Review question

# Can we simplify $\cos 3\theta \sin 2\theta - \cos 4\theta \sin \theta$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6807

## Suggestion

1. 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…

1. 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 …