Q.Prove that $\cos 3x = 4\cos^3x - 3\cos x$.

Setting $y=x$ in the sum formulas gives the double-angle

identities $\sin2x=2\sin x\cos x$ and $\cos2x=\cos^2x-\sin^2x$, which -- using the fundamental

identity to eliminate one of the two squared terms -- also take the equivalent forms

$\cos2x=2\cos^2x-1=1-2\sin^2x$, each useful in a different situation. Writing $3x=2x+x$ and

applying the sum formula again with the double-angle results already known gives the

triple-angle identities $\sin3x=3\sin x-4\sin^3x$ and $\cos3x=4\cos^3x-3\cos x$, together with …