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Question

Mathematics Question on Inverse Trigonometric Functions

The period of sin2θ\sin^2 \theta is

A

π2\pi^2

B

π\pi

C

2π2 \pi

D

π/2\pi / 2

Answer

π\pi

Explanation

Solution

sin2θ=1cos2θ2\sin^{2} \theta = \frac{1- \cos2\theta}{2}; Period =2π2=π= \frac{2\pi}{2} = \pi