Problem Statement
Trigonometric compound angle identities provide analytical methods for expanding multiple angles.
**(a)** [4 marks] Using the compound angle formulas for $\tan(A + B)$, show that:
$$\tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$$
**(b)** [3 marks] Hence, solve the equation $\frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = \sqrt{3}$ for $\theta \in [0, \pi]$. Give your answers in exact form in terms of $\pi$.
Verified Solution & Marking Scheme
Part (a) [M1]
Express $3\theta = 2\theta + \theta$: $\tan(3\theta) = \frac{\tan 2\theta + \tan\theta}{1 - \tan 2\theta\tan\theta}$
Part (a) [M1]
Substitute double angle identity $\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$
Part (a) [A1]
Numerator $= \frac{2\tan\theta}{1 - \tan^2\theta} + \tan\theta = \frac{2\tan\theta + \tan\theta - \tan^3\theta}{1 - \tan^2\theta} = \frac{3\tan\theta - \tan^3\theta}{1 - \tan^2\theta}$
Part (a) [A1]
Denominator $= 1 - \left(\frac{2\tan\theta}{1 - \tan^2\theta}\right)\tan\theta = \frac{1 - \tan^2\theta - 2\tan^2\theta}{1 - \tan^2\theta} = \frac{1 - 3\tan^2\theta}{1 - \tan^2\theta}$. Dividing gives $\frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$ (AG)
Part (b) [M1]
Use result from part (a): $\tan(3\theta) = \sqrt{3}$
Part (b) [M1]
Base reference angle: $3\theta = \frac{\pi}{3} + k\pi \implies \theta = \frac{\pi}{9} + \frac{k\pi}{3}$ for $k \in \mathbb{Z}$
Part (b) [A1]
For $\theta \in [0, \pi]$: $k = 0 \implies \theta = \frac{\pi}{9}$; $k = 1 \implies \theta = \frac{\pi}{9} + \frac{3\pi}{9} = \frac{4\pi}{9}$; $k = 2 \implies \theta = \frac{\pi}{9} + \frac{6\pi}{9} = \frac{7\pi}{9}$