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.

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. For the purposes of this article, all we care about is the property $$e^{x + y} = e^x \times e^y$$ for real numbers $$x$$ and $$y$$. It turns out this property is true even for complex numbers $$x$$ and $$y$$. As such, we can see that

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

for real numbers $$\theta$$ and $$\phi$$. Using Euler’s formula now, we get

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

Expanding the left hand side and considering the real and imaginary parts, we get

\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*}

This is what we want!

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*}

Expanding out terms, one eventually deduces

\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*}