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Using sin²θ + cos²θ = 1

Using 2sinθcosθ = sin2θ

y = sin^–1(sin2θ)

Considering the limits,

For

Now, y = sin^–1(sin2θ)

y = 2θ

y = 2cos^–1x

Differentiating w.r.t x, we get

For

Now, y = sin^–1(sin2θ)

y = –2θ

y = –2cos^–1x

Differentiating w.r.t x, we get