can any one help me verify these problems

tanx(1+cos2x)=sin2x

and

sin3x=3sinx-4sinx

For the first, recall that cos(2x) = 2cos^2(x) -1. Sub that in, then use the fact that tan(x) = sin(x)/cos(x), and finally the identity sin(2x) = 2sin(x)cos(x) to simplify.

The second one is mis-typed, please try again. If you substitute x = pi/6, you get 3sin(pi/6)-4sin(pi/6) = 3(1/2) - 4(1/2) = -1/2, which is not the same as sin(3pi/6)= sin(pi/2) = 1.

The second case:

Correct identity is: Sin 3x = 3 sin x - 4 sin^3 x

Start with sin (2x + x), use sin (A + B) = sin A cos B + cos A sin B

Then use cos 2x = 1 - 2 sin^2 x

sin 2x = 2 sin x cos x

etc.