4100.41 – That's Impossible!

Answer: $\tan 5x$ !

Use the addition formulae:

$$\begin{aligned}\sin(a + b) &= \sin(a) \cos(b) + \sin(b) \cos(a) \\\sin(a - b) &= \sin(a) \cos(b) - \sin(b) \cos(a)\end{aligned}$$

Adding these gives

$$\sin(a - b) + \sin(a + b) = 2 \sin(a) cos(b)$$

Similar identities for cosine yield,

$$\cos(a - b) + \cos(a + b) = 2 \cos(a) cos(b)$$

These lead to

$$\begin{aligned}\sin(2x) + \sin(8x) &= 2 \sin(5x) \cos(3x) \\\sin(4x) + \sin(6x) &= 2 \sin(5x) \cos(x)\end{aligned}$$

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

Similarly,

$$\begin{aligned}\cos(2x) + \cos(8x) &= 2 \cos(5x) \cos(3x) \\\cos(4x) + \cos(6x) &= 2 \cos(5x) \cos(x)\end{aligned}$$

Combining all of these results,

$$\begin{aligned}\dfrac{\sin(2x) + \sin(4x) + \sin(6x) + \sin(8x)}{\cos(2x) + \cos(4x) + \cos(6x) + \cos(8x)} &= \dfrac{2 \sin(5x)\cos(3x) + 2 \sin(5x) \cos(x)}{2 \cos(5x)\cos(3x) + 2 \cos(5x)\cos(x)} \\&= \dfrac{\sin(5x)}{\cos(5x)} \cdot \dfrac{2 \cos(3x) + 2 \cos(x)}{2 \cos(3x) + 2 \cos(x)} \\&= \tan(5x)\end{aligned}$$

(with thanks to Alexandra Du).