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UPSC NDA Math Model Paper 9

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Question : 106 of 120

Marks: +1, -0

If π / 2 < x < π , then write the value of

\sqrt{2 + \sqrt{2 + 2\cos 2x}}

sin 2x
1 – sin 2x
1 - sin (x / 2)
2 sin (x / 2)

Solution:

Given: π / 2 < x < π

i.e x lies in the 2

^{\mathrm{nd}}

quadrant where all trigonometric ratios are negative except sin and cosec.

\Rightarrow\sqrt{2 + \sqrt{2 + 2\cos 2x}} = \sqrt{2 + \sqrt{2(1 + \cos 2x)}}

As we know that,

\cos 2x = 2 \cos^2 x - 1

or

1 + \cos 2x = 2 \cos^2 x

\Rightarrow\sqrt{2 + \sqrt{2 + 2\cos 2x}} = \sqrt{2 + \sqrt{2 \times 2 \cos^2 x}}

\Rightarrow\sqrt{2 + \sqrt{2 + 2\cos 2x}} = \sqrt{2 \pm 2\cos x}

As x lies in 2

^{\mathrm{nd}}

quadrant

\Rightarrow\sqrt{2 + \sqrt{2 + 2\cos 2x}} = \sqrt{2 - 2\cos x}

\Rightarrow\sqrt{2 + \sqrt{2 + 2\cos 2x}} = \sqrt{2\;(1 - \cos x)}

As we know that,

\cos x = 1 - 2 \sin^2\left(\frac{x}{2}\right)

or

2 \sin^2\left(\frac{x}{2}\right) = 1 - \cos x

\Rightarrow\sqrt{2 + \sqrt{2 + 2\cos 2x}}

= \sqrt{2 \times 2 \sin^2\left(\frac{x}{2}\right)}

As x lies in the 2

^{\mathrm{nd}}

quadrant

\Rightarrow\sqrt{2 + \sqrt{2 + 2\cos 2x}} = 2\sin\left(\frac{x}{2}\right)

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