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MHTCET 16 May 2024 Shift 2 Paper

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Question : 125 of 150

Marks: +1, -0

The value of

\cos 20^{\circ}+2 \sin ^{2} 55^{\circ}-\sqrt{2} \sin 65^{\circ}

is

0
1
-1
$\frac{1}{2}$

Solution:

Concept:

Use the double-angle identity

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

and the sum-to-product formula to simplify the expression.

Explanation:

Start with the given expression:

\cos 20^{\circ} + 2\sin^2 55^{\circ} - \sqrt{2}\sin 65^{\circ}

Replace

2\sin^2 55^{\circ}

using

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

:

= \cos 20^{\circ} + 1 - \cos 110^{\circ} - \sqrt{2}\sin 65^{\circ}

Now

\cos 20^{\circ} - \cos 110^{\circ} = 2\sin 65^{\circ}\sin 45^{\circ}

Since

\sin 45^{\circ} = \frac{1}{\sqrt{2}}

, this becomes:

= 2\sin 65^{\circ}\left(\frac{1}{\sqrt{2}}\right) - \sqrt{2}\sin 65^{\circ} + 1

= \sqrt{2}\sin 65^{\circ} - \sqrt{2}\sin 65^{\circ} + 1

= 1

Answer:

1

(Option B)

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