Question

Rewrite with only sin x and cos x.
cos 3x
(please explain everything)

Answer 1

\[I\quad\cos3x=\cos(2x+x)=\cos2x\cos{x}-\sin2x\sin{x}=(\cos^2x-\sin^2x)\cos{x}-2\sin^2{x}\cos{x}=\]\[=\cos^3{x}-3\sin^2{x}\cos{x}\]

Answer 2

@nikvist i dont get how the cos2xcosx -sin2xsinx =(cos^2x-sin^2x)

Answer 3

\[\cos3x=\cos(2x+x)=\cos2x\cos{x}-\sin2x\sin{x}=\]\[=(\cos^2x-\sin^2x)\cos{x}-2\sin^2x\cos{x}=\]\[=\cos^3{x}-3\sin^2x\cos{x}\]

Answer 4

or \[\cos^3x-3\sin^2x\cos{x}=\cos^3x-3(1-\cos^2x)\cos{x}=\]\[=4\cos^3x-3\cos{x}\]