Expansion of sin and cos using De Moivre's theorem

We saw in the last section that:

(cosq+isinq)ⁿ=(cosnq+isinnq)

Alternatively, we could say that:

(cosnq+isinnq)=(cosq+isinq)ⁿ

Now if we write c for cosq and s for sinq, for the sake of brevity, we can expand that right-hand side using the binomial theorem (see MathHelp notebook on Series if you're not familiar with that).

For example, suppose n=3. The right-hand side becomes:

(c+is)³=c³+3c²is+3c(is)² +(is)³.

This simplifies to:

(c+is)³=c³+3isc²-3cs²- is³

Now here's the point of doing all that, we can now compare the real and imaginary parts of the original equation. Here's what I mean. Here's the original equation again (with n=3), but with this new version of the right-hand side:

(cos3q+isin3q)=c³+3isc²-3cs²- is³.

The real part of the left-hand side must be equal to the real part of the right-hand side and similarly for the imaginary parts, so:

cos3q=c³-3cs²

and

sin3q=3sc²-s³.

What we have done is find a quick way to expand cos3q and sin3q in terms of powers of cosq and sinq.

This works for larger powers too. Here's another example, what is the expansion of cos(5q) in powers of cosq and sinq?

You have a go and then check your answer.