- Prove that \[\cos 3\theta \sin 2\theta - \cos 4\theta \sin \theta = \cos 2\theta \sin \theta.\]

I don’t know any formulae that I think will help me directly to answer this question.

Perhaps I can use some of the formulae that I do know to get me started, at least? That might then give me some ideas for how to continue…

- Starting from the relations
\[\begin{align*}
\sin(A+B) &{}= \sin A \cos B + \cos A \sin B,\\
\cos(A+B) &{}= \cos A \cos B - \sin A \sin B,
\end{align*}\]
obtain an expression for \(\tan(A+B)\) in terms of \(\tan A\) and \(\tan B\).

Without the use of a calculator or tables, show that the sum of the three acute angles whose tangents are \(\dfrac{2}{9}\), \(\dfrac{1}{4}\) and \(\dfrac{1}{3}\) is \(\dfrac{\pi}{4}\).

For the second part, I don’t know the angles themselves, and working out \(\arctan\frac{2}{9}\) and so on seems difficult (especially as I’m not allowed to use a calculator). Perhaps the question is hinting that I should use the formula for \(\tan(A+B)\) that I’ve just derived?

But there are three angles to add. Is there some way I can break down the problem?

Also, I know that the three acute angles are each less than …, so their sum is less than …