Proving Trigonometric Identities Using (cos θ + i sin θ)^n

Besides finding roots, De Moivre's theorem is a genuine proof technique for trigonometric identities, especially ones involving multiple angles like $\cos n\theta$ or $\sin n\theta$. The idea rests on a simple but powerful observation: since $(\cos\theta+i\sin\theta)^n$ equals $\cos n\theta+i\sin n\theta$ by the theorem, and it can also be expanded directly using the binomial theorem, comparing the real part of one expansion with the real part of the other (and similarly the imaginary parts) hands us a formula for $\cos n\theta$ and $\sin n\theta$ purely in terms of powers of $\cos\theta$ and $\sin\theta$ — no calculus, no separate multiple-angle memorisation needed.

A second, related trick handles identities built from both $(\cos\theta+i\sin\theta)^n$ and its conjugate $(\cos\theta-i\sin\theta)^n = \cos n\theta - i\sin n\theta$: adding the two cancels every imaginary term and leaves $2\cos n\theta$, while subtracting them cancels every real term and leaves $2i\sin n\theta$. This is exactly how expressions like $(1+i)^n+(1-i)^n$ collapse to a clean cosine, once $1+i$ and $1-i$ are first rewritten in $cis$ form.

Worked example 1 (deriving a multiple-angle formula). Derive formulas for $\cos3\theta$ and $\sin3\theta$ in terms of $\cos\theta$ and $\sin\theta$.

By De Moivre's theorem, $(\cos\theta+i\sin\theta)^3=\cos3\theta+i\sin3\theta$. Expand the left side by the binomial theorem, writing $c=\cos\theta,\ s=\sin\theta$ for brevity:

Group real and imaginary parts: real part $=c^3-3cs^2$, imaginary part $=3c^2s-s^3$. Matching these against $\cos3\theta+i\sin3\theta$ gives

Using $\sin^2\theta=1-\cos^2\theta$ and $\cos^2\theta=1-\sin^2\theta$ these simplify to the familiar single-function forms $\cos3\theta=4\cos^3\theta-3\cos\theta$ and $\sin3\theta=3\sin\theta-4\sin^3\theta$.

Worked example 2 (a conjugate-pair identity). If $n$ is a positive integer, show that $(\sqrt3+i)^n+(\sqrt3-i)^n = 2^{n+1}\cos\!\left(\dfrac{n\pi}{6}\right)$.

First convert $\sqrt3+i$ to $cis$ form: its modulus is $\sqrt{3+1}=2$ and its argument is $\arctan(1/\sqrt3)=\pi/6$, so $\sqrt3+i = 2\,cis(\pi/6)$, and by symmetry $\sqrt3-i = 2\,cis(-\pi/6)$. By De Moivre's theorem, …