The angle between the line y= 2 x and y= 2 x is

Correct Answer is : tan⁡−122^{-1} 2 {2}tan−122

Correct Answer is : tan⁡−122^{-1} 2 {2}tan−122

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TS EAMCET 5 May 2018 Shift 2 Solved Paper

Section: Mathematics

Question : 69 of 160

Marks: +1, -0

y=\sin 2 x

y=\cos 2 x

Solution:

Consider the curves,

y=\sin 2 x

And,

y=\cos 2 x

Therefore,

\sin 2 x =\cos 2 x

\tan 2 x =1

2 x = \frac{\pi}{4}

x = \frac{\pi}{8}

Therefore,

y=\sin 2 x

= \sin \frac{\pi}{4}

= \frac{1}{\sqrt{2}}

And,

y =\cos 2 x

=\cos \frac{\pi}{4}

= \frac{1}{\sqrt{2}}

Therefore,

m_{1}= \left( \frac{d y}{d x} \right) \left( \frac{\pi}{8} \cdot \frac{1}{\sqrt{2}} \right)

=2 \cos 2 x

=2 \cos \frac{\pi}{4}

=\sqrt{2}

And,

m_{2}= \left( \frac{d y}{d x} \right) \left( \frac{\pi}{8}, \frac{1}{\sqrt{2}} \right)

=-2 \sin 2 x

=-2 \sin \frac{\pi}{4}

=-\sqrt{2}

The angle between the curve is given by,

\tan^{-1} \left( \frac{m_{2}-m_{1}}{1+m_{1} m_{2}} \right)

=\tan^{-1} \left( \frac{-\sqrt{2}-\sqrt{2}}{1-2} \right)

=\tan^{-1} \left( \frac{2\sqrt{2}}{1} \right)

=\tan^{-1}(2\sqrt{2})

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