Compound angle formulae are one of the most useful tools you have in high school trigonometry, yet reciting them could be quite difficult. It turns out one could deduce them from the more straightforward Euler’s formula $$e^{i \theta} = \cos \theta + i \sin \theta$$.

Hopefully this reduces the amount of facts you have to recite while providing some intuition behind the identities. We would not stress on the rigour or the logical flow of the arguments.

## Compound angle formulae

To begin, let’s introduce Euler’s formula as fact, which states that

$e^{i \theta} = \cos \theta + i \sin \theta$

Where $$i$$ is the imaginary unit and $$\theta$$ is a real number. The Euler’s formula may seem very confusing at first, and I could recommend great videos by 3b1b that explain it in great detail.

It turns out that the property $$e^{x + y} = e^x \times e^y$$ is true even for complex numbers $$x$$ and $$y$$. (As an aside, this property leads to a very geometric way of visualizing complex multiplication) Taking it as fact from now on, we can see that

$e^{i \theta} \times e^{i \phi} = e^{i (\theta + \phi)}$

and so

$(\cos \theta + i \sin \theta) (\cos \phi + i \sin \phi) = \cos (\theta + \phi) + i \sin (\theta + \phi)$

By expanding the left hand side and considering the real and imaginary parts, we get that

\begin{align*} \cos (\theta + \phi) &= \cos \theta \cos \phi - \sin \theta \sin \phi \\ \sin (\theta + \phi) &= \sin \theta \cos \phi + \cos \theta \sin \phi \end{align*}

which is what we want!

## Aside: On complex exponentials

You can now see that for complex number $$a+bi$$ with $$a, b$$ real,

$e^{a+bi} = e^a ( \cos b + i \sin b )$

so we can make sense of exponential of complex numbers as well! By considering the complex plane, you can think of $$e^a$$ as “scaling” the coordinate / vector and $$e^{bi}$$ determining the angle of the coordinate / vector. For more on this interesting idea you can check out the video by 3b1b.

## Other identities

By our previous identity we can see that for positive integer $$n$$,

$e^{ni\theta} = (e^{i\theta})^n$

This is called De Moivre’s formula. Combining this with the Binomial Theorem, it gives us a way of finding formulae for powers of trigonometric functions.

For example,

\begin{align*} e^{3i\theta} &= (e^{i\theta})^3 \\ \text{so }\cos (3\theta) + i \sin (3\theta ) &= (\cos \theta + i \sin \theta)^3 \end{align*}

By expanding out the terms, one would eventually deduce

\begin{align*} \cos (3\theta ) &= \cos^3 \theta - 3 \cos \theta \sin^2 \theta \\ \sin (3\theta ) &= 3 \cos^2 \theta \sin \theta - \sin^3 \theta \end{align*}