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Algebra Difficulty 4.7 AIME Prove it Slovenia

Let xx be a real number such that cos⁡(2x)+cos⁡(3x)=1\cos(2x) + \cos(3x) = 1. Show that 2sin⁡(2x)+2sin⁡(3x)=sin⁡(4x)+2sin⁡(5x)+sin⁡(6x)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
sin⁡(4x)+2sin⁡(5x)+sin⁡(6x)==2sin⁡(2x)cos⁡(2x)+2(sin⁡(2x)cos⁡(3x)+sin⁡(3x)cos⁡(2x))+2sin⁡(3x)cos⁡(3x)=2cos⁡(2x)(sin⁡(2x)+sin⁡(3x))+2cos⁡(3x)(sin⁡(2x)+sin⁡(3x))=2(cos⁡(2x)+cos⁡(3x))(sin⁡(2x)+sin⁡(3x)). \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))2(\sin(2x) + \sin(3x)), which was to be shown.

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