Prove $$\sin(x)+ \sin(x+a) + \sin(x+2a) + \sin(x+3a) + \sin(x+4a)=0 $$ given $a=72^\circ$
From what I know, I have to apply $\sin(\alpha+\beta)= \sin\alpha\cos\beta+\cos\alpha\sin\beta$
But this alone is not enough, am I lacking a crucial formula somewhere?
Hint: $y(1+z+z^2+z^3+z^4)=y(z^5-1)/(z-1)$.
What if $y=e^{ix}$ and $z=e^{ia}$?
Adding the first and the fifth, the second and the fourth, and finally the third, you will have
\begin{align} \sin x + \sin (x+4a) + \sin (x+a) + \sin (x+3a) +\sin (x+2a) &=0\\ 2\sin\frac{2x+4a}{2}\cos\frac{-4a}{2}+2\sin\frac{2x+4a}{2}\cos\frac{-2a}{2}+\sin (x+2a) &=0\\ \sin(x+2a)(2 \cos 2a + 2 \cos a + 1) &=0 \end{align}
As $x$ is unknown, then the second factor that must be zero. I will use a magic non-zero factor $\cos a -1$ here.
\begin{align} 2 \cos 2a + 2 \cos a + 1 &=0\\ 4\cos^2 a +2\cos a -1&=0\\ (\cos a -1)(4\cos^2 a +2\cos a -1)&=0\\ 4\cos^3 a - 2\cos^2 a - 3\cos a +1 &=0\\ 2\cos^2 a -1 &= 4\cos^3 a - 3\cos a \\ \cos 2a &=\cos 3a\\ \cos 2a &= \cos (360^\circ -3a)\\ 2a &=360^\circ -3a\\ 5a&=360^\circ\\ a&=72^\circ \end{align}
