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MHTCET 10 May 2024 Shift 1 Paper

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Question : 126 of 150

Marks: +1, -0

The value of

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

is

$\frac{1}{8}$
$\frac{-1}{8}$
$\frac{1}{16}$
$\frac{-1}{16}$

Solution:

Concept:

Use the identity

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

to pair terms and simplify the product.

Explanation:

Pair

\frac{7\pi}{8}

with

\frac{\pi}{8}

and

\frac{5\pi}{8}

with

\frac{3\pi}{8}

.

Since

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

and

\cos\frac{5\pi}{8} = -\cos\frac{3\pi}{8}

, the expression becomes

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

.

Using

\left(1+\cos A\right)\left(1-\cos A\right) = 1 - \cos^2 A = \sin^2 A

, we get

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

.

Rewrite this as

\frac{1}{4}\left(2\sin\frac{\pi}{8}\sin\frac{3\pi}{8}\right)^2

.

Apply the product-to-sum formula

2\sin A \sin B = \cos(A-B) - \cos(A+B)

:

\frac{1}{4}\left(\cos\frac{\pi}{4} - \cos\frac{\pi}{2}\right)^2

.

Since

\cos\frac{\pi}{4} = \frac{1}{\sqrt2}

and

\cos\frac{\pi}{2} = 0

, the value is

\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}

.

Answer:

\frac{1}{8}

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