Class 11 Chapter 3 Miscellaneous Exercise trigonometry

Chapter 3:Miscellaneous Exercise

Prove that(Class 11 Chapter 3 Miscellaneous Exercise trigonometry )

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

Solution: Given, 2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+\cos\frac{3\pi}{13}+\cos \frac{5\pi}{13}

Using formula \cos x+\cos y = 2\cos\frac{x+y}{2}\cos \frac{x-y}{2}

= 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})

=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})

= 2\cos \frac{\pi}{13}\left[2\cos(\frac{\frac{9\pi}{13}+\frac{4\pi}{13}}{2})\cos(\frac{\frac{9\pi}{13}-\frac{4\pi}{13}}{2})\right]

= 2\cos \frac{\pi}{13}\left[2\cos \frac{\pi}{2}.\cos \frac{5\pi}{26}\right]

=4\cos\frac{\pi}{13}\times 0\times \cos \frac{5\pi}{26}

=0=RHS

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

Solution: Taking LHS = (\sin 3x + \sin x)\sin x + (\cos 3x - \cos x) \cos x

Using formula: \sin x+\sin y=2\sin\frac{x+y}{2}\cos\frac{x-y}{2} and

\cos x-\cos y=-2\sin\frac{x+y}{2}\sin\frac{x-y}{2}

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

=2\sin 2x \cos x\sin x-2\sin 2x \sin x \cos x

=0=RHS

Question 3: (\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4 \cos^2\frac{x+y}{2}

Solution: Taking LHS= (\cos x + \cos y)^2 + (\sin x - \sin y)^2

Using formula: \sin x-\sin y=2\cos\frac{x+y}{2}\sin\frac{x-y}{2} and

\cos x+\cos y=2\cos \frac{x+y}{2}\cos\frac{x-y}{2}

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

= 4\cos^2 \frac{x+y}{2}\cos^2\frac{x-y}{2}+4\cos^2 \frac{x+y}{2}\sin^2\frac{x-y}{2}

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

=4\cos^2\frac{x+y}{2}=RHS

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

Solution: Taking LHS= (\cos x - \cos y)^2 + (\sin x - \sin y)^2

Using formulae: \sin x-\sin y=2\cos\frac{x+y}{2}\sin\frac{x-y}{2} and

\cos x-\cos y=-2\sin \frac{x+y}{2}\sin\frac{x-y}{2}

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

= 4\sin^2 \frac{x+y}{2}\sin^2\frac{x-y}{2}+4\cos^2 \frac{x+y}{2}\sin^2\frac{x-y}{2}

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

=4\cos^2\frac{x-y}{2}=RHS

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

Solution: Taking LHS=\sin x + \sin 3x + \sin 5x + \sin 7x

=\sin 7x+\sin x+\sin 5x+\sin x

Using formulae: \sin x+\sin y=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}

=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

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

Again using formula: \cos x+\cos y=2\cos \frac{x+y}{2}\cos\frac{x-y}{2}

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

=4\sin 4x\cos 2x\cos x=RHS

Question 6:  \frac{(\sin 7x+\sin 5x)+(\sin 9x+\sin 3x)}{(\cos 7x+\cos5x)+(\cos 9x+\cos 3x)}=\tan 6x

Solution: Taking LHS= \frac{(\sin 7x+\sin 5x)+(\sin 9x+\sin 3x)}{(\cos 7x+\cos5x)+(\cos 9x+\cos 3x)}

Using formula: \sin x+\sin y=2\sin\frac{x+y}{2}\cos\frac{x-y}{2} and

\cos x+\cos y=2\cos \frac{x+y}{2}\cos\frac{x-y}{2}

=\frac{(2\sin\frac{7x+5x}{2}\cos\frac{7x-5x}{2})+(2\sin\frac{9x+3x}{2}\cos\frac{9x-3x}{2})}{(2\cos \frac{7x+5x}{2}\cos\frac{7x-5x}{2})+(2\cos \frac{9x+3x}{2}\cos\frac{9x-3x}{2})}

=\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(\cos x+\cos 3x)}{2\cos 6x(\cos x+\cos 3x)}

=\tan 6x=RHS

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

Solution: Taking LHS = \sin 3x + \sin 2x - \sin x

= \sin 3x - \sin x +\sin 2x

Using formula: \sin x-\sin y=2\cos\frac{x+y}{2}\sin\frac{x-y}{2} and

\sin 2x = 2\sin x\cos x

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

=2\cos 2x\sin x+2\sin x\cos x

=2\sin x(\cos 2x+\cos x)

Again using formula:\cos x+\cos y=2\cos \frac{x+y}{2}\cos\frac{x-y}{2}

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

=4\sin x \cos\frac{x}{2}\cos \frac{3x}{2}=RHS

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

Question 8:  if \tan x=-\dfrac{\text{4}}{\text{3}} , \text{x} lies in 2nd quadrant.

Solution:  Here, x is in 2nd quadrant.

Therefore , \frac{\pi}{2}<x<\pi

\Rightarrow \frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}

hence \frac{x}{2} lies in 1st quadrant.

Therefore, \text{sin}\dfrac{\text{x}}{\text{2}}\text{,cos}\dfrac{\text{x}}{\text{2}}\,\, and \text{tan}\dfrac{\text{x}}{\text{2}} are positive.

Given that \text{tan x=-}\dfrac{\text{4}}{\text{3}}

We know that \sec^2x=1+\tan^2x

\Rightarrow \sec^2x =1+(-\frac{4}{3})^2

\Rightarrow \sec^2 x =1+\frac{16}{9}=\frac{25}{9}

\Rightarrow \sec x= -\frac{5}{3} x lies in second quadrant

Since, \cos x = \frac{1}{\sec x} =-\frac{3}{5}

From the formula: \cos x = 2\cos^2\frac{x}{2}-1

-\frac{3}{5} = 2\cos^2\frac{x}{2}-1

\Rightarrow 1-\frac{3}{5}=2\cos^2\frac{x}{2}

\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}} =\frac{\sqrt{5}}{5},\frac{x}{2} lies in Ist quadrant

From formula: \sin^2\frac{x}{2}=1-\cos^2\frac{x}{2}

\Rightarrow \sin^2\frac{x}{2}=1- \frac{1}{5}

\Rightarrow \sin^2\frac{x}{2}=\frac{4}{5}

\Rightarrow \sin\frac{x}{2}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5},\frac{x}{2} lies in Ist quadrant

Since,\tan \frac{x}{2} = \frac{ \sin\frac{x}{2}}{ \cos\frac{x}{2}}

\Rightarrow \tan \frac{x}{2}= \frac{\frac{2}{5}}{\frac{1}{\sqrt{5}}}

\Rightarrow \tan \frac{x}{2}= 2,\frac{x}{2} lies in Ist quadrant

Hence the 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},2

Question 9: \cos x = −\frac{1}{3}, x in quadrant III

Solution: –Here, x is in IIIrd quadrant.

Therefore , \pi<x<\frac{\pi}{2}

\Rightarrow \frac{\pi}{2}<\frac{x}{2}<\frac{3\pi}{4}

hence \frac{x}{2} lies in IInd quadrant.

Since, \cos x = -\frac{1}{3}

From the formula: \cos x = 2\cos^2\frac{x}{2}-1

-\frac{1}{3} = 2\cos^2\frac{x}{2}-1

\Rightarrow 1-\frac{1}{3}=2\cos^2\frac{x}{2}

\Rightarrow 2\cos^2\frac{x}{2}=\frac{2}{3}

\Rightarrow \cos^2\frac{x}{2}=\frac{1}{3}

\Rightarrow \cos\frac{x}{2}=-\frac{1}{\sqrt{3}} ,\frac{x}{2} lies in IInd quadrant

From formula: \sin^2\frac{x}{2}=1-\cos^2\frac{x}{2}

\Rightarrow \sin^2\frac{x}{2}=1- \frac{1}{3}

\Rightarrow \sin^2\frac{x}{2}=\frac{2}{3}

\Rightarrow \sin\frac{x}{2}=\frac{\sqrt{2}}{\sqrt{3}},\frac{x}{2} lies in IInd quadrant

Since,\tan \frac{x}{2} = \frac{ \sin\frac{x}{2}}{ \cos\frac{x}{2}}

\Rightarrow \tan \frac{x}{2}= \frac{\frac{\sqrt{2}}{3}}{-\frac{1}{\sqrt{3}}}

\Rightarrow \tan \frac{x}{2}= -\sqrt{2},\frac{x}{2} lies in Ist quadrant

Hence the value of \sin\frac{x}{2}, \cos\frac{x}{2} and \tan\frac{x}{2} are \frac{\sqrt{2}}{\sqrt{3}},\frac{1}{\sqrt{3}},-\sqrt{2}

Question 10: \sin x =\frac{1}{4}, x in quadrant II

Solution: Here, x is in IInd quadrant.

Therefore , \frac{\pi}{2}<x<\pi

\Rightarrow \frac{\pi}{4}<\frac{x}{2}<\frac{3\pi}{2}

hence \frac{x}{2} lies in Ist quadrant.

Formula : \cos^2x=1-\sin^2x

\Rightarrow \sin x = 1-(\frac{1}{4})^2=1-\frac{1}{16}

\Rightarrow \cos^2 x = \frac{15}{16}

\Rightarrow \cos x = -\frac{\sqrt{15}}{4} x lies in IInd quadrant

From the formula: \cos x = 2\cos^2\frac{x}{2}-1

-\frac{\sqrt{15}}{4} = 2\cos^2\frac{x}{2}-1

\Rightarrow 1-\frac{\sqrt{15}}{4}=2\cos^2\frac{x}{2}

\Rightarrow 2\cos^2\frac{x}{2}=\frac{4-\sqrt{15}}{4}

\Rightarrow \cos^2\frac{x}{2}=\frac{4-\sqrt{15}}{8}

\Rightarrow \cos\frac{x}{2}=\sqrt{\frac{4-\sqrt{15}}{8}} ,

\Rightarrow \cos \frac{x}{2}=\sqrt{(\frac{4-\sqrt{15}}{8})\frac{2}{2}}

\Rightarrow \cos \frac{x}{2}=\sqrt{\frac{8-2\sqrt{15}}{16}}=\frac{\sqrt{8-2\sqrt{15}}}{4}, \frac{x}{2} lies in Ist quadrant

From formula: \sin^2\frac{x}{2}=1-\cos^2\frac{x}{2}

\Rightarrow \sin^2\frac{x}{2}=1- (\frac{\sqrt{8-2\sqrt{15}}}{4})^2

\Rightarrow \sin^2\frac{x}{2}=1-\frac{8-2\sqrt{15}}{16}

\Rightarrow \sin^2\frac{x}{2} =\frac{16-8+2\sqrt{15}}{16}=\frac{8+2\sqrt{15}}{16}

\Rightarrow \sin\frac{x}{2} = \frac{\sqrt{8+2\sqrt{15}}}{4},\frac{x}{2} lies in IInd quadrant

Since,\tan \frac{x}{2} = \frac{ \sin\frac{x}{2}}{ \cos\frac{x}{2}}

\Rightarrow \tan \frac{x}{2}= \frac{\frac{\sqrt{8+2\sqrt{15}}}{4}}{\frac{\sqrt{8-2\sqrt{15}}}{4}}

\Rightarrow \tan \frac{x}{2}= \frac{\sqrt{8+2\sqrt{15}}}{\sqrt{8-2\sqrt{15}}}.\frac{\sqrt{8+2\sqrt{15}}}{\sqrt{8+2\sqrt{15}}}

= \frac{8+2\sqrt{15}}{\sqrt{64-60}}

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

Hence the 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},4+\sqrt{15}

 

https://gmath.in/ex-3-4-trigonomety-ncert-maths-solution-class-11/

 

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