11th NCERT/CBSE Trigonometric Functions Miscellaneous Exercise Questions 10

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Do or do not
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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

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

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

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

Question (7)

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

Solution

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

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
\[\frac{{\sqrt {8 + 2\sqrt {15} } }}{4},\frac{{\sqrt {8 - 2\sqrt {15} } }}{4},and4 + \sqrt {15} \]