Let $C$ be the matrix with
$$c(k,n)=\dfrac{1}{\sqrt{N}},\text{ if }k=0,\ 0\leq n \leq N-1,$$
$$c(k,n)=\sqrt{\dfrac{2}{N}}\cos\left(\dfrac{\pi (2n+1)k}{2N}\right),\text{ if }1\leq k \leq N-1,\ 0\leq n \leq N-1.$$
How can I prove that $C$ is orthogonal?
I've begun using the $k$-row and $k'$-column then multiply these:
$$\sqrt{\frac{2}{N}}\cos\left(\frac{\pi (2(0)+1)k'}{2N}\right)+\frac{2}{N}\cos\left(\frac{\pi (2(1)+1)k}{2N}\right)\cos\left(\frac{\pi (2(1)+1)k'}{2N}\right)+\frac{2}{N}\cos\left(\frac{\pi (2(2)+1)k}{2N}\right)\cos\left(\frac{\pi (2(2)+1)k'}{2N}\right)+\cdots +\frac{2}{N}\cos\left(\frac{\pi (2(N-1)+1)k}{2N}\right)\cos\left(\frac{\pi (2(N-1)+1)k'}{2N}\right).$$
