# Double-angle Formulae

1. Nov 4, 2012

### odolwa99

Having some trouble with this one. Can anyone help me out?

Many thanks.

1. The problem statement, all variables and given/known data

Prove that $\tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan\theta}$

2. Relevant equations

3. The attempt at a solution

$\frac{3\sin\theta}{\cos\theta}-\frac{\sin^3\theta}{\cos^3\theta}/1-\frac{3\sin^2\theta}{\cos^2\theta}$
$\frac{3\sin\theta\cos\theta-\sin^3\theta}{\cos^2\theta-3\sin^2\theta}$
$\frac{\sin\theta(3\cos\theta-\sin^2\theta)}{\cos^2\theta-3\sin^2\theta}$
...

2. Nov 4, 2012

### rock.freak667

It will be easier to use the left side to prove the right side.

tan3θ = tan(2θ+θ)

Expand out tan(2θ+θ), what do you get?

3. Nov 4, 2012

### vela

Staff Emeritus
The tangent in the denominator should be squared.

Use parentheses! That should be
$$\left( \frac{3\sin\theta}{\cos\theta} - \frac{\sin^3\theta}{\cos^3\theta}\right) / \left(1-\frac{3\sin^2\theta}{\cos^2\theta}\right)$$