11th NCERT/CBSE Trigonometric Functions Miscellaneous Exercise Questions 10
Do or do not
There is no try

### Prove that

Question (1)

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

Solution

$LHS = 2\cos \frac{\pi }{{13}}\cos \frac{{9\pi }}{{13}} + \cos \frac{{3\pi }}{{13}} + \cos \frac{{5\pi }}{{13}}$ $= \cos \left( {\frac{\pi }{{13}} + \frac{{9\pi }}{{13}}} \right) + \cos \left( {\frac{\pi }{{13}} - \frac{{9\pi }}{{13}}} \right) + \cos \left( {\frac{{3\pi }}{{13}}} \right) + \cos \left( {\frac{{5\pi }}{{13}}} \right)$ $= \cos \left( {\frac{{10\pi }}{{13}}} \right) + \cos \left( {\frac{{ - 8\pi }}{{13}}} \right) + \cos \left( {\frac{{3\pi }}{{13}}} \right) + \cos \left( {\frac{{5\pi }}{{13}}} \right)$ $= \left[ {\cos \left( {\frac{{10\pi }}{{13}}} \right) + \cos \left( {\frac{{3\pi }}{{13}}} \right)} \right] + \left[ {\cos \left( {\frac{{8\pi }}{{13}}} \right) + \cos \left( {\frac{{5\pi }}{{13}}} \right)} \right]$ $= \cos \left( {\pi - \frac{{3\pi }}{{13}}} \right) + \cos \left( {\frac{{3\pi }}{{13}}} \right) + \cos \left( {\pi - \frac{{5\pi }}{{13}}} \right) + \cos \left( {\frac{{5\pi }}{{13}}} \right)$ $= - \cos \left( {\frac{{3\pi }}{{13}}} \right) + \cos \left( {\frac{{3\pi }}{{13}}} \right) - \cos \left( {\frac{{5\pi }}{{13}}} \right) + \cos \left( {\frac{{5\pi }}{{13}}} \right)$ $= 0$ $= RHS$

Question (2)

$\left( {\sin 3x + \sin x} \right)\sin x + \left( {\cos 3x - \cos x} \right)\cos x = 0$

Solution

$LHS = \left( {\sin 3x + \sin x} \right)\sin x + \left( {\cos 3x - \cos x} \right)\cos x$ $= 2\sin 2x\cos x\sin x + \left( { - 2\sin 2x\sin x} \right)\cos x$ $= 2\sin 2x\cos x\sin x - 2\sin 2x\sin x\cos x$ $= 0$ $= RHS$

Question (3)

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

Solution

$LHS = {\left( {\cos x + \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2}$ $= {\left[ {2\cos \left( {\frac{{x + y}}{2}} \right)\cos \left( {\frac{{x - y}}{2}} \right)} \right]^2} + {\left[ {2\cos \left( {\frac{{x + y}}{2}} \right)\sin \left( {\frac{{x - y}}{2}} \right)} \right]^2}$ $= 4{\cos ^2}\left( {\frac{{x + y}}{2}} \right){\cos ^2}\left( {\frac{{x - y}}{2}} \right) + 4{\cos ^2}\left( {\frac{{x + y}}{2}} \right){\sin ^2}\left( {\frac{{x - y}}{2}} \right)$ $= 4{\cos ^2}\left( {\frac{{x + y}}{2}} \right)\left[ {{{\cos }^2}{{\left( {\frac{{x - y}}{2}} \right)}^2} + {{\sin }^2}\left( {\frac{{x - y}}{2}} \right)} \right]$ $= 4{\cos ^2}\left( {\frac{{x + y}}{2}} \right)$ $= RHS$

Question (4)

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

Solution

$LHS = {\left( {\cos x - \cos y} \right)^2} + {\left( {\sin x - sin\;y} \right)^2}$ $= {\left[ { - 2\sin \left( {\frac{{x + y}}{2}} \right)\sin \left( {\frac{{x - y}}{2}} \right)} \right]^2} + {\left[ {2\cos \left( {\frac{{x + y}}{2}} \right)\sin \left( {\frac{{x - y}}{2}} \right)} \right]^2}$ $= 4{\sin ^2}\left( {\frac{{x + y}}{2}} \right){\sin ^2}\left( {\frac{{x - y}}{2}} \right) + 4{\cos ^2}\left( {\frac{{x + y}}{2}} \right){\sin ^2}\left( {\frac{{x - y}}{2}} \right)$ $= 4{\sin ^2}\left( {\frac{{x - y}}{2}} \right)\left[ {{{\sin }^2}\left( {\frac{{x + y}}{2}} \right) + {{\cos }^2}{{\left( {\frac{{x + y}}{2}} \right)}^2}} \right]$ $= 4{\sin ^2}\left( {\frac{{x - y}}{2}} \right)$ $= RHS$

Question (5)

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

Solution

$LHS = \sin x + \sin 3x + \sin 5x + \sin 7x$ $= \left( {\sin x + \sin 5x} \right) + \left( {\sin 3x + \sin 7x} \right)$
Use following formula
$\sin A + \sin B = 2\sin \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)$
$= 2\sin \left( {\frac{{x + 5x}}{2}} \right)\cos \left( {\frac{{x - 5x}}{2}} \right) + 2\sin \left( {\frac{{3x + 7x}}{2}} \right)\cos \left( {\frac{{3x - 7x}}{2}} \right)$ $= 2\sin 3x\cos \left( { - 2x} \right) + 2\sin 5x\cos \left( { - 2x} \right)$ $= 2\sin 3x\cos 2x + 2\sin 5x\cos 2x$ $= 2\cos 2x\left[ {\sin 3x + \sin 5x} \right]$ $= 2\cos 2x\left[ {2\sin \left( {\frac{{3x + 5x}}{2}} \right) \cdot \cos \left( {\frac{{3x - 5x}}{2}} \right)} \right]$ $= 2\cos 2x\left[ {2\sin 4x \cdot \cos \left( { - x} \right)} \right]$ $= 4\cos 2x\sin 4x\cos x$ $= RHS$

Question (6)

$\frac{{\left( {\sin 7x + \sin 5x} \right) + \left( {\sin 9x + \sin 3x} \right)}}{{\left( {\cos 7x + \cos 5x} \right) + \left( {\cos 9x + \cos 3x} \right)}} = \tan 6x$

Solution

$LHS = \frac{{\left( {\sin 7x + \sin 5x} \right) + \left( {\sin 9x + \sin 3x} \right)}}{{\left( {\cos 7x + \cos 5x} \right) + \left( {\cos 9x + \cos 3x} \right)}}$
$\sin A + \sin B = 2\sin \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)$ and $\cos A + \cos B = 2\cos \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)$
$= \frac{{\left[ {2\sin \left( {\frac{{7x + 5x}}{2}} \right)\cos \left( {\frac{{7x - 5x}}{2}} \right)} \right] + \left[ {2\sin \left( {\frac{{9x + 3x}}{2}} \right)\cos \left( {\frac{{9x - 3x}}{2}} \right)} \right]}}{{\left[ {2\cos \left( {\frac{{7x + 5x}}{2}} \right).\cos \left( {\frac{{7x - 5x}}{2}} \right)} \right] + \left[ {2\cos \left( {\frac{{9x + 3x}}{2}} \right)\cos \left( {\frac{{9x - 3x}}{2}} \right)} \right]}}$ $= \frac{{\left[ {2\sin 6x \cdot \cos x} \right] + \left[ {2\sin 6x \cdot \cos 3x} \right]}}{{\left[ {2\sin 6x \cdot \cos x} \right] + \left[ {2\sin 6x \cdot \cos 3x} \right]}}$ $= \frac{{2\sin 6x\left[ {\cos x + \cos 3x} \right]}}{{2\cos 6x\left[ {\cos x + \cos 3x} \right]}}$ $= \tan 6x = RHS$

Question (7)

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

Solution

$LHS = \sin 3x + \sin 2x - \sin x$ $= \sin 3x + \left( {\sin 2x - \sin x} \right)$
$\sin A - \sin B = 2\cos \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)$
$= \sin 3x + \left[ {2\cos \left( {\frac{{2x + x}}{2}} \right)\sin \left( {\frac{{2x - x}}{2}} \right)} \right]$ $= \sin 3x + \left[ {2\cos \left( {\frac{3}{2}} \right)\sin \left( {\frac{x}{2}} \right)} \right]$ $= \sin 3x + 2\cos \frac{{3x}}{2}\sin \frac{x}{2}$
$\sin 2A = 2\sin A\cos B$
$= 2\sin \frac{{3x}}{2}\cos \frac{{3x}}{2} + 2\cos \frac{{3x}}{2}\sin \frac{x}{2}$ $= 2\cos \left( {\frac{{3x}}{2}} \right)\left[ {\sin \left( {\frac{{3x}}{2}} \right) + \sin \left( {\frac{x}{2}} \right)} \right]$
$\sin A + \sin B = 2\sin \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)$
$= 2\cos \left( {\frac{{3x}}{2}} \right)\left[ {2\sin \left\{ {\frac{{\left( {\frac{{3x}}{2}} \right) + \left( {\frac{x}{2}} \right)}}{2}} \right\}\cos \left\{ {\frac{{\left( {\frac{{3x}}{2}} \right) - \left( {\frac{\pi }{2}} \right)}}{2}} \right\}} \right]$ $= 2\cos \left( {\frac{{3x}}{2}} \right)2\sin x\cos \left( {\frac{x}{2}} \right)$ $= 4\sin x\cos \left( {\frac{x}{2}} \right)\cos \left( {\frac{x}{2}} \right) = RHS$

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

Question (8)

$\tan x = - \frac{4}{3} \text{,x in quadrant II}$

Solution

Here, x is in quadrant II. $\frac{\pi }{2} < x < \pi$ $\Rightarrow \frac{\pi }{4} < \frac{x}{2} < \frac{\pi }{2}$ $\therefore\sin \frac{x}{2},\cos \frac{x}{2}and\tan \frac{x}{2} \text{are all positive}$ $\text{It is given that } \tan x = - \frac{4}{3}$ ${\sec ^2}x = 1 + {\tan ^2}x = 1 + {\left( {\frac{{ - 4}}{3}} \right)^2} = 1 + \frac{{19}}{9} = \frac{{25}}{9}$ $\therefore {\cos ^2}x = \frac{9}{{25}}$ $\Rightarrow \cos x = \pm \frac{3}{5}$ As x is in quadrant II, cos x is negative
$\therefore \cos x = \frac{{ - 3}}{5}$ $text{Now} \; \;\cos x = 2{\cos ^2}\frac{x}{2} - 1$ $\Rightarrow \frac{{ - 3}}{5} = 2{\cos ^2}\frac{x}{2} - 1$ $\Rightarrow 2{\cos ^2}\frac{x}{2} = 1 - \frac{3}{5}$ $\Rightarrow 2{\cos ^2}\frac{x}{2} = \frac{2}{5}$ $\Rightarrow {\cos ^2}\frac{x}{2} = \frac{1}{5}$ $\Rightarrow \cos \frac{x}{2} = \frac{1}{{\sqrt 5 }}\left[ \therefore {\cos \frac{x}{2} \text{is positive}} \right]$ $\therefore \cos \frac{x}{2} = \frac{{\sqrt 5 }}{5}$ ${\sin ^2}\frac{x}{2} + {\cos ^2}\frac{x}{2} = 1$ $\Rightarrow {\sin ^2}\frac{x}{2} + {\left( {\frac{1}{{\sqrt 5 }}} \right)^2} = 1$ $\Rightarrow {\sin ^2}\frac{x}{2} = 1 - \frac{1}{5} = \frac{4}{5}$ $\Rightarrow \sin \frac{x}{2} = \frac{2}{{\sqrt 5 }}\left[ {\sin \frac{x}{2} \text{is positive}} \right]$ $\sin \frac{x}{2} = \frac{{2\sqrt 5 }}{5}$ $\tan \frac{x}{2} = \frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}} = \frac{{\left( {\frac{2}{{\sqrt 5 }}} \right)}}{{\left( {\frac{1}{{\sqrt 5 }}} \right)}} = 2$ Thus, the respective value of $\sin \frac{x}{2},\cos \frac{x}{2}and\tan \frac{x}{2}$ are $\frac{{2\sqrt 5 }}{5},\frac{{\sqrt 5 }}{5},and2$

Question (9)

$\cos x = - \frac{1}{3} \text{,x in quadrant III}$

Solution

Here, x is in quadrant III. $\pi < x < \frac{{3\pi }}{2}$ $\Rightarrow \frac{\pi }{2} < \frac{x}{2} < \frac{{3\pi }}{4}$ Therefore, $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are negative, where as $\sin \frac{x}{2}$ is positive
It is given that $\cos x = - \frac{1}{3}$
$\cos x = 1 - 2{\sin ^2}\frac{x}{2}$ 0965432345$\Rightarrow {\sin ^2}\frac{{x}}{2} = \frac{{1 - \cos x}}{2}$ $\Rightarrow {\sin ^2}\frac{x}{2} = \frac{{1 - \left( { - \frac{1}{3}} \right)}}{2} = \frac{2}{3}$ $\Rightarrow \sin \frac{x}{2} = \frac{{\sqrt 2 }}{{\sqrt 3 }}\left[ {\sin \frac{x}{2} \text{is positive}} \right]$ $\sin \frac{x}{2} = \frac{{\sqrt 2 }}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} = \frac{{\sqrt 6 }}{3}$ $\text{Now} \; \; \cos x = 2{\cos ^2}\frac{x}{2} - 1$ $\Rightarrow {\cos ^2}\frac{x}{2} = \frac{{1 + \cos x}}{2} = \frac{{1 + \left( { - \frac{1}{3}} \right)}}{2} = \frac{1}{3}$ $\Rightarrow \cos \frac{x}{2} = -\frac{1}{{\sqrt 3 }}\left[\ text{As}\;\; {\cos \frac{x}{2}isnegative} \right]$ $\cos \frac{x}{2} = - \frac{1}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} = \frac{{ - \sqrt 3 }}{3}$ $\tan \frac{x}{2} = \frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}} = \frac{{\left( {\frac{{\sqrt 2 }}{{\sqrt 3 }}} \right)}}{{\left( {\frac{{ - 1}}{{\sqrt 3 }}} \right)}} = - \sqrt 2$ Thus, the respective values of $\sin \frac{x}{2},\cos \frac{x}{2}\;and\; \tan \frac{x}{2} \;\;are$ $\frac{{\sqrt 6 }}{3},\frac{{ - \sqrt 3 }}{3},and - \sqrt 2$

Question (10)

$\sin x = \frac{1}{4} \text{,x in quadrant II}$

Solution

Here, x is in quadrant II. $\frac{\pi }{2} < x < \pi$ $\Rightarrow \frac{\pi }{4} < \frac{x}{2} < \frac{\pi }{2}$ Therefore, $\sin \frac{x}{2},\cos \frac{x}{2},and\tan \frac{x}{2}$ are all positive
It is given that $\sin x = \frac{1}{4}$
${\cos ^2}x = 1 - {\sin ^2}x = 1 - {\left( {\frac{1}{4}} \right)^2} = 1 - \frac{1}{{16}} = \frac{{15}}{{16}}$ $\Rightarrow \cos x = - \frac{{\sqrt {15} }}{4}\left[ {\cos x \text{is negative in quadrant II} \right]$ ${\sin ^2}\frac{x}{2} = \frac{{1 - \cos x}}{2}$ ${\sin ^2}\frac{x}{2} = \frac{{1 - \left( { - \frac{{\sqrt {15} }}{4}} \right)}}{2} = \frac{{4 + \sqrt {15} }}{8}$ $\sin \frac{x}{2} = \sqrt {\frac{{4 + \sqrt {15} }}{8}} \left[ {As\; \;\sin \frac{x}{2} text{is positive}} \right]$ $= \sqrt {\frac{{4 + \sqrt {15} }}{8} \times \frac{2}{2}}$ $= \sqrt {\frac{{8 + 2\sqrt {15} }}{{16}}}$ $= \frac{{\sqrt {8 + 2\sqrt {15} } }}{4}$ $\Rightarrow \cos \frac{x}{2} = \sqrt {\frac{{4 - \sqrt {15} }}{8}} \left[ {As\cos \frac{x}{2}ispositive} \right]$ $= \sqrt {\frac{{4 - \sqrt {15} }}{8} \times \frac{2}{2}}$ $= \sqrt {\frac{{8 - 2\sqrt {15} }}{{16}}} = \frac{{\sqrt {8 - 2\sqrt {15} } }}{4}$ $\tan \frac{x}{2} = \frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}} = \frac{{\left( {\frac{{\sqrt {8 + 2\sqrt {15} } }}{4}} \right)}}{{\left( {\frac{{\sqrt {8 - 2\sqrt {15} } }}{4}} \right)}} = \frac{{\sqrt {8 + 2\sqrt {15} } }}{{\sqrt {8 - 2\sqrt {15} } }}$ $= \sqrt {\frac{{8 + 2\sqrt {15} }}{{8 - 2\sqrt {15} }} \times \frac{{8 + 2\sqrt {15} }}{{8 + 2\sqrt {15} }}}$ $= \sqrt {\frac{{{{\left( {8 + 2\sqrt {15} } \right)}^2}}}{{64 - 60}}} = \frac{{8 + 2\sqrt {15} }}{2} = 4 + \sqrt {15}$ Thus, the respective value of $\sin \frac{x}{2},\cos \frac{x}{2}and\tan \frac{x}{2}$ are
$\frac{{\sqrt {8 + 2\sqrt {15} } }}{4},\frac{{\sqrt {8 - 2\sqrt {15} } }}{4},and4 + \sqrt {15}$