TITLE: Why is $\frac{\cos(\alpha + \beta) + \cos(-\alpha)}{\sin(\alpha + \beta) + \sin(-\alpha)}$ independent of $\alpha$?
QUESTION [0 upvotes]: By accident I found (numerically) that the expression 
$$\frac{\cos(\alpha + \beta) + \cos(-\alpha)}{\sin(\alpha + \beta) + \sin(-\alpha)}$$ 
only depends on $\beta$. This looks like it shouldn't be hard to prove, but I didn't manage. Can someone think of a nice proof?
Additionally, I would be interested in any geometric interpretation of this result.

REPLY [5 votes]: $$\cos(\alpha+\beta)+\cos(-\alpha)=2\cos{\frac{\alpha+\alpha+\beta}{2}}\cos{\frac{\alpha+\beta-\alpha}{2}}=2\cos{\frac{\beta}{2}}\cos\left(\alpha+\frac{\beta}{2}\right)$$
$$\sin(\alpha+\beta)+\sin(-\alpha)=2\sin{\frac{\alpha+\beta-\alpha}{2}}\cos{\frac{\alpha+\beta-(-\alpha)}{2}}=2\sin{\frac{\beta}{2}}\cos\left(\alpha+\frac{\beta}{2}\right)$$
{"set_name": "stack_exchange", "score": 0, "question_id": 3103153}