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MHTCET 15 Apr 2026 Shift 2 Paper

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Question : 106 of 150

Marks: +1, -0

If

33 \theta=\pi

, then the value of

\cos \theta \cos 2 \theta \cos 4 \theta \cos 8 \theta \cos 16 \theta

is

$\frac{1}{16}$
$\frac{1}{32}$
$\frac{1}{64}$
$\frac{-1}{32}$

Solution:

Concept:

The product of cosines can be evaluated by using the multiple-angle identity

\sin 2x = 2\sin x \cos x

repeatedly.

Explanation:

Given

33\theta = \pi

, we have:

\theta = \frac{\pi}{33}

We need the value of

\cos \theta \cos 2\theta \cos 4\theta \cos 8\theta \cos 16\theta

.

Using the identity

\sin 2x = 2\sin x \cos x

repeatedly, we get:

\sin 32\theta = 32 \sin \theta \cos \theta \cos 2\theta \cos 4\theta \cos 8\theta \cos 16\theta

Therefore,

\cos \theta \cos 2\theta \cos 4\theta \cos 8\theta \cos 16\theta = \frac{\sin 32\theta}{32 \sin \theta}

Since

33\theta = \pi

, we can write

32\theta = \pi - \theta

.

Thus,

\sin 32\theta = \sin(\pi - \theta) = \sin \theta

Substituting this value:

\cos \theta \cos 2\theta \cos 4\theta \cos 8\theta \cos 16\theta = \frac{\sin \theta}{32 \sin \theta} = \frac{1}{32}

Answer:

The value is

\frac{1}{32}

, which corresponds to option B.

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