Triple-angle formula for the cosine

Problem

Prove that $\cos(3\alpha) = 4\cos^3\alpha - 3\cos\alpha$ (hint: $\cos(3\alpha) = \cos(2\alpha+\alpha)$).

Solution

We write $3\alpha = 2\alpha + \alpha$ and apply the addition formula for the cosine: $\cos(3\alpha) = \cos(2\alpha+\alpha) = \cos(2\alpha)\cos\alpha - \sin(2\alpha)\sin\alpha.$ We substitute the duplication formulae $\cos(2\alpha) = 2\cos^2\alpha - 1$ and $\sin(2\alpha) = 2\sin\alpha\cos\alpha$: $= (2\cos^2\alpha - 1)\cos\alpha - (2\sin\alpha\cos\alpha)\sin\alpha = 2\cos^3\alpha - \cos\alpha - 2\sin^2\alpha\cos\alpha.$ Finally we use $\sin^2\alpha = 1-\cos^2\alpha$: $= 2\cos^3\alpha - \cos\alpha - 2(1-\cos^2\alpha)\cos\alpha = 2\cos^3\alpha - \cos\alpha - 2\cos\alpha + 2\cos^3\alpha = \boxed{4\cos^3\alpha - 3\cos\alpha}. \qquad \blacksquare$

Links

Topics: Trigonometric formulae
Concepts: Addition formulae · Duplication formulae
Functions: Cosine
Methods: Addition formulae · Duplication formulae
Skills: Proving · Using formulae
Exercise types: Proof