3 Marks Question

Question 13 Marks

Find sin $\frac{x}{2},\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ in the $\cos\;x\;=-\frac13$, x in quadrant III.

Answer

Here $\cos x= \frac{{ - 1}}{3}$, x in quadrant III
Now $\pi < x < \frac{{3\pi }}{2} \Rightarrow \frac{\pi }{2} < \frac{x}{2} < \frac{{3\pi }}{4}$
So $\frac{x}{2}$ lies in second quadrant.
$\therefore \sin \frac{x}{2}$ is positive and $\cos \frac{x}{2},\tan \frac{x}{2}$ are negative.
Now $\cos \frac{x}{2} = - \sqrt {\frac{{1 + \cos x}}{2}} = - \sqrt {\frac{{1 - \frac{1}{3}}}{2}} $$ = - \frac{1}{{\sqrt 3 }} $
$\sin \frac{x}{2} = \sqrt {\frac{{1 - \cos x}}{2}} = \sqrt {\frac{{1 + \frac{1}{3}}}{2}} $$ = \sqrt {\frac{2}{3}} $
$\tan \frac{x}{2} = \frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}} = \frac{{\sqrt {\frac{2}{3}} }}{{ - \frac{1}{{\sqrt 3 }}}} = - \sqrt 2 $

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Question 23 Marks

Find sin $\frac{x}{2},\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ in the $\tan x = - \frac{4}{3}$, x in quadrant II.

Answer

Here $\tan x = - \frac{4}{3}$, x in quadrant II.
$\therefore \;{\sec ^2}x = 1 + {\tan ^2}x$$ \Rightarrow {\sec ^2}x = 1 + {\left( {\frac{{ - 4}}{3}} \right)^2} = 1 + \frac{{16}}{9} = \frac{{25}}{9}$
$\therefore \sec x = \pm \frac{5}{3} \Rightarrow \cos x = \pm \frac{3}{5}$
But x lies in IInd quadrant.
$\therefore \cos x = \frac{{ - 3}}{5}$
Also $\frac{\mathrm\pi}2<\mathrm x<\mathrm\pi\Rightarrow\frac{\mathrm\pi}4<\frac{\mathrm x}2<\frac{\mathrm\pi}2$
So $\frac{x}{2}$ lies in first quadrant.
$\therefore \sin \frac{x}{2},\cos \frac{x}{2}$ and $\\tan \frac{x}{2}$ are all positive.
Now $\cos \frac{x}{2} = \sqrt {\frac{{1 + \cos x}}{2}} = \sqrt {\frac{{1 - \frac{3}{5}}}{2}} $$\;=\sqrt{\frac2{5\times2}}\;=\frac1{\sqrt5}$
$\sin \frac{x}{2} = \sqrt {\frac{{1 - \cos x}}{2}} = \sqrt {\frac{{1 + \frac{3}{5}}}{2}} $$\;=\sqrt{\frac8{5\times2}}\;=\sqrt{\frac45}\;=\;\frac2{\sqrt5}$
$\tan \frac{x}{2} = \frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}} = \frac{{\frac{2}{{\sqrt 5 }}}}{{\frac{1}{{\sqrt 5 }}}} = 2$

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Question 33 Marks

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

Answer

We have L.H.S. = sin 3x + sin 2x - sinx
$=\left[2 \cos \left(\frac{3 x+x}{2}\right) \sin \left(\frac{3 x-x}{2}\right)\right]+2 \sin x \cos x$
$\left[\because \sin C-\sin D=2 \cos \frac{C+D}{2} \cdot \sin \frac{C-D}{2}\right]$
$=2 \cos 2 x \sin x+2 \sin x \cos x$
$=2 \sin x[\cos 2 x+\cos x]$
$=2 \sin x\left[2 \cos \left(\frac{2 x+x}{2}\right) \cos \left(\frac{2 x-x}{2}\right)\right] \quad=2 \sin x\left[2 \cos \left(\frac{3 x}{2}\right) \cos \left(\frac{x}{2}\right)\right]$
$=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}=R . H . S$

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Question 43 Marks

Find sin $\frac{x}{2},\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ in the $\sin x = \frac{1}{4}$, x in quadrant II.

Answer

Here $\sin x = \frac{1}{4}$, x in quadrant II
$\therefore {\cos ^2}x = 1 - {\sin ^2}x \Rightarrow {\cos ^2}x =$$1 - {\left( {\frac{1}{4}} \right)^2} = 1 - \frac{1}{{16}} = \frac{{15}}{{16}}$
$\therefore \cos x = \pm \frac{{\sqrt {15} }}{4}$
But x lies in second quadrant.
$\therefore \cos x = - \frac{{\sqrt {15} }}{4}$
Also $\frac{\pi }{2} < x < \pi \Rightarrow \frac{\pi }{4} < \frac{x\ }{2} < \frac{\pi }{2}$
So $\frac{x}{2}$ lies in first quadrant.
$\therefore \sin \frac{x}{2},\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are all positive.
Now $\cos \frac{x}{2} = \sqrt {\frac{{1 + \cos x}}{2}} = \sqrt {\frac{{1 -\frac{{\sqrt {15} }}{4}}}{2}} $$ = \sqrt {\frac{{4 - \sqrt {15} }}{8}} = \frac{{\sqrt {8 - 2\sqrt 5 } }}{4}$
$\sin \frac{x}{2} = \sqrt {\frac{{1 - \cos x}}{2}} = \sqrt {\frac{{1 + \frac{{\sqrt {15} }}{4}}}{2}} $$ = \sqrt {\frac{{4 + \sqrt {15} }}{8}} = \frac{{\sqrt {8 + 2\sqrt {15} } }}{4}$
$\tan \frac{x}{2} = \frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}} = \frac{{\frac{{\sqrt {8 + 2\sqrt {15} } }}{4}}}{{\frac{{\sqrt {8 - 2\sqrt {15} } }}{4}}}$$ = \frac{{\sqrt {8 + 2\sqrt 5 } }}{{\sqrt {8 - 2\sqrt 5 } }} = \frac{{\sqrt {4 + \sqrt {15} } }}{{\sqrt {4 - \sqrt {15} } }}$
$ = \frac{{\sqrt {4 + \sqrt 5 } }}{{\sqrt {4 - \sqrt 5 } }} \times \frac{{\sqrt {4 + \sqrt {15} } }}{{\sqrt {4 + \sqrt {15} } }} = \frac{{4 + \sqrt {15} }}{{\sqrt {16 - 15} }}$$ = 4 + \sqrt {15} $

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Question 53 Marks

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

Answer

To prove: $\cos 6x = 32 \cos^6 x - 48 \cos^2 x + 18 \cos^2 x - 1$
Take L.H.S we have
LHS $= \cos 6 x = \cos 3(2x)$
Using the formula $\cos 3 A = 4 \cos^3 A - 3 \cos A$
$= 4 \cos^3 2x - 3 \cos 2x$
Again by using formula $\cos 2x = 2 \cos^2 x - 1$
$= 4[(2 \cos^2 x - 1)^3 - 3(2 \cos^2 x - 1)$
By further simplification
$= 4[(2 \cos^2 x)^3 - (1)^3 - 3(2 \cos^2 x)^2 + 3(2 \cos^2 x)] - 6 \cos^2 x + 3$
We get
$= 4[8 \cos^6 x - 1 - 12 \cos^4 x + 6 \cos^2 x] - 6 \cos^2 x + 3$
By multiplication
$= 32 \cos^6 x - 4 - 48 \cos^4 x + 24 \cos^2 x - 6 \cos^2 x + 3$
On further calculation
$= 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1$
$=$ RHS

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Question 63 Marks

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

Answer

Taking L.H.S, we have
$\text { L.H.S. }=\frac{1+\cos 2 x}{2}+\frac{1+\cos \left(2 x+\frac{2 \pi}{3}\right)}{2}+\frac{1+\cos \left(2 x-\frac{2 \pi}{3}\right)}{2}$
$=\frac{1}{2}\left[3+\cos 2 x+\cos \left(2 x+\frac{2 \pi}{3}\right)+\cos \left(2 x-\frac{2 \pi}{3}\right)\right]$
$=\frac{1}{2}\left[3+\cos 2 x+2 \cos 2 x \cos \frac{2 \pi}{3}\right]$
$=\frac{1}{2}\left[3+\cos 2 x+2 \cos 2 x \cos \left(\pi-\frac{\pi}{3}\right)\right]$
$=\frac{1}{2}\left[3+\cos 2 x-2 \cos 2 x \cos \frac{\pi}{3}\right]$
$=\frac{1}{2}[3+\cos 2 x-\cos 2 x]=\frac{3}{2}=\mathrm{R.H.S.}$

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Question 73 Marks

If tan $x=\frac{3}{4}, \pi<x<\frac{3 \pi}{2}$, find the value of $\sin \frac{x}{2}, \cos \frac{x}{2} \text { and } \tan \frac{x}{2}$

Answer

Here we have, tan $x=\frac{3}{4}, \pi<x<\frac{3 \pi}{2}$
Since $\pi<x<\frac{3 \pi}{2}$, cos x is negative
Also $\frac{\pi}{2}<\frac{x}{2}<\frac{3 \pi}{4}$
Therefore, sin $\frac{x}{2}$ is positive and cos $\frac{x}{2}$ is negative.
Now $\sec ^{2} x=1+\tan ^{2} x=1+\frac{9}{16}=\frac{25}{16}$
Therefore $\cos ^{2} x=\frac{16}{25} \text { or } \cos x=-\frac{4}{5}$
Now $2 \sin ^{2} \frac{x}{2}=1-\cos x=1+\frac{4}{5}=\frac{9}{5}$
Therefore $\sin ^{2} \frac{x}{2}=\frac{9}{10}$
or $\sin \frac{x}{2}=\frac{3}{\sqrt{10}}$
Again $2 \cos ^{2} \frac{x}{2}=1+\cos x=1-\frac{4}{5}=\frac{1}{5}$
Therefore $\cos ^{2} \frac{x}{2}=\frac{1}{10}$
or $\cos \frac{x}{2}=-\frac{1}{\sqrt{10}}$
Hence $\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{3}{\sqrt{10}} \times\left(\frac{-\sqrt{10}}{1}\right)=-3$

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Question 83 Marks

Find the value of tan $\frac { \pi } { 8 }$.

Answer

Suppose, $x =$ $\frac { \pi } { 8 }$
Then, $2x =$ $\frac { \pi } { 4 }$
Now, $ tan 2x =$ tan $\frac { \pi } { 4 }$
$\Rightarrow \quad \frac { 2 \tan x } { 1 - \tan ^ { 2 } x } = \tan \frac { \pi } { 4 }$ [$\because$$ tan 2x =$ $\frac { 2 \tan x } { 1 - \tan ^ { 2 } x }$]
Putting x = $\frac { \pi } { 8 }$,
$\Rightarrow \frac { 2 \tan \frac { \pi } { 8 } } { 1 - \tan ^ { 2 } \frac { \pi } { 8 } }$$ = 1$
Let y = tan $\frac { \pi } { 8 }$
Then, $\frac { 2 y } { 1 - y ^ { 2 } }$ = 1
$\Rightarrow$ $2y = 1 - y^2$
$\Rightarrow$ $y^2 + 2y - 1 = 0$
$\therefore$ y = $\frac { - 2 \pm \sqrt { ( 2 ) ^ { 2 } + 4 } } { 2 } = \frac { - 2 \pm \sqrt { 8 } } { 2 } = \frac { - 2 \pm 2 \sqrt { 2 } } { 2 }$
$\Rightarrow$ y = - 1 $\pm \sqrt { 2 }$
But $\frac { \pi } { 8 }$ lies in I quadrant, so y = tan $\frac { \pi } { 8 }$ is positive.
Hence, tan $\frac { \pi } { 8 }$ = - 1 + $\sqrt { 2 }$

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Question 93 Marks

Prove that: $\cos 2 x \cos \frac{x}{2}-\cos 3 x \cos \frac{9 x}{2}$ $=\sin 5 x \sin \frac{5 x}{2}$

Answer

LHS $=\cos 2 x \cos \frac{x}{2}-\cos 3 x \cos \frac{9 x}{2}$
$\Rightarrow \mathrm{LHS}=\frac{1}{2}\left\{2 \cos 2 x \cos \frac{x}{2}-2 \cos 3 x \cos \frac{9 x}{2}\right\}$
$\Rightarrow \mathrm{LHS}=\frac{1}{2}\left[\cos \left\{\left(2 x+\frac{x}{2}\right)+\cos \left(2 x-\frac{x}{2}\right)\right\}\right.$ $-\left\{\cos \left(3 x+\frac{9 x}{2}\right)+\cos \left(\frac{9 x}{2}-3 x\right)\right\} ]$[Using: 2 cos A cos B = cos (A + B) + cos (A - B)]
$\Rightarrow \mathrm{LHS}=\frac{1}{2}\left\{\cos \frac{5 x}{2}+\cos \frac{3 x}{2}-\cos \frac{15 x}{2}-\cos \frac{3 x}{2}\right\}$
$\Rightarrow \mathrm{LHS}=\frac{1}{2}\left\{\cos \frac{5 x}{2}-\cos \frac{15 x}{2}\right\}$
$\left. { \Rightarrow \quad {\text{LHS}} = \frac{1}{2}\left\{ {2\sin \left( {\frac{{\frac{{5x}}{2} + \frac{{15x}}{2}}}{2}} \right)\sin \left( {\frac{{\frac{{15x}}{2} - \frac{{5x}}{2}}}{2}} \right)} \right.} \right\}$ $\left[\because \cos C-\cos D=2 \sin \frac{C+D}{2} \sin \frac{D-C}{2}\right]$
$\Rightarrow $ LHS = sin (5x) sin ($\frac {5x} {2}$) = RHS
Hence proved.

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Question 103 Marks

If $ \sin x = \frac{3}{5} , \cos y = - \frac{12}{13}$, where $x$ and $y$ both lie in second quadrant, find the value of $\sin (x + y)$

Answer

Given:$ \sin x = \frac{3}{5} , \cos y = - \frac{12}{13}$
We know that $\sin (x + y) = \sin x \cos y + \cos x \sin y ........ (i)$
Now $\cos^2 x = 1 – \sin^2 x = 1 -\frac{9}{25}=\frac{16}{25}$
Therefore $\cos x=\pm \frac{4}{5}$
Since $x$ lies in second quadrant, $\cos x$ is negative.
Hence $ \cos x = -\frac{4}{5}$
Now $\sin^2y = 1 - \cos^2y = 1-\frac{144}{169}=\frac{25}{169}$
i.e. $\sin y=\pm \frac{5}{13}$
Since $y$ lies in second quadrant, hence $\sin y$ is positive.
Therefore,$ \sin y = \frac{5}{13}$ Substituting the values of $\sin x, \sin y, \cos x $ and $\cos y$ in (i), we get
$\sin (x+y)=\frac{3}{5} \times\left(-\frac{12}{13}\right)+\left(-\frac{4}{5}\right) \times \frac{5}{13} \quad=-\frac{36}{65}-\frac{20}{65}=-\frac{56}{65}$

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Question 113 Marks

Solve: $2 \cos^2 x + 3 \sin x = 0$

Answer

$2 \cos^2 x + 3 \sin x = 0$
$\Rightarrow 2 (1 - \sin^2 x) + 3 \sin x = 0$
$\Rightarrow 2 \sin^2 x - 3 \sin x - 2 = 0$
$\Rightarrow 2 \sin^2\ x - 4 \sin x + \sin x - 2 = 0$
$\Rightarrow 2 \sin x (\sin x - 2) +1 (\sin x - 2 ) = 0$
$\Rightarrow (\sin x - 2) (2 \sin x +1) = 0$
$\Rightarrow 2 \sin x + 1 = 0 [\because \sin x \neq 2 \quad \therefore \sin x-2 \neq 0]$
$\Rightarrow \quad \sin x=-\frac{1}{2}$
$\Rightarrow \quad \sin x=\sin \left(-\frac{\pi}{6}\right) \Rightarrow x=n \pi+(-1)^{n}\left(-\frac{\pi}{6}\right), n \in Z \Rightarrow x=n \pi+(-1)^{n+1} \frac{\pi}{6}, \quad n \in Z.$

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Question 123 Marks

Solve: Sin 2x + sin 4x + sin 6x = 0.

Answer

We have,
$sin 2 x+sin 4 x+sin 6x = 0$
$\Rightarrow$ $sin 4 x+(sin 2 x+sin 6x) = 0$
$\Rightarrow$ $sin 4 x+2 sin 4 x\ cos 2x = 0$
$\Rightarrow$ $sin 4 x(1+2 cos 2x) = 0$
$\Rightarrow$ sin 4 x=0 or ,1+2 cos 2 x=0 $\Rightarrow$ sin 4x = 0 or $\cos 2 x=-\frac{1}{2}$
Now, $\sin 4 x=0 \Rightarrow 4 x=n \pi, n \in Z \Rightarrow x=\frac{n \pi}{4}, n \in Z$
And, $\cos 2 x=-\frac{1}{2}$
$\Rightarrow \quad \cos 2 x=\cos \frac{2 \pi}{3}$
$\Rightarrow \quad 2 x=2 m \pi \pm \frac{2 \pi}{3}, m \in Z$
$\Rightarrow \quad x=m \pi \pm \frac{\pi}{3}, m \in Z$
Hence, $x=\frac{n \pi}{4} or, x=m \pi \pm \frac{\pi}{3}$ where m, n $\in$Z.

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