QuestionReportMathematics Question on Inverse Trigonometric FunctionsThe period of sin2θ\sin^2 \thetasin2θ isAπ2\pi^2π2Bπ\piπC2π2 \pi 2πDπ/2\pi / 2 π/2Answerπ\piπExplanationSolutionsin2θ=1−cos2θ2\sin^{2} \theta = \frac{1- \cos2\theta}{2}sin2θ=21−cos2θ; Period =2π2=π= \frac{2\pi}{2} = \pi=22π=π