JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

\sin ^2\left(\frac{\pi}{8}+\frac{A}{2}\right)-\sin ^2\left(\frac{\pi}{8}-\frac{A}{2}\right)

is equal to:

View answer and explanation

Correct option: A

(1 / \sqrt{2}) \sin A

Option A (

\frac{1}{\sqrt{2}} \sin A

) is correct because using the identity

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

, the expression simplifies to

\frac{1}{\sqrt{2}} \sin A

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