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UPSC NDA I 2022 Mathematics Solved Paper

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Question : 32 of 120

Marks: +1, -0

What is the value of

\cos^4\frac{7\pi}{8}+\cos^4\frac{5\pi}{8}?

$\frac{3}{2}$
$\frac{3}{4}$
$\frac{3}{8}$
$\frac{3}{16}$

Solution:

Concept:

Use the half-angle identity

\cos^2 \theta = \frac{1 + \cos 2\theta}{2}

and simplify trigonometric values.

Explanation:

Let

y = \cos^4\frac{7\pi}{8} + \cos^4\frac{5\pi}{8}

.

Write

y = \left(\cos^2\frac{7\pi}{8}\right)^2 + \left(\cos^2\frac{5\pi}{8}\right)^2

.

Apply

\cos^2\theta = \frac{1+\cos2\theta}{2}

to each term:

y = \left(\frac{1 + \cos\frac{14\pi}{8}}{2}\right)^2 + \left(\frac{1 + \cos\frac{10\pi}{8}}{2}\right)^2

.

Simplify angles:

\frac{14\pi}{8} = \frac{7\pi}{4}

and

\frac{10\pi}{8} = \frac{5\pi}{4}

.

So

y = \left(\frac{1 + \cos\frac{7\pi}{4}}{2}\right)^2 + \left(\frac{1 + \cos\frac{5\pi}{4}}{2}\right)^2

.

Now evaluate:

\cos\frac{7\pi}{4} = \cos\left(2\pi - \frac{\pi}{4}\right) = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}

.

\cos\frac{5\pi}{4} = \cos\left(\pi + \frac{\pi}{4}\right) = -\cos\frac{\pi}{4} = -\frac{1}{\sqrt{2}}

.

Substitute:

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

.

Simplify each fraction:

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

and

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

.

Square:

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

.

Expand:

(\sqrt{2}+1)^2 = 2 + 1 + 2\sqrt{2} = 3 + 2\sqrt{2}

and

(\sqrt{2}-1)^2 = 2 + 1 - 2\sqrt{2} = 3 - 2\sqrt{2}

.

Sum:

(3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6

.

Thus

y = \frac{6}{8} = \frac{3}{4}

.

Answer:

\frac{3}{4}

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