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

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

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

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

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

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

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

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

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

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
$(\sin 3x+\sin x)\sin x+(\cos 3x-\cos x)\cos x$
= $(3\sin x-4{\sin }^{3}x+sinx)sinx$ + $(4{\cos }^{3}x-3\cos x-cosx)cosx$
= (4sinx - 4 ${\sin }^{3}x$ )sinx + (4 ${\cos }^{3}x$ - 4cos x)cosx
now take the 4sinx common from 1st term and -4cosx from 2nd term
= 4 ${\sin }^{2}x$ (1 - ${\sin }^{2}x$ ) - 4 ${\cos }^{2}x$ (1 - ${\cos }^{2}x$ )
= 4 ${\sin }^{2}x$ ${\cos }^{2}x$ - 4 ${\cos }^{2}x$ ${\sin }^{2}x$ $$\begin{gathered} \because {\cos }^{2}x=1-{\sin }^{2}x \\ and \\ {\sin }^{2}x=1-{\cos }^{2}x \end{gathered}$$
= 0 = R.H.S.

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

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

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

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

= R.H.S.

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

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

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

so

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

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

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

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

$(sin7x+sinx)+(sin5x+sin3x)=2\sin \frac{7x+x}{2}\cos \frac{7x-x}{2}$ $+2\sin \frac{5x+3x}{2}\cos \frac{5x-3x}{2}$
$=$ $2\sin 4x\cos 3x+2\sin 4x\cos x$
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
$cos3x+cosx=2\cos \frac{3x+x}{2}\cos \frac{3x-x}{2}$
= $2\cos 2x\cos x$
= 2sin4x( $2\cos 2x\cos x$ )
= 4cosxcos2xsin4x = R.H.S.

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

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 = $2\sin \frac{7x+5x}{2}\cos \frac{7x-5x}{2}$ = $2\sin 6x\cos x$

sin 9x + sin 3x = $2\sin \frac{9x+3x}{2}\cos \frac{9x-3x}{2}$ = $2\sin 6x\cos 3x$

cos 7x + cos5x = $2\cos \frac{7x+5x}{2}\cos \frac{7x-5x}{2}$ = $2\cos 6x\cos x$

cos 9x + cos3x = $2\cos \frac{9x+3x}{2}\cos \frac{9x-3x}{2}$ = $2\cos 6x\cos 3x$


$\frac{(\sin 7x+\sin 5x)+(\sin 9x+\sin 3x)}{(\cos 7x+\cos 5x)+(\cos 9x+\cos 3x)}$ = $\frac{(2\sin 6x\cos x)+(2\sin 6x\cos 3x)}{(2\cos 6xcosx)+(2\cos 6xcos3x)}$

= $\frac{2\sin 6x(\cos x+\cos 3x)}{2\cos 6x(cosx+cos3x)}=\tan 6x$ = R.H.S. $(\because \frac{\sin x}{\cos x}=\tan x)$

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

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
$sin3x-sinx=2\cos \frac{3x+x}{2}\sin \frac{3x-x}{2}$

$=2\cos 2x\sin x$

sin2x + $2\cos 2x\sin x$ = 2sinx cosx + $2\cos 2x\sin x$
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 $\sin \frac{x}{2},\cos \frac{x}{2},and\tan \frac{x}{2}$ in $\tan x=-\frac{4}{3}$ , x in quadrant II

Answer:

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

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

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

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

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

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

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

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

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

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

Answer:

$$\begin{gathered} \pi <x<\frac{3\pi }{2} \\ \frac{\pi }{2}<\frac{x}{2}<\frac{3\pi }{4} \end{gathered}$$

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

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

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

$$\begin{gathered} 2{\sin }^2\frac{x}{2}=\frac{4}{3} \\ {\sin }^2\frac{x}{2}=\frac{2}{3} \\ \sin \frac{x}{2}=\sqrt{\frac{2}{3}}=\pm \sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3} \end{gathered}$$
Because $\sin \frac{x}{2}$ is +ve in given quadrant

$\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{\frac{\sqrt{6}}{3}}{\frac{-\sqrt{3}}{3}}=-\sqrt{2}$

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

Answer:

$$\begin{gathered} \frac{\pi }{2}<x<\pi \\ \frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2} \end{gathered}$$ 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 = $\sqrt{\frac{15}{16}}=\pm \frac{\sqrt{1}5}{4}=-\frac{\sqrt{1}5}{4}$ (cos x is -ve in II quadrant)

We know that
cosx = $2{\cos }^2\frac{x}{2}-1$
$2{\cos }^2\frac{x}{2}=\cos x+1=-\frac{\sqrt{1}5}{4}+1=\frac{-\sqrt{1}5+4}{4}$

${\cos }^2\frac{x}{2}=\frac{-\sqrt{1}5+4}{8}$
$\cos \frac{x}{2}=\pm \sqrt{\frac{-\sqrt{1}5+4}{8}}=\frac{\sqrt{-\sqrt{1}5+4}}{2\sqrt{2}}=\frac{\sqrt{8-2\sqrt{1}5}}{4}$ (because all functions are posititve in given range)

similarly,
cos x = $1-2{\sin }^2\frac{x}{2}$
$$\begin{gathered} 2{\sin }^2\frac{x}{2}=1-\cos x \\ 2{\sin }^2\frac{x}{2}=1-\left(\frac{-\sqrt{1}5}{4}\right)=\frac{4+\sqrt{1}5}{4} \end{gathered}$$
$\sin \frac{x}{2}=\pm \sqrt{\frac{\sqrt{1}5+4}{8}}=\frac{\sqrt{\sqrt{1}5+4}}{2\sqrt{2}}=\frac{\sqrt{8+2\sqrt{1}5}}{4}$ (because all functions are posititve in given range)
$\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{\frac{\sqrt{8+2\sqrt{1}5}}{4}}{\frac{\sqrt{8-2\sqrt{1}5}}{4}}=\frac{8+2\sqrt{1}5}{\sqrt{64-15\times 4}}=\frac{8+2\sqrt{1}5}{\sqrt{4}}=4+\sqrt{1}5$