trigonometry identity?

Question

How do I get (sin x)^3 = sinx = sinx*(cos x)^2

and square root of 4-4*(sin x)^2 = 2cos x

Answer 1

Use the identity (sinx)^2 + (cosx)^2 = 1.

(sin x)^3 = sinx(1 - (cosx)^2) =
sinx - sinx(cosx)^2

sqrt(4-4*(sin x)^2) = 2*sqrt(1 - (sinx)^2) =
2*sqrt((cosx)^2) = 2cosx

Answer 2

0=x

Answer 3

kIt appears that there is some mistake in your first problem
4-4*(sin x)^2
=4(sin^2 x +cos^2 x)-4sin^2 x
=4 sin^2x+4 cos^ x-4sin^2 x
=4cos^2x
Therfore square root of 4-4*(sin x)^2
=square root of 4cos^2 x
=2cos x (Proved)