14.7 Cool stuff with trigonometry

We can write $\cos(a+b)$ as the real part of $e^{(a+b)i}$. This because $\cos(a+b)$ is equal to the real part of $\cos(a+b)+i\sin(a+b)$, which is equal to $e^{(a+b)i}$.

We can then apply De Moivre’s theorem⁵⁵ 5 $\cos(nx)+i\sin(nx)=(\cos(x)+i\sin(x))^{n}$ to rewrite the expression in terms of $\sin(x)$ and $\cos(x)$

We can now expand the binomial obtained, which leads to the result that

Then, we can tidy this up a bit, leading to the expression

We are only interested in the imaginary parts of the expansion, so it is therefore equal to just

We want $\sin(3x)$ in terms of $\sin(x)$, however! There’s a rogue gatecrasher ⁶⁶ 6 A handy way to remember whether $\cos(x)$ or $\sin(x)$ shows a certain property is (as previously mentioned, TODO: mention) that $\sin(x)$ generally behaves ”better” than $\cos(x)$. in the previous expression - the $\cos(x)$! Fortunately we can remove the $\cos^{2}(x)$ without too much difficulty using the Pythagorean identity.

Thus, we can say that

Using Euler’s formula, it is possible to write both $\cos(\theta)$ and $\sin(\theta)$ in terms of $e^{x}$. As $e^{i\theta}=\cos(\theta)+i\sin(\theta)$, and $e^{-ix}=\cos(-\theta)+i\sin(-\theta)=\cos(\theta)-i\sin(\theta)$ we can either add or subtract these two quantities in order to write both trigonometric functions in terms of $e$.

For $\cos(x)$, we can add $e^{ix}$ and $e^{-ix}$.

Thus we can say that

for all values of x. ⁷⁷ 7 Which looks remarkably like a hyperbolic function!.

We can do a similar thing for $\sin(x)$.

Which means that

Proof:

Proof: