Question

Can you show me the steps of this identity cos(3X) ?

Answer 1

Cos3x=4cos^3x-3cosx?

Answer 2

M i right?

Answer 3

\[\cos(3x)=\cos(2x+x)=\cos(2x)\cos(x)-\sin(2x)\sin(x)\]\[\cos(2x)=\cos(x+x)=\cos(x)\cos(x)-\sin(x)\sin(x)=\cos^2x-\sin^2x\]\[\sin(2x)=\sin(x+x)=\sin(x)\cos(x)+\sin(x)\cos(x)=2\sin(x)\cos(x)\]
So, you get,

Answer 4

\[\cos(3x)=\cos(2x)\cos(x)-\sin(2x)\sin(x)\]\[\cos(3x)=(\cos^2x-\sin^2x)\cos(x)-2\sin(x)\cos(x)\sin(x)\]\[\cos(3x)=\cos^3(x)-\sin^2(x) \cos(x)-2\sin^2(x)\cos(x)\]

Answer 5

THen I suppose you can play with the answer if you want to....

Answer 6

yes, i've just figure the answer!
Thank you guys.