TITLE: Show that $\tan 3x =\frac{ \sin x + \sin 3x+ \sin 5x }{\cos x + \cos 3x + \cos 5x}$
QUESTION [9 upvotes]: I was able to prove this but it is too messy and very long. Is there a better way of proving the identity? Thanks.

REPLY [4 votes]: The identities for the sum of sines and the sum of cosines yield
$$
\frac{\sin(x)+\sin(y)}{\cos(x)+\cos(y)}=\frac{2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)}{2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)}=\tan\left(\frac{x+y}{2}\right)\tag{1}
$$
Equation $(1)$ implies that
$$
\frac{\sin(x)+\sin(5x)}{\cos(x)+\cos(5x)}=\tan(3x)=\frac{\sin(3x)}{\cos(3x)}\tag{2}
$$
We also have that if $b+d\not=0$, then
$$
\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\tag{3}
$$
Combining $(2)$ and $(3)$ yields
$$
\tan(3x)=\frac{\sin(x)+\sin(3x)+\sin(5x)}{\cos(x)+\cos(3x)+\cos(5x)}\tag{4}
$$