Question

show that....................​

Answer 1

Put x = cosY.

Problem solved :)

$sin^{-1}(2cosYsinY) = 2cos^{-1}cosY$ (since $\sqrt{1 - cos^2x} = sinx$)

$sin^{-1}(sin2Y) = 2Y$ (since 2sinxcosx = sin2x)

The final statement is true, hence proved/shown.

Tiny missed details:

1. Putting x = sinY won't solve the problem.

2. The domain of x here is important - notice that in context that we resolved $sin^{-1}sin2Y = 2Y$, which is only true for a certain domain of Y.