4100.41 – That's Impossible!

Simplify

sin⁡2x+sin⁡4x+sin⁡6x+sin⁡8xcos⁡2x+cos⁡4x+cos⁡6x+cos⁡8x.\frac{\sin 2x + \sin 4x + \sin 6x + \sin 8x}{\cos 2x + \cos 4x + \cos 6x + \cos 8x}.

Solution

Answer: tan⁡5x\tan 5x !

Use the addition formulae:

sin⁡(a+b)=sin⁡acos⁡b+sin⁡bcos⁡a,sin⁡(a−b)=sin⁡acos⁡b−sin⁡bcos⁡a. \begin{aligned} \sin(a + b) &=& \sin a \cos b + \sin b \cos a, \\ \sin(a - b) &=& \sin a \cos b - \sin b \cos a. \end{aligned}

Adding these gives

sin⁡(a−b)+sin⁡(a+b)=2sin⁡(a)cos(b).\sin(a - b) + \sin(a + b) = 2 \sin(a) cos(b).

Similar identities for cosine yield,

cos⁡(a−b)+cos⁡(a+b)=2cos⁡(a)cos(b).\cos(a - b) + \cos(a + b) = 2 \cos(a) cos(b).

These lead to

sin⁡2x+sin⁡8x=2sin⁡5xcos⁡3x,sin⁡4x+sin⁡6x=2sin⁡5xcos⁡x, \begin{aligned} \sin 2x + \sin 8x &=& 2 \sin 5x \cos 3x, \\ \sin 4x + \sin 6x &=& 2 \sin 5x \cos x, \end{aligned}

where the first uses a=5 and b=3a = 5 \text{ and } b = 3 and the second uses a=5 and b=1a = 5 \text{ and } b = 1.

Similarly,

cos⁡2x+cos⁡8x=2cos⁡5xcos⁡3x,cos⁡4x+cos⁡6x=2cos⁡5xcos⁡x. \begin{aligned} \cos 2x + \cos 8x &=& 2 \cos 5x \cos 3x, \\ \cos 4x + \cos 6x &=& 2 \cos 5x \cos x. \end{aligned}

Combining all of these results,

sin⁡2x+sin⁡4x+sin⁡6x+sin⁡8xcos⁡2x+cos⁡4x+cos⁡6x+cos⁡8x=2sin⁡5xcos⁡3x+2sin⁡5xcos⁡x2cos⁡5xcos⁡3x+2cos⁡5xcos⁡x=sin⁡5xcos⁡5x⋅2cos⁡3x+2cos⁡x2cos⁡3x+2cos⁡x=tan⁡5x \begin{aligned} \frac{\sin 2x + \sin 4x + \sin 6x + \sin 8x}{\cos 2x + \cos 4x + \cos 6x + \cos 8x} &=& \frac{2 \sin 5x \cos 3x + 2 \sin 5x \cos x}{2 \cos 5x \cos 3x + 2 \cos 5x \cos x} \\ &=& \frac{\sin 5x}{\cos 5x} \cdot \frac{2 \cos 3x + 2 \cos x}{2 \cos 3x + 2 \cos x} \\ &=& \tan 5x \end{aligned}

(with thanks to Alexandra Du).