### Multiple angles

The double angle formulas can be extended to larger multiples. For example, to find $$\cos 3\theta$$, we write $$3\theta$$ as $$2\theta + \theta$$ and expand:

$\cos 3\theta = \cos(2\theta + \theta) = \cos2\theta\,\cos\theta - \sin2\theta\,\sin\theta.$

We can now apply the double angle formulas to obtain

$\cos 3\theta = \bigl(\cos^2\theta - \sin^2\theta\bigr)\,\cos\theta - 2\sin\theta\,\cos\theta\,\sin\theta = \cos^3\theta - 3\sin^2\theta\,\cos\theta.$

Replacing $$\sin^2\theta$$ with $$1-\cos^2\theta$$, we have

$\cos 3\theta = 4\cos^3\theta - 3\cos\theta.$

Exercise 18

Use the method above to find a formula for $$\sin 3\theta$$.

A more general approach is obtained using complex numbers. The complex number $$i$$ is defined by $$i^2=-1$$.

De Moivre's theorem says that, if $$n$$ is a positive integer, then

$\bigl(\cos \theta + i\sin \theta\bigr)^n = \cos n\theta + i\sin n\theta.$

Hence, for any given $$n$$, we can expand $$(\cos \theta + i\sin \theta)^n$$ using the binomial theorem and equate the real and imaginary parts to find formulas for $$\cos n\theta$$ and $$\sin n\theta$$.

For example, in the case of $$n=3$$,

$\bigl(\cos \theta + i\sin \theta\bigr)^3 = \cos 3\theta +i\sin 3\theta$ and $\bigl(\cos \theta + i\sin \theta\bigr)^3 = \cos^3\theta + 3i\cos^2\theta \sin\theta - 3\cos\theta \sin^2\theta - i\sin^3\theta.$

Equating real and imaginary parts, plus some algebraic manipulation, will produce the triple angle formulas.

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