# Simplify $\frac{\sin(2022) + \sin(2022+\alpha)}{\cos(2022)-\cos(2022+\alpha)}$ as much as possible

Let $$\alpha\in\mathbb{R}$$ be so that $$\cos\alpha\neq 1$$. Simplify as much as possible $$\dfrac{\sin(2022) + \sin(2022+\alpha)}{\cos(2022)-\cos(2022+\alpha)}$$. To clarify, if for a finite number of values of $$\alpha,$$ the expression takes a numerical value, find that numeric value for each of those values of $$\alpha$$. If not, find a function of $$\alpha$$ that is as simplified as possible.

Clearly $$\alpha$$ is not an integer multiple of $$2\pi.$$ Let $$f(\alpha)$$ denote the expression that needs to be simplified. We have that $$f(\alpha + 2k\pi) = f(\alpha)$$ for any integer $$k$$, so we may assume WLOG that $$0 < \alpha < 2\pi$$. It might be easier to generalize and replace $$2022$$ with $$m$$. Perhaps we only need the fact that $$2022$$ is an integer? Using the addition law, we get that $$f(\alpha) = \dfrac{\sin(m)(1+ \cos\alpha) + \sin(\alpha)\cos m}{\cos(m)(1 - \cos(\alpha)) + \sin(m)\sin(\alpha)}$$. I'm not sure if Euler's formula will help simplify $$f(\alpha)$$. Multiply the denominator and numerator of $$f(\alpha)$$ by $$\cos m(1+\cos \alpha) - \sin m \sin \alpha$$. The denominator becomes $$\cos^2 m \sin^2 \alpha + 2\sin m \cos m \sin \alpha \cos \alpha - \sin^2m\sin^2\alpha = \sin\alpha( \cos(2m) \sin \alpha + \sin(2m)\cos \alpha) = \sin \alpha \sin(2m+\alpha)$$. The numerator becomes $$\sin m \cos m (1+\cos \alpha)^2 -\sin^2 m \sin \alpha(1+\cos \alpha) + \cos^2 m \sin \alpha(1+\cos \alpha) - \sin m \cos m \sin^2 \alpha = \cos(2m) \sin (\alpha) ( 1+\cos \alpha) + \sin m \cos m(2\cos \alpha + 2\cos^2 \alpha) = (1+\cos \alpha)(\cos (2m) \sin (\alpha) + \sin(2m)\cos (\alpha)) = (1+\cos\alpha)\sin(2m+\alpha).$$ Hence $$f(\alpha) = \dfrac{1+\cos \alpha}{\sin \alpha}.$$ Note that $$m$$ could've actually been any real number by the same proof as above.

Let $$x = 2022$$ and $$y = 2022 + \alpha$$. Observe that using sum-to-product identities \begin{align*} \frac{\sin(x) + \sin(y)}{\cos(x) - \cos(y)} &= \frac{2\sin\left(\frac{x + y}{2}\right) \cos\left(\frac{x - y}{2}\right)}{-2\sin\left(\frac{x + y}{2}\right) \sin\left(\frac{x - y}{2}\right)}\\ &=\cot\left (\frac{y - x}{2} \right )\\ &= \cot\left ( \frac{\alpha}{2}\right ) \end{align*}