Choosing convenient limits of integration on Basel problem I have recently discovered $$\sum_{k=1}^{\infty}\frac{\cos\left(k\alpha\right)}{k^{2}}-\sum_{k=1}^{\infty}\frac{\cos\left(k\beta\right)}{k^{2}}=\int_{\alpha}^{\beta}\tan^{-1}\left(\cot\left(\theta/2\right)\right)\,\mathrm{d}\theta$$ which seems to be a nice way to attack Basel problem. The integral nicely reduces to $$\int_{0}^{\beta}\frac{\pi}{2}-\frac{\theta}{2}\,\mathrm{d}\theta=\frac{2\pi\beta-\beta^{2}-2\pi\alpha+\alpha^{2}}{4}=\frac{1}{4}\left(\alpha-\beta\right)\left(\alpha+\beta-2\pi\right)$$ when $\alpha,\,\beta\in\left[0,\,2\pi\right]$. The question: how do I choose $\alpha$ and $\beta$ such that $\cos\left(k\alpha\right)-\cos\left(k\beta\right)=1$ for all $k\in\mathbb{N}$? Does this $k$ even exist?

Sadly, there's no such $\alpha$ and $\beta$. We'll try and solve the more general equation - $\cos(k\alpha)-\cos(k\beta)$ is a non-zero constant, say $C$.
Assume there are such $\alpha, \beta$. Let $x=\cos \alpha, y = \cos \beta$. 
Note that $\cos(2t) = 2\cos(t)^2 - 1, \cos(3t) = 4\cos(t)^{3} - 3\cos(t)$. Then the following 2 equalities hold:
$$x-y = C$$
$$2x^2-2y^2 = C$$
Factoring $2x^2-2y^2$ as $2(x-y)(x+y)$ and plugging the first equation, we find that $x+y = 0.5$ This leads to $x = \frac{C+0.5}{2}, y=\frac{0.5 - C}{2}$.
Now we choose $k=3$:
$$4x^3 - 3x - (4y^3 - 3y) = C$$
The LHS factors as $(x-y)(4(x^2+xy+y^2)-3)=(x-y)(4(x+y)^2-4xy-3)$, which is 
$$C(4(0.5)^2-4(\frac{1}{16} - \frac{C^2}{4})-3)=C$$
Cancelling the $C$s, it becomes
$$C^2 = 3.25$$
So $C\neq 1$!
Now we note that $\cos(4t) = 8(\cos(t)^4-\cos(t)^2)+1$, which gives the following equation:
$$8(x^4-x^2-(y^4-y^2)) = 1$$
The LHS factors as $8(x^2-y^2)(x^2+y^2-1)=1$. We can plug $2x^2-2y^2 = C$ and this becomes 
$$x^2+y^2 = \frac{1}{4C}+1$$
But $x^2+y^2 = \frac{C^2}{2}+\frac{1}{8} = 1.75$, so it follows that $C=\frac{1}{3}$, a contradiction to $C^2 = 3.25$.
