\cos\frac{5\pi }{26}$ = 0 = R.H.S.

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

Answer:

We know that
$sin3x=3\sin x - 4\sin^{3}x$
and
$cos3x=4\cos^{3}x - 3\cos x$
We use this in our problem
$\small (\sin 3x + \sin x)\sin x + (\cos 3x - \cos x )\cos x$
= $(3\sin x - 4\sin^{3}x+ sin x) sinx$ + $(4\cos^{3}x - 3\cos x- cos x)cos x$
= (4sinx - 4 $\small \sin^{3}x$ )sinx + (4 $\small \cos^{3}x$ - 4cos x)cosx
now take the 4sinx common from 1st term and -4cosx from 2nd term
= 4 $\small \sin^{2}x$ (1 - $\small \sin^{2}x$ ) - 4 $\small \cos^{2}x$ (1 - $\small \cos^{2}x$ )
= 4 $\small \sin^{2}x$ $\small \cos^{2}x$ - 4 $\small \cos^{2}x$ $\small \sin^{2}x$ $\small \because \ \ \ \cos^{2}x = 1 - \sin^2x\\ and\\ \sin^{2}x = 1 -\cos^{2}x$
= 0 = R.H.S.

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

Answer:

We know that $(a+b)^{2} = a^{2} + 2ab + b^{2}$
and
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
We use these two in our problem

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

$\small (\cos x + \cos y)^{2} + (\sin x - \sin y)^{2}$ = $\cos^{2}x + 2\cos x\cos y + \cos^{2}y$ + $\sin^{2}x - 2\sin x\sin y + \sin^{2}y$
= 1 + 2cosxcosy + 1 - 2sinxsiny $\left ( \because \sin^{2}x + \cos^{2}x = 1\ and \ \sin^{2}y + \cos^{2}y = 1 \right )$
= 2 + 2(cosxcosy - sinxsiny)
= 2 + 2cos(x + y)
= 2(1 + cos(x + y) )
Now we can write
$cos(x + y) =2cos^{2}\frac{(x + y)}{2} - 1$ $\left ( \because \cos2x = 2cos^{2}x - 1 \ \Rightarrow \cos x = 2\cos^{2}\frac{x}{2} - 1\right )$

= $2(1 + 2cos^{2}\frac{(x + y)}{2} - 1)$
$=4cos^{2}\frac{(x + y)}{2}$

= R.H.S.

Question 4: Prove that $\small (\cos x-\cos y)^{2} + (\sin x - \sin y)^{2} = 4\sin^{2}\left ( \frac{x-y}{2} \right )$

Answer:

We know that $(a+b)^{2} = a^{2} + 2ab + b^{2}$
and
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
We use these two in our problem

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

$\small (\cos x - \cos y)^{2} + (\sin x - \sin y)^{2}$ = $\cos^{2}x - 2\cos x\cos y + \cos^{2}y$ + $\sin^{2}x - 2\sin x\sin y + \sin^{2}y$
= 1 - 2cosxcosy + 1 - 2sinxsiny $\left ( \because \sin^{2}x + \cos^{2}x = 1\ and \ \sin^{2}y + \cos^{2}y = 1 \right )$
= 2 - 2(cosxcosy + sinxsiny)
= 2 - 2cos(x - y) $\small (\because \cos(x-y) =\cos x \cos y + \sin x \sin y)$
= 2(1 - cos(x - y) )
Now we can write
$cos(x + y) = 1 -2sin^{2}\frac{(x + y)}{2}$ $\left ( \because \cos2x = 1 - 2\sin^{2}x \ \Rightarrow \cos x = 1 - 2\sin^{2}\frac{x}{2} \right )$

so

$2(1 - cos(x - y) ) = 2(1 - ( 1 -2sin^{2}\frac{(x + y)}{2}))$

$= 4sin^{2}\frac{(x - y)}{2}$ = R.H.S.

Question 5: Prove that $\small \sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x\cos2x \sin4x$

Answer:
we know that
$sinA + sinB =2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$
We use this identity in our problem
If we notice we need sin4x in our final result so it is better if we made a combination of sin7x and sin x , sin3x and sin5x tp get sin4x

$(sin7x + sinx) + (sin5x + sin3x) = 2\sin\frac{7x+x}{2}\cos\frac{7x-x}{2}$ $+2\sin\frac{5x+3x}{2}\cos\frac{5x-3x}{2}$
$=$ $2\sin4x\cos3x + 2\sin4x\cos x$
take 2sin4x common
= 2sin4x(cos3x + cosx)
Now,
We know that
$cosA + cosB =2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
We use this
$cos3x + cosx =2\cos\frac{3x+x}{2}\cos\frac{3x-x}{2}$
= $2\cos2x\cos x$
= 2sin4x( $2\cos2x\cos x$ )
= 4cosxcos2xsin4x = R.H.S.

Question 6: Prove that $\small \frac{(\sin 7x + \sin 5x) + (\sin9x + \sin 3x)}{(\cos7x + \cos5x) + (\cos9x + \cos3x)} = \tan6x$

Answer:

We know that

$sinA + sinB = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$
and
$cosA + cosB =2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$

We use these two identities in our problem

sin7x + sin5x = $2\sin\frac{7x+5x}{2}\cos\frac{7x-5x}{2}$ = $2\sin6x\cos x$

sin 9x + sin 3x = $2\sin\frac{9x+3x}{2}\cos\frac{9x-3x}{2}$ = $2\sin6x\cos 3x$

cos 7x + cos5x = $2\cos\frac{7x+5x}{2}\cos\frac{7x-5x}{2}$ = $2\cos6x\cos x$

cos 9x + cos3x = $2\cos\frac{9x+3x}{2}\cos\frac{9x-3x}{2}$ = $2\cos6x\cos 3x$


$\small \frac{(\sin 7x + \sin 5x) + (\sin9x + \sin 3x)}{(\cos7x + \cos5x) + (\cos9x + \cos3x)}$ = $\small \frac{(2\sin 6x\cos x) + (2\sin6x \cos3x)}{(2\cos6x cos x) + (2\cos6x cos3x)}$

= $\small \frac{2\sin6x(\cos x + \cos3x)}{2\cos6x (cos x + cos3x)} = \tan6x$ = R.H.S. $\small \left ( \because \frac{\sin x}{\cos x} = \tan x\right )$

Question 7: Prove that $\small \sin3x + \sin2x - \sin x = 4\sin x \cos\frac{x}{2}\cos\frac{3x}{2}$

Answer:

We know that
$cosA + cosB = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
$sinA - sinB = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$

we use these identities
$sin3x - sinx = 2\cos\frac{3x+x}{2}\sin\frac{3x-x}{2}$

$= 2\cos2x\sin x$

sin2x + $2\cos2x\sin x$ = 2sinx cosx + $2\cos2x\sin x$
take 2 sinx common
$2sinx ( cosx + cos2x) = 2sinx(2\cos\frac{2x+x}{2}\cos\frac{2x-x}{2})$

$= 2sinx(2\cos\frac{3x}{2}\cos\frac{x}{2})$
$= 4sinx\cos\frac{3x}{2}\cos\frac{x}{2}$

= R.H.S.

Question 8: Find $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$ in $\small \tan x = - \frac{4}{3}$ , x in quadrant II

Answer:

tan x = $-\frac{4}{3}$
We know that ,
$\sec^{2}x = 1 + \tan^{2}x$
$= 1 +\left ( -\frac{4}{3} \right )^{2}$
$= 1 + \frac{16}{9}$ = $\frac{25}{9}$
$sec x = \sqrt{\frac{25}{9}}$ = $\pm\frac{5}{3}$
x lies in II quadrant thats why sec x is -ve
So,

$sec x =-\frac{5}{3}$
Now, $cos x = \frac{1}{\sec x}$ = $-\frac{3}{5}$
We know that,
$cos x = 2\cos^{2}\frac{x}{2}- 1$ ( $\because \cos2x = 2\cos^{2}x - 1 \Rightarrow \cos x = 2\cos^{2}\frac{x}{2} - 1$ )
$-\frac{3}{5}+ 1 = 2$ $\cos^{2}\frac{x}{2}$

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

$\frac{2}{5}$ = $2\cos^{2}\frac{x}{2}$
$\cos^{2}\frac{x}{2}$ = $\frac{1}{5}$
$\cos\frac{x}{2}$ = $\sqrt{\frac{1}{5}}$ = $\pm\frac{1}{\sqrt5}$
x lies in II quadrant so value of $\cos\frac{x}{2}$ is +ve

$\cos\frac{x}{2}$ = $\frac{1}{\sqrt5} = \frac{\sqrt5}{5}$
we know that
$cos x =1 - 2\sin^{2}\frac{x}{2}$

$2\sin^{2}\frac{x}{2}$ = 1 - $(-\frac{3}{5})$ = $\frac{8}{5}$

$\sin^{2}\frac{x}{2} = \frac{4}{5}\\ \\=\sin\frac{x}{2} = \sqrt{ \frac{4}{5}} = \pm \frac{2}{\sqrt{5}}$
x lies in II quadrant So value of sin x is +ve

$\sin\frac{x}{2} = \frac{2}{\sqrt{}5} = \frac{2\sqrt5}{5}$

$\tan \frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{2\sqrt5}{5}}{\left ( \frac{\sqrt5}{5} \right )} = 2$

Question 9: Find $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$ in $\small \cos x = -\frac{1}{3}$ , x in quadrant III

Answer:

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

We know that
cos x = $2\cos^{2}\frac{x}{2} - 1$
$2\cos^{2}\frac{x}{2} =$ cos x + 1
= $\left ( -\frac{1}{3} \right )$ + 1 = $\left ( \frac{-1+3}{3} \right )$ = $\frac{2}{3}$

$\cos\frac{x}{2} = \sqrt{ \frac{1}{3}} = \pm \frac{1}{\sqrt3}$

$\cos\frac{x}{2} = - \frac{1}{\sqrt3} = - \frac{\sqrt3}{3}$
Now,
we know that
cos x = $1 - 2\sin^{2}\frac{x}{2}$
$2\sin^{2}\frac{x}{2} = 1 - \cos x$
= 1 - $\left ( -\frac{1}{3} \right )$ = $\frac{3+1}{3}$ = $\frac{4}{3}$

$2\sin^{2}\frac{x}{2} = \frac{4}{3} \\ \\ \sin^{2}\frac{x}{2} = \frac{2}{3}\\ \sin\frac{x}{2} = \sqrt{ \frac{2}{3}} = \pm \sqrt{ \frac{2}{3}} = \frac{\sqrt6}{3}$
Because $\sin\frac{x}{2}$ is +ve in given quadrant

$\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{\sqrt6}{3}}{\frac{-\sqrt3}{3}} = - \sqrt2$

Question 10: Find $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$ in $\small \sin x = \frac{1}{4}$ ,x in quadrant II

Answer:

$\frac{\pi}{2} < x < \pi\\ \\ \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$ all functions are positive in this range
We know that
$\cos^{2}x = 1 - \sin^{2}x$
= 1 - $\left ( \frac{1}{4} \right )^{2}$ = $1 - \frac{1}{16}$ = $\frac{15}{16}$

cos x = $\sqrt\frac{15}{16} = \pm \frac{\sqrt15}{4} = - \frac{\sqrt15}{4}$ (cos x is -ve in II quadrant)

We know that
cosx = $2\cos^{2}\frac{x}{2} - 1$
$2\cos^{2}\frac{x}{2} = \cos x + 1 = -\frac{\sqrt15}{4} + 1 = \frac{-\sqrt15+4}{4}$

$\cos^{2}\frac{x}{2} = \frac{-\sqrt15+4}{8}$
$\cos\frac{x}{2} = \pm \sqrt\frac{-\sqrt15+4}{8} = \frac{\sqrt{-\sqrt15+4}}{2\sqrt2} = \frac{\sqrt{8-2\sqrt15}}{4}$ (because all functions are posititve in given range)

similarly,
cos x = $1-2\sin^{2}\frac{x}{2}$
$2\sin^{2}\frac{x}{2} = 1 - \cos x\\ \\ 2\sin^{2}\frac{x}{2} = 1 -\left (\frac{-\sqrt15}{4} \right ) = \frac{4+\sqrt15}{4}$
$\sin\frac{x}{2} = \pm \sqrt\frac{\sqrt15+4}{8} = \frac{\sqrt{\sqrt15+4}}{2\sqrt2} = \frac{\sqrt{8+2\sqrt15}}{4}$ (because all functions are posititve in given range)
$\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{\sqrt{8+2\sqrt15}}{4}}{\frac{\sqrt{8-2\sqrt15}}{4}} = \frac{{8+2\sqrt15}}{\sqrt{64 - 15\times4}} = \frac{{8+2\sqrt15}}{\sqrt{4}} = 4 + \sqrt15$