Prove…

Prove: $\ \displaystyle\sum_{k=1}^n\cos^2kx=\frac{2n-1}4+\frac{\sin(2n+1)x}{4\sin x}$.

Step 1 · Reduce the power

$$ \cos^2kx=\left(\frac{e^{ikx}+e^{-ikx}}2\right)^2=\frac12+\frac12\cos2kx $$

so:

$$ \sum_{k=1}^n\cos^2kx=\frac n2+\frac12\sum_{k=1}^n\cos2kx $$

Step 2 · Sum the cosines via a geometric series

$$ \sum_{k=1}^n\cos2kx=\text{Re}\left(\sum_{k=1}^ne^{2ikx}\right), \qquad \sum_{k=1}^ne^{2ikx}=e^{2ix}\cdot\frac{1-e^{2nix}}{1-e^{2ix}}=\frac{e^{2ix}-e^{2(n+1)ix}}{1-e^{2ix}} $$

Step 3 · Simplify using the sine form of a difference of exponentials

Factor $e^{ix}$ out of both the numerator and the denominator:

$$ e^{2ix}-e^{2(n+1)ix} = -e^{ix}\big(e^{i(2n+1)x}-e^{ix}\big), \qquad 1-e^{2ix} = -e^{ix}\cdot2i\sin x $$

so the factors of $-e^{ix}$ cancel:

$$ \sum_{k=1}^ne^{2ikx} = \frac{e^{i(2n+1)x}-e^{ix}}{2i\sin x} $$

Now use $e^{i(2n+1)x}-e^{ix}=e^{ix}(e^{2inx}-1)=e^{ix}\cdot2ie^{inx}\sin nx=2ie^{i(n+1)x}\sin nx$:

$$ \sum_{k=1}^ne^{2ikx} = \frac{2ie^{i(n+1)x}\sin nx}{2i\sin x} = \frac{e^{i(n+1)x}\sin nx}{\sin x} $$

Step 4 · Take the real part

$$ \sum_{k=1}^n\cos2kx = \text{Re}\left(\frac{e^{i(n+1)x}\sin nx}{\sin x}\right) = \frac{\cos(n+1)x\,\sin nx}{\sin x} $$

Using the product-to-sum identity $\sin(nx)\cos((n+1)x)=\frac12\big[\sin(2n+1)x-\sin x\big]$:

$$ \sum_{k=1}^n\cos2kx=\frac{\sin(2n+1)x-\sin x}{2\sin x} $$

Step 5 · Assemble the final sum

$$ \sum_{k=1}^n\cos^2kx=\frac n2+\frac12\cdot\frac{\sin(2n+1)x-\sin x}{2\sin x}=\frac n2-\frac14+\frac{\sin(2n+1)x}{4\sin x} $$

$$ \boxed{\sum_{k=1}^n\cos^2kx=\frac{2n-1}4+\frac{\sin(2n+1)x}{4\sin x}} $$

Step 6 · Verification, $n=1$

The formula gives $\cos^2x=\frac14+\dfrac{\sin3x}{4\sin x}$. Using $\sin3x=\sin x(4\cos^2x-1)$:

$$ \frac14+\frac{4\cos^2x-1}4=\cos^2x \quad\checkmark $$

The general identity was also checked symbolically for $n=1,2,3,5$.