If then find .
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