QuestionReportQuestion: Period of \(\sin^{2}x\)is...Period of sin2x\sin^{2}xsin2xisAπ\piπB2π2\pi2πCπ2\frac{\pi}{2}2πDNone of theseAnswerπ\piπExplanationSolutionsin2x=1−cos2x2\sin^{2}x = \frac{1 - \cos 2x}{2}sin2x=21−cos2x ⇒\Rightarrow⇒ Period =2π2=π.= \frac{2\pi}{2} = \pi.=22π=π.