Can we simplify $\cos 3\theta \sin 2\theta - \cos 4\theta \sin \theta$?
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Ref: R6807

Question

Prove that \[\cos 3\theta \sin 2\theta - \cos 4\theta \sin \theta = \cos 2\theta \sin \theta.\]
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 $\dfrac{\pi}{4}$.