Miscellaneous Exercise Trigonometric Functions Solution - NCERT Class 11

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Chapter 3 Ex.3.ME Question 1

Prove that:

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

Solution

\[\begin{align} &LHS=2\cos \frac{\pi }{13}\cos \frac{9\pi }{13}+\cos \frac{3\pi }{13}+\cos \frac{5\pi }{13} \\ &=2\cos \frac{\pi }{13}\cos \frac{9\pi }{13}+2\cos \left( \frac{\frac{3\pi }{13}+\frac{5\pi }{13}}{2} \right)\cos \left( \frac{\frac{3\pi }{13}-\frac{5\pi }{13}}{2} \right)\ \\&\qquad \left[ \because \cos A+\cos B=2\cos \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right) \right] \\ &=2\cos \frac{\pi }{13}\cos \frac{9\pi }{13}+2\cos \frac{4\pi }{13}\cos \left( -\frac{\pi }{13} \right) \\ &=2\cos \frac{\pi }{13}\cos \frac{9\pi }{13}+2\cos \frac{4\pi }{13}\cos \frac{\pi }{13} \\ &=2\cos \frac{\pi }{13}\left[ \cos \frac{9\pi }{13}+\cos \frac{4\pi }{13} \right] \\ &=2\cos \frac{\pi }{13}\left[ 2\cos \left( \frac{\frac{9\pi }{13}+\frac{4\pi }{13}}{2} \right)\cos \left( \frac{\frac{9\pi }{13}-\frac{4\pi }{13}}{2} \right) \right] \\&\qquad \left[ \because \cos A+\cos B=2\cos \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right) \right] \\ &=2\cos \frac{\pi }{13}\left[ 2\cos \frac{\pi }{2}\cos \frac{5\pi }{26} \right] \\ &=2\cos \frac{\pi }{13}\times 2\times 0\times \cos \frac{5\pi }{26} \qquad \quad \left[ \because \cos \frac{\pi }{2}=0 \right] \\ &=0 \\ &=RHS \end{align}\]

Chapter 3 Ex.3.ME Question 2

Prove that:

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

Solution

\[\begin{align} LHS&=\left( \sin 3x+\sin x \right)\sin x+\left( \cos 3x-\cos x \right)\cos x \\ &=\sin 3x\sin x+{{\sin }^{2}}x+\cos 3x\cos x-{{\cos }^{2}}x \\ &=\cos 3x\cos x+\sin 3x\sin x-\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right) \\ &=\cos \left( 3x-x \right)-\cos 2x\\& \qquad  \left[ \begin{array}{l} \because \cos \left( A-B \right)=\cos A\cos B+\sin A\sin B \\ \&\cos 2A={{\cos }^{2}}A-{{\sin }^{2}}A \\ \end{array} \right] \\ &=\cos 2x-\cos 2x \\ &=0 \\& =RHS \end{align}\]

Chapter 3 Ex.3.ME Question 3

Prove that:

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

Solution

\[\begin{align} LHS&={{\left( \cos x+\cos y \right)}^{2}}+{{\left( \sin x-\sin y \right)}^{2}} \\ &={{\cos }^{2}}x+{{\cos }^{2}}y+2\cos x\cos y+{{\sin }^{2}}x+{{\sin }^{2}}y-2\sin x\sin y\ \qquad\\&\qquad \ \left[ \begin{array}{l} \because {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\ \& {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\ \end{array} \right] \\ &=\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)+\left( {{\cos }^{2}}y+{{\sin }^{2}}y \right)+2\left( \cos x\cos y-\sin x\sin y \right) \\ &=1+1+2\cos \left( x+y \right) \\&\qquad \quad \left[ \begin{array}{l} \because \left( {{\cos }^{2}}A+{{\sin }^{2}}A \right)=1 \\ \& \cos \left( A+B \right)=\cos A\cos B-\sin A\sin B \\ \end{array} \right] \\ &=2+2\cos \left( x+y \right) \\ &=2\left[ 1+\cos 2\left( \frac{x+y}{2} \right) \right] \\ &=2\left[ 1+2{{\cos }^{2}}\left( \frac{x+y}{2} \right)-1 \right]\ \\&\qquad\ \left[ \because \cos 2A=2{{\cos }^{2}}A-1 \right] \\ &=4{{\cos }^{2}}\frac{x+y}{2} \\ &=RHS \end{align}\]

Chapter 3 Ex.3.ME Question 4

Prove that:

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

Solution

\[\begin{align} LHS&={{\left( \cos x-\cos y \right)}^{2}}+{{\left( \sin x-\sin y \right)}^{2}} \\ &={{\cos }^{2}}x+{{\cos }^{2}}y-2\cos x\cos y+{{\sin }^{2}}x+{{\sin }^{2}}y-2\sin x\sin y \qquad\\&\qquad \left[ \begin{array}{l} \because {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\ \& {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\ \end{array} \right] \\ &=\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)+\left( {{\cos }^{2}}y+{{\sin }^{2}}y \right)-2\left( \cos x\cos y+\sin x\sin y \right) \\ &=1+1-2\cos \left( x-y \right)\qquad \quad \left[ \begin{array}{l} \because \left( {{\cos }^{2}}A+{{\sin }^{2}}A \right)=1 \\ \& \cos \left( A-B \right)=\cos A\cos B+\sin A\sin B \\ \end{array} \right] \\ &=2-2\cos \left( x-y \right) \\ &=2\left[ 1-\cos 2\left( \frac{x-y}{2} \right) \right] \\ &=2\left[ 1-\left\{ 1-2{{\sin }^{2}}\left( \frac{x-y}{2} \right) \right\} \right] \qquad \left[ \because \cos 2A=2{{\cos }^{2}}A-1 \right] \\ &=2\left[ 1-1+2{{\sin }^{2}}\left( \frac{x-y}{2} \right) \right] \\ &=4{{\sin }^{2}}\frac{x-y}{2} \\ &=RHS \end{align}\]

Chapter 3 Ex.3.ME Question 5

Prove that:

\(\sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x\cos 2x\sin 4x\)

Solution

\[\begin{align} LHS&=\sin x+\sin 3x+\sin 5x+\sin 7x \\ &=\left( \sin 5x+\sin x \right)+\left( \sin 7x+\sin 3x \right) \\ &=\left[ 2\sin \left( \frac{5x+x}{2} \right)\cos \left( \frac{5x-x}{2} \right) \right]+\left[ 2\sin \left( \frac{7x+3x}{2} \right)\cos \left( \frac{7x-3x}{2} \right) \right] \\ & \qquad \quad \left[ \because \sin A+\sin B=2\sin \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right) \right] \\ &=2\sin 3x\cos 2x+2\sin 5x\cos 2x \\ &=2\cos 2x\left( \sin 5x+\sin 3x \right) \\ &=2\cos 2x\left[ 2\sin \left( \frac{5x+3x}{2} \right)\cos \left( \frac{5x-3x}{2} \right) \right]\ \ \ \ \ \ \ \\&\qquad\left[ \because \sin A+\sin B=2\sin \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right) \right] \\ &=2\cos 2x\left[ 2\sin 4x\cos x \right] \\ &=4\cos x\cos 2x\sin 4x \\ &=RHS \end{align}\]

Chapter 3 Ex.3.ME Question 6

Prove that:

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

Solution

\[\begin{align} LHS&=\frac{\left( \sin 7x+\sin 5x \right)+\left( \sin 9x+\sin 3x \right)}{\left( \cos 7x+\cos 5x \right)+\left( \cos 9x+\cos 3x \right)} \\ &=\frac{2\sin \left( \frac{7x+5x}{2} \right)\cos \left( \frac{7x-5x}{2} \right)+2\sin \left( \frac{9x+3x}{2} \right)\cos \left( \frac{9x-3x}{2} \right)}{2\cos \left( \frac{7x+5x}{2} \right)\cos \left( \frac{7x-5x}{2} \right)+2\cos \left( \frac{9x+3x}{2} \right)\cos \left( \frac{9x-3x}{2} \right)} \\ & \qquad \quad \left[ \begin{array}{l} \because \sin A+\sin B=2\sin \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right) \\ \& \cos A+\cos B=2\cos \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right) \\ \end{array} \right] \\ &=\frac{2\sin 6x\cos x+2\sin 6x\cos 3x}{2\cos 6x\cos x+2\cos 6x\cos 3x} \\ &=\frac{2\sin 6x\left( \cos x+\cos 3x \right)}{2\cos 6x\left( \cos x+\cos 3x \right)} \\ &=\frac{\sin 6x}{\cos 6x} \\ &=\tan 6x \\ &=RHS \end{align}\]

Chapter 3 Ex.3.ME Question 7

Prove that:

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

Solution

\[\begin{align} &LHS=\sin 3x+\sin 2x-\sin x \\ &=\sin 3x+\left( \sin 2x-\sin x \right) \\ &=\sin 2\left( \frac{3x}{2} \right)+\left[ 2\cos \left( \frac{2x+x}{2} \right)\sin \left( \frac{2x-x}{2} \right) \right] \quad \\&\qquad\left[ \because \sin A-\sin B=2\cos \left( \frac{A+B}{2} \right)\sin \left( \frac{A-B}{2} \right) \right] \\ &=2\sin \frac{3x}{2}\cos \frac{3x}{2}+2\cos \frac{3x}{2}\sin \frac{x}{2} \qquad \quad \left[ \because \sin 2A=2\sin A\cos A \right] \\ &=2\cos \frac{3x}{2}\left( \sin \frac{3x}{2}+\sin \frac{x}{2} \right) \\ &=2\cos \frac{3x}{2}\left[ 2\sin \left( \frac{\frac{3x}{2}+\frac{x}{2}}{2} \right)\cos \left( \frac{\frac{3x}{2}-\frac{x}{2}}{2} \right) \right] \qquad \\&\qquad\left[ \because \sin A+\sin B=2\sin \left( \frac{A+B}{2} \right)\cos \left( \frac{A-B}{2} \right) \right] \\ &=2\cos \frac{3x}{2}\left[ 2\sin x\cos \frac{x}{2} \right] \\ &=4\sin x\cos \frac{x}{2}\cos \frac{3x}{2} \\ &=RHS \end{align}\]

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

Chapter 3 Ex.3.ME Question 8

Prove that:

\(\tan x = - \frac{4}{3},\;x\) in quadrant II.

Solution

Since \(x\) lies in quadrant II

\(\frac{\pi }{2} < x < \pi \)

Therefore,

\(\frac{\pi }{4} < \frac{x}{2} < \frac{\pi }{2}\)

Hence, \(\sin \frac{x}{2}\), \(\cos \frac{x}{2}\)and \(\tan \frac{x}{2}\)lie in quadrant I and all are positive.

It is given that \(\tan x = - \frac{4}{3}\)

\[\begin{align}{\sec ^2}x &= 1 + {\tan ^2}x\\&= 1 + {\left( { - \frac{4}{3}} \right)^2}\\&= 1 + \frac{{16}}{9}\\&= \frac{{25}}{9}\\\sec x &= \pm \sqrt {\frac{{25}}{9}} \\\frac{1}{{\cos x}}& = \pm \frac{5}{3}\\\cos x& = \pm \frac{3}{5}\end{align}\]

As \(x\) is in quadrant II, \(\cos x\) is negative.

\[\begin{align}\cos x &= - \frac{3}{5}\\\cos 2\left( {\frac{x}{2}} \right) &= - \frac{3}{5}\\2{\cos ^2}\frac{x}{2} - 1 &= - \frac{3}{5}\\2{\cos ^2}\frac{x}{2} &= 1 - \frac{3}{5}\\{\cos ^2}\frac{x}{2} &= \frac{2}{5} \times \frac{1}{2}\\{\cos ^2}\frac{x}{2} &= \frac{1}{5}\\\cos \frac{x}{2} &= \pm \sqrt {\frac{1}{5}} \end{align}\]

Since, \(\cos \frac{x}{2}\) lies in quadrant I and positive, \(\cos \frac{x}{2} = \frac{1}{{\sqrt 5 }}\)

Now,

\[\begin{align} {{\sin }^{2}}\frac{x}{2}&=1-{{\cos }^{2}}\frac{x}{2} \qquad \quad \left[ \because {{\sin }^{2}}A+{{\cos }^{2}}A=1 \right] \\ &=1-{{\left( \frac{1}{\sqrt{5}} \right)}^{2}} \\ &=1-\frac{1}{5} \\ &=\frac{4}{5} \\ &\sin \frac{x}{2}=\pm \sqrt{\frac{4}{5}} \\ &\sin \frac{x}{2}=\pm \frac{2}{\sqrt{5}} \end{align}\]

Since, \(\sin \frac{x}{2}\) lies in quadrant I and positive, \(\sin \frac{x}{2} = \frac{2}{{\sqrt 5 }}\)

Now,

\[\begin{align}\tan \frac{x}{2} &= \frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}}\\&= \frac{{\left( {\frac{2}{{\sqrt 5 }}} \right)}}{{\left( {\frac{1}{{\sqrt 5 }}} \right)}}\\&= \frac{2}{{\sqrt 5 }} \times \frac{{\sqrt 5 }}{1}\\&= 2\end{align}\]

Therefore, \(\sin \frac{x}{2} = \frac{2}{{\sqrt 5 }}\), \(\cos \frac{x}{2} = \frac{1}{{\sqrt 5 }}\) and \(\tan \frac{x}{2} = 2\).

Chapter 3 Ex.3.ME Question 9

Prove that:

\(\cos x = - \frac{1}{3},\;x\) in quadrant III.

Solution

Since \(x\) lies in quadrant III

\[\pi < x < \frac{{3\pi }}{2}\]

Therefore,

\[\frac{\pi }{2} < \frac{x}{2} < \frac{{3\pi }}{4}\]

Hence, \(\cos \frac{x}{2}\) and \(\tan \frac{x}{2}\)are negative while \(\sin \frac{x}{2}\) is positive as all lie in quadrant II.

It is given that \(\cos x = - \frac{1}{3}\)

\[\begin{align}\cos 2\left( {\frac{x}{2}} \right) &= - \frac{1}{3}\\2{\cos ^2}\frac{x}{2} - 1 &= - \frac{1}{3}\\2{\cos ^2}\frac{x}{2} &= 1 - \frac{1}{3}\\{\cos ^2}\frac{x}{2} &= \frac{2}{3} \times \frac{1}{2}\\{\cos ^2}\frac{x}{2} &= \frac{1}{3}\\\cos \frac{x}{2} &= \pm \sqrt {\frac{1}{3}} \end{align}\]

Since, \(\cos \frac{x}{2}\) is negative

So,

\[\begin{align}\cos \frac{x}{2} &= - \frac{1}{{\sqrt 3 }}\\&= - \frac{1}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }}\\&= - \frac{{\sqrt 3 }}{3}\end{align}\]

Now,

\[\begin{align} {{\sin }^{2}}\frac{x}{2}&=1-{{\cos }^{2}}\frac{x}{2} \qquad \quad \left[ \because {{\sin }^{2}}A+{{\cos }^{2}}A=1 \right] \\ &=1-{{\left( -\frac{\sqrt{3}}{3} \right)}^{2}} \\ &=1-\frac{1}{3} \\ &=\frac{2}{3} \\ &\sin \frac{x}{2}=\pm \sqrt{\frac{2}{3}} \end{align}\]

Since, \(\sin \frac{x}{2}\) positive,

So,

\[\begin{align}\sin \frac{x}{2}&= \sqrt {\frac{2}{3}} \\&= \frac{{\sqrt 2 }}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }}\\&= \frac{{\sqrt 6 }}{3}\end{align}\]

Now,

\[\begin{align}\tan \frac{x}{2}&= \frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}}\\&= \frac{{\frac{{\sqrt 6 }}{3}}}{{\left( { - \frac{{\sqrt 3 }}{3}} \right)}}\\&= \frac{{\sqrt 6 }}{3} \times \left( { - \frac{3}{{\sqrt 3 }}} \right)\\&= - \sqrt 2 \end{align}\]

Therefore, \(\sin \frac{x}{2} = \frac{{\sqrt 6 }}{3}\), \(\cos \frac{x}{2} = - \frac{{\sqrt 3 }}{3}\) and \(\tan \frac{x}{2} = - \sqrt 2 \).

Chapter 3 Ex.3.ME Question 10

Prove that:

\(\sin x = \frac{1}{4},\;x\) in quadrant II.

Solution

Since \(x\) lies in quadrant II

\[\frac{\pi }{2} < x < \pi \]

Therefore,

\[\frac{\pi }{4} < \frac{x}{2} < \frac{\pi }{2}\]

Hence, \(\sin \frac{x}{2}\),  \(\cos \frac{x}{2}\) and \(\tan \frac{x}{2}\) lie in quadrant I and all are positive.

It is given that \(\sin x = \frac{1}{4}\)

Therefore,

\[\begin{align} {{\cos }^{2}}x&=1-{{\sin }^{2}}x \qquad \quad \left[ \because {{\sin }^{2}}A+{{\cos }^{2}}A=1 \right] \\ &=1-{{\left( \frac{1}{4} \right)}^{2}} \\ &=1-\frac{1}{16} \\ &=\frac{15}{16} \\ \cos x&=\pm \sqrt{\frac{15}{16}} \end{align}\]

Since, \(\cos x\) lies in quadrant II and negative

So,

\[\begin{align}\cos x& = - \sqrt {\frac{{15}}{{16}}} \\\cos 2\left( {\frac{x}{2}} \right) &= - \frac{{\sqrt {15} }}{4}\\2{\cos ^2}\frac{x}{2} - 1 &= - \frac{{\sqrt {15} }}{4}\\2{\cos ^2}\frac{x}{2} &= 1 - \frac{{\sqrt {15} }}{4}\\2{\cos ^2}\frac{x}{2} &= \frac{{4 - \sqrt {15} }}{4}\\{\cos ^2}\frac{x}{2} &= \frac{{4 - \sqrt {15} }}{8}\\\cos \frac{x}{2} &= \pm \sqrt {\frac{{4 - \sqrt {15} }}{8}} \end{align}\]

Since \(\cos \frac{x}{2}\), lies in quadrant I and is positive.

So,

\[\begin{align}\cos \frac{x}{2} &= \sqrt {\frac{{4 - \sqrt {15} }}{8}} \\&= \sqrt {\frac{{4 - \sqrt {15} }}{8} \times \frac{2}{2}} \\&= \frac{{\sqrt {8 - 2\sqrt {15} } }}{4}\end{align}\]

Now,

\[\begin{align} {{\sin }^{2}}\frac{x}{2}&=1-{{\cos }^{2}}\frac{x}{2} \qquad \quad \left[ \because {{\sin }^{2}}A+{{\cos }^{2}}A=1 \right] \\ &=1-{{\left( \frac{\sqrt{8-2\sqrt{15}}}{4} \right)}^{2}} \\ &=1-\frac{8-2\sqrt{15}}{16} \\ &=\frac{16-8+2\sqrt{15}}{16} \\ &=\frac{8+2\sqrt{15}}{16} \\ \sin \frac{x}{2}&=\pm \sqrt{\frac{8+2\sqrt{15}}{16}} \end{align}\]

Since, \(\sin \frac{x}{2}\)positive,

So,

\(\sin \frac{x}{2} = \frac{{\sqrt {8 + 2\sqrt {15} } }}{4}\)

Now,

\[\begin{align}\tan \frac{x}{2} &= \frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}}\\&= \frac{{\left( {\frac{{\sqrt {8 + 2\sqrt {15} } }}{4}} \right)}}{{\left( {\frac{{\sqrt {8 - 2\sqrt {15} } }}{4}} \right)}}\\&= \left( {\frac{{\sqrt {8 + 2\sqrt {15} } }}{4}} \right) \times \left( {\frac{4}{{\sqrt {8 - 2\sqrt {15} } }}} \right)\\&= \sqrt {\frac{{2\left( {4 + \sqrt {15} } \right)}}{{2\left( {4 - \sqrt {15} } \right)}}} \\A&= \sqrt {\frac{{4 + \sqrt {15} }}{{4 - \sqrt {15} }} \times \frac{{4 + \sqrt {15} }}{{4 + \sqrt {15} }}} \\&= \sqrt {\frac{{{{\left( {4 + \sqrt {15} } \right)}^2}}}{{16 - 15}}} \\&= 4 + \sqrt {15} \end{align}\]

Therefore, \(\sin \frac{x}{2} = \frac{{\sqrt {8 + 2\sqrt {15} } }}{4}\), \(\cos \frac{x}{2}=\frac{\sqrt{8-2\sqrt{15}}}{4}\) and \(\tan \frac{x}{2} = 4 + \sqrt {15} \).

  
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