Closed form of $\sum_{n=1}^N\cos^2(n x)$

Question

From the Lagrange identity we have: $$\sum_{n=0}^N\cos(n x)=\frac12+\frac{\sin[(2N+1)\frac x2]}{2\sin\frac x2}$$

What is the closed form of: $$\sum_{n=1}^N\cos^2(n x)$$

Mathematica tells me that it is: $$\sum_{n=1}^N\cos^2(n x)=\frac N2+\frac{\sin Nx \cos(Nx+x)}{2\sin x}$$ I actually want to see the process. If you could include the derivation of the original identity, that would be appreciated!

Also, am I correct to say that: $$\sum_{k=0}^{n}\cos(kx)=\sum_{k=1}^{n}\cos(kx)+1$$

Answer 1

Hint 1: (Telescoping)

$$2\sin\frac{x}{2}\sum_{n=0}^N\cos(n x)=\sum_{n=0}^N\sin\left(nx+\frac{1}{2}x\right)-\sum_{n=0}^N\sin\left(nx-\frac{1}{2}x\right)$$

Hint 2: $$\cos^2(nx)=\frac{1}{2}+\frac{1}{2}\cos(2nx)$$

Answer 2

Hint:

Just use

$$\cos^2nx=\frac12+\frac12\cos2nx$$ and apply the Lagrange identity.