# Why is $\frac{\cos(\alpha + \beta) + \cos(-\alpha)}{\sin(\alpha + \beta) + \sin(-\alpha)}$ independent of $\alpha$?

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.

• I don't believe that this is true. If $\alpha=0$ and $\beta=\pi/2$, then the expression is $1$. If $\alpha=\pi/2$ and $\beta=0$, then the expression is $0$. – Josh B. Feb 6 at 22:42
• Did you try using en.wikipedia.org/wiki/… – N74 Feb 6 at 22:44
• @JoshB. I think you misread – doetoe Feb 6 at 22:53
• You're absolutely right, I did misread. Sorry for the confusion. – Josh B. Feb 6 at 23:05
• mathworld.wolfram.com/ProsthaphaeresisFormulas.html – lab bhattacharjee Feb 7 at 1:07

$$\cos(\alpha+\beta)+\cos(-\alpha)=2\cos{\frac{\alpha+\alpha+\beta}{2}}\cos{\frac{\alpha+\beta-\alpha}{2}}=2\cos{\frac{\beta}{2}}\cos\left(\alpha+\frac{\beta}{2}\right)$$
$$\sin(\alpha+\beta)+\sin(-\alpha)=2\sin{\frac{\alpha+\beta-\alpha}{2}}\cos{\frac{\alpha+\beta-(-\alpha)}{2}}=2\sin{\frac{\beta}{2}}\cos\left(\alpha+\frac{\beta}{2}\right)$$
• I think you should put parentheses around $\alpha + \frac \beta 2$ – J. W. Tanner Feb 6 at 22:49