## 4100.41 – That's Impossible!

Simplify

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

Solution

Answer: $\tan 5x$ !

\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}

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

Similar identities for cosine yield,

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

\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} \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).