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

Notice that $\tan 3x = \sin 3x/\cos 3x$, and if $a/b=c/d$ then $a/b=c/d=(a+c)/(b+d)$, so it's enough to prove
$$
\tan 3x = \frac{\sin x + \sin 5x}{\cos x + \cos 5x}.
$$
So generally, how does one prove
$$
\tan\left(\frac{\alpha+\beta}{2}\right) = \frac{\sin\alpha+\sin\beta}{\cos\alpha+\cos\beta}\  ?
$$
One can say that
$$
\sin \alpha = \sin\left( \frac{\alpha+\beta}{2} + \frac{\alpha-\beta}{2} \right)
$$
$$
\sin \beta = \sin\left( \frac{\alpha+\beta}{2} - \frac{\alpha-\beta}{2} \right)
$$
and do the same for the two cosines, then apply the formulas for sine of a sum and cosine of sum.  After that, it's trivial simplification.
