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Slovenia — National Math Olympiad

Algebra Difficulty 4.7 AIME Prove it Slovenia

Let $x$ be a real number such that $\cos(2x) + \cos(3x) = 1$ . Show that $2\sin(2x) + 2\sin(3x) = \sin(4x) + 2\sin(5x) + \sin(6x)$ .

Solution

The double-angle formula and the angle sum identity for sines and cosines imply
$\begin{aligned} \sin(4x) + 2\sin(5x) + \sin(6x) &= \\ &= 2\sin(2x)\cos(2x) + 2(\sin(2x)\cos(3x) + \sin(3x)\cos(2x)) + 2\sin(3x)\cos(3x) \\ &= 2\cos(2x)(\sin(2x) + \sin(3x)) + 2\cos(3x)(\sin(2x) + \sin(3x)) \\ &= 2(\cos(2x) + \cos(3x))(\sin(2x) + \sin(3x)). \end{aligned}$
and the latter is equal to $2(\sin(2x) + \sin(3x))$ , which was to be shown.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.