If $\sin(\beta+\gamma)= \sin\epsilon$ and $\cos\beta\cos\gamma<\sin\beta\sin\gamma$, then $\beta+\gamma=\pi-\epsilon$

Question

I have the following system:

$$\begin{align} \sin\beta \,\cos\gamma + \cos\beta\,\sin\gamma &= \sin\epsilon \\ \cos\beta\,\cos\gamma &< \sin\beta\,\sin\gamma \end{align}$$

In other terms:

$$\begin{align} \sin(\beta+\gamma) &= \sin \epsilon \\ \cos\beta \,\cos\gamma &< \sin\beta\,\sin\gamma \end{align}$$

Can you explain me why $\beta+\gamma=\pi-\epsilon$, please?

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EDIT: I know also that $\epsilon$ belongs to $[-\pi /2, \pi /2]$.

Thank you so much for your time.

Answer 1

The inequality says $\cos(\beta+\gamma)<0$. So your problem is that of finding the relations between $A$ and $B$ in the following: \begin{align*} \sin A & =\sin B\\ \cos A &<0. \end{align*} From the first equation, we get $A=n\pi+(-1)^nB$. In the interval $[0,2\pi)$ we have $A=B$ or $A=\pi-B$.

If we further know something about the angle $B$ (for example, if it is in $[0,\pi/2]$), then with $\cos A <0$, we can conclude that $A=\pi-B$.

So in your problem, depending on what you know about $\epsilon$, the conclusion can be inferred from that.