Question

Derive a formula for cos 3θ, in terms of cos θ and sin θ or just cos θ.

Answer 1

We have to find an expression for cos(3θ) in function of cos(θ) and/or sin(θ).

This is known as the cosine of the triple angle, but in this case we have to derive it from other identities.

We can start writing it as:

`\begin{gathered} \cos(3\theta)=\cos(2\theta+\theta) \\ \cos(3\theta)=\cos(2\theta)\cos(\theta)-\sin(2\theta)\sin(\theta) \end{gathered}`

We can apply again the identity for sum of angles, now with equal angles, and get:

`\cos(3\theta)=(\cos^2(\theta)-\sin^2(\theta))\cos(\theta)-(2\sin(\theta)\cos(\theta))\sin(\theta)`

We can rearrange the terms and use other identities to write the expression in a simpler way:

`\begin{gathered} \cos(3\theta)=(2\cos^2(\theta)-1)\cos(\theta)-2\sin^2(\theta)\cos(\theta) \\ \cos(3\theta)=2\cos^3(\theta)-\cos(\theta)-2(1-\cos^2(\theta))\cos(\theta) \\ \cos(3\theta)=2\cos^3(\theta)-\cos(\theta)-2\cos(\theta)+2\cos^3(\theta) \\ \cos(3\theta)=4\cos^3(\theta)-3\cos(\theta) \end{gathered}`

The final expression for the cosine of a triple angle is:

`\cos(3\theta)=4\cos^3(\theta)-3\cos(\theta)`