$\displaystyle (\sin(x)+\sin(2x)+\sin(3x)+\sin(4x)+\sin(5x))^2$$\displaystyle +(\cos(x)+\cos(2x)+\cos(3x)+\cos(4x)+\cos(5x)+1)^2$

I was able to do with the brute force way, that is, multiply everything out then use the identity: $\displaystyle cos(a-b)=cos(a)cos(b)+sin(a)sin(b)$ to reduce everything to:

$\displaystyle 6+10 cos(x)+8 cos(2x)+6 cos(3x)+4 cos(4x)+2 cos(5 x)$

My question is: is there a more elegant way to do this that doesn't take 5 YEARS.