Q.Prove that $\tan 3x\,\tan 2x\,\tan x = \tan 3x - \tan 2x - \tan x$.

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

$\tan2x$ and $\tan3x$ built the same way from the tangent sum formula. These identities let an

expression in $2x$ or $3x$ be rewritten entirely in powers of the single angle $x$ (or vice

versa), which is essential for solving equations that mix $\sin x$ with $\sin2x$, factoring

identities, and, later, for integrating powers of sine and cosine.