Prove that $f(x) = \frac{\sin(x+\alpha)}{\sin(x+\beta)}$ is monotonic in any interval of its domain

Question

Prove that $f(x) = \frac{\sin(x+\alpha)}{\sin(x+\beta)}$ is monotonic in any interval of its domain, where $\alpha, \beta \in \mathbb R$

Obviously $x + \beta \ne \pi n$. Then i've tried to split the problem into three parts: $\alpha = \beta$, $\alpha > \beta$, $\beta > \alpha$.

For the first case it's obvious that the function turns into a constant and therefore is monotonic.

Since $f(x)$ is periodic we may use that fact and consider some interval for example $(0, {\pi\over2})$ or $(0, \pi)$. But since $\alpha$ and $\beta$ may take any values i didn't manage to figure out any inequality to start from.

What steps should i take to proceed?

Answer 1

Using $$ \sin((x+\beta) + (\alpha - \beta)) = \sin(x+\beta) \cos (\alpha - \beta) + \cos(x+\beta) \sin (\alpha - \beta) $$ one gets $$ f(x) = \cos (\alpha - \beta) + \cot(x+\beta) \sin (\alpha - \beta) $$ and the cotangent is decreasing on each interval of its domain.

You could also attack the problem by substituting $x + \alpha = y$ first, then the expression becomes $$ \frac{\sin(y + \alpha - \beta)}{\sin(y)} $$ Now substitute $\gamma = \alpha - \beta$, and you are left with $$ g(y) = \frac{\sin(y + \gamma)}{\sin(y)} $$ which might be easier to handle.

Answer 2

Hint: Differentiating with respect to $x$ we get $$f'(x)=\sin (\alpha -\beta ) \left(-\csc ^2(\beta +x)\right)$$