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MHTCET 25 Apr 2025 Shift 2 Paper

Question : 148 of 150

Marks: +1, -0

\cos ^{4} \frac{\pi}{8}+\cos ^{4} \frac{3 \pi}{8}+\cos ^{4} \frac{5 \pi}{8}+\cos ^{4} \frac{7 \pi}{8}=

Solution:

Concept:

Pair angles using

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

, then apply the double-angle identity

2\cos^2\theta = 1 + \cos 2\theta

Explanation:

Observe the complementary angle relations:

\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}

Since fourth powers eliminate the negative sign, the given sum reduces to:

2\left[\cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8}\right]

Now use

2\cos^2\theta = 1 + \cos 2\theta

to rewrite each term:

2\left[\cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8}\right] = \frac{1}{2}\left[\left(2\cos^2 \frac{\pi}{8}\right)^2 + \left(2\cos^2 \frac{3\pi}{8}\right)^2\right]

= \frac{1}{2}\left[\left(1 + \cos\frac{\pi}{4}\right)^2 + \left(1 + \cos\frac{3\pi}{4}\right)^2\right]

Substitute

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

and

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

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

Expanding both squares and cancelling the cross terms

\frac{2}{\sqrt{2}}

= \frac{1}{2}\left[1 + \frac{2}{\sqrt{2}} + \frac{1}{2} + 1 - \frac{2}{\sqrt{2}} + \frac{1}{2}\right]

= \frac{1}{2}\left[1 + 1 + \frac{1}{2} + \frac{1}{2}\right] = \frac{1}{2}(3) = \frac{3}{2}

Answer:

\frac{3}{2}

Therefore, the correct option is B.

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