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UPSC NDA 2 2016 Mathematics Solved Paper

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Question : 44 of 120

Marks: +1, -0

What is the value of

\left( {1 + \cos \frac{{\rm{\pi }}}{8}} \right)\left( {1 + \cos \frac{{3{\rm{\pi }}}}{8}} \right)\left( {1 + \cos \frac{{5{\rm{\pi }}}}{8}} \right)\left( {1 + \cos \frac{{7{\rm{\pi }}}}{8}} \right)?

$\frac{1}{2}$
$\frac{1}{2} + \frac{1}{{2\sqrt 2 }}$
$\frac{1}{2} - \frac{1}{{2\sqrt 2 }}$
$\frac{1}{8}$

Solution:

Concept:

Use trigonometric identities:

\cos(\pi - \theta) = -\cos\theta

,

1 - \cos^2\theta = \sin^2\theta

, and

1 - \cos 2\theta = 2\sin^2\theta

.

Explanation:

Step 1: Rewrite

\cos\frac{5\pi}{8}

and

\cos\frac{7\pi}{8}

using

\cos(\pi - \theta) = -\cos\theta

.

\cos\frac{5\pi}{8} = \cos\left(\pi - \frac{3\pi}{8}\right) = -\cos\frac{3\pi}{8}

.

\cos\frac{7\pi}{8} = \cos\left(\pi - \frac{\pi}{8}\right) = -\cos\frac{\pi}{8}

.

Thus the product becomes

(1+\cos\frac{\pi}{8})(1+\cos\frac{3\pi}{8})(1-\cos\frac{\pi}{8})(1-\cos\frac{3\pi}{8})

.

Step 2: Pair each

1+\cos

with its

1-\cos

counterpart.

(1+\cos\frac{\pi}{8})(1-\cos\frac{\pi}{8}) = 1 - \cos^2\frac{\pi}{8} = \sin^2\frac{\pi}{8}

.

Similarly

(1+\cos\frac{3\pi}{8})(1-\cos\frac{3\pi}{8}) = \sin^2\frac{3\pi}{8}

.

So the product equals

\sin^2\frac{\pi}{8} \cdot \sin^2\frac{3\pi}{8}

.

Step 3: Express each

\sin^2

using

1-\cos 2\theta = 2\sin^2\theta

.

\sin^2\frac{\pi}{8} = \frac{1-\cos\frac{\pi}{4}}{2}

,

\sin^2\frac{3\pi}{8} = \frac{1-\cos\frac{3\pi}{4}}{2}

.

Step 4: Substitute known values:

\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}

,

\cos\frac{3\pi}{4} = -\frac{1}{\sqrt{2}}

.

Therefore product

= \frac{1}{4}\left(1-\frac{1}{\sqrt{2}}\right)\left(1+\frac{1}{\sqrt{2}}\right) = \frac{1}{4}\left(1-\frac{1}{2}\right) = \frac{1}{8}

.

Answer:

\frac{1}{8}

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