Multiple angle identities

We can derive multiple angle identities for trigonometric functions fairly easily using complex analysis and the binomial theorem.

For $n\in \mathbb{N}$, we can integrate fairly trivially for $n=1,n=2$, with the latter involving some manipulation using trigonometric identities. For $n\ge 3$, we can do it term-by-term using integration by parts, but this is computationally intensive for large $n$ and frankly dull as shit. Instead, we turn to De Moivre’s theorem and the binomial theorem.

From De Moivre’s theorem, we have:

$$z^n+\frac{1}{z^n}=2\cos (n\theta )$$ $$z^n-\frac{1}{z^n}=2i\sin (n\theta )$$

So our central problem examines $\int {\sin }^n(x)dx$ and $\int {\cos }^n(x)dx$ for $\{n\ge 3\mid n\in \mathbb{N}\}$.

The steps

Let $z=\text{cis}\theta$, and say we want to integrate ${\sin }^4\theta$. We expand $(z-\frac{1}{z}{)}^4$ using the binomial theorem, since our identity above with subtraction has an output with $\sin$. Expanding and grouping some terms together, we get:

$$(z-\frac{1}{z}{)}^4=z^4+\frac{1}{z^4}-4(z^2+\frac{1}{z^2})+6$$

Then explicitly using the identity above, we rewrite our expansion:

$$(2i\sin \theta {)}^4=2\cos (4\theta )-4(2\cos (2\theta ))+6$$ $$16{\sin }^4\theta =2\cos (4\theta )-8\cos (2\theta )+6$$ $${\sin }^4\theta =\frac{1}{8}\cos (4\theta )-\frac{1}{2}\cos (2\theta )+\frac{3}{8}$$

So we just derived a multiple angle formula for ${\sin }^4\theta$ fairly quickly, which is in an easier form to integrate. How neat!

The inverse

The reverse holds true too! For some multiple angle input of $\sin$ or $\cos$, we can expand $(\text{cis}\theta {)}^n$ and set it equal to $\text{cis}(n\theta )$. The real and imaginary parts respectively make up some equivalent identity.