Prove this math problem

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sin3x = 3sinx - 4sin^3x

May 6th, 2015

Recall: sin (A + B) = sin A cos B + cos A sin B 
sin 2x = 2 sin x cos x 
cos 2x = cos^2 x - sin^2 x = 2 cos^2 x - 1 = 1 - 2 sin^2 x 
sin^2 x + cos^2 x = 1 

sin 3x 
= sin (2x + x) 
= sin 2x. cos x + cos 2x. sin x 
= 2 sin x cos x. cos x + (1 - 2 sin^2 x) . sin x 
= 2 sin x cos x. cos x + sin x - 2 sin^3 x 
= 2 sin x cos^2 x + sin x - 2 sin^3 x 
= 2 sin x (1 - sin^2 x) + sin x - 2 sin^3 x 
= 2 sin x - 2 sin^3 x + sin x - 2 sin^3 x 
= 3 sin x - 4 sin^3 x. 

I hope that helps. :) 

May 6th, 2015

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