Alvin Pugh

2022-01-29

If $\mathrm{sin}\left(\beta +\gamma \right)=\mathrm{sin}ϵ$ and $\mathrm{cos}\beta \mathrm{cos}\gamma <\mathrm{sin}\beta \mathrm{sin}\gamma$ then $\beta +\gamma =\pi -ϵ$
I have the following system:
$\mathrm{sin}\beta ,\mathrm{cos}\gamma +\mathrm{cos}\beta ,\mathrm{sin}\gamma =\mathrm{sin}ϵ$
$\mathrm{cos}\beta ,\mathrm{cos}\gamma <\mathrm{sin}\beta ,\mathrm{sin}\gamma$
In other terms:
$\mathrm{sin}\left(\beta +\gamma \right)=\mathrm{sin}ϵ$
$\mathrm{cos}\beta ,\mathrm{cos}\gamma <\mathrm{sin}\beta ,\mathrm{sin}\gamma$
Can you explain me why $\beta +\gamma =\pi -ϵ$ please?

marzembreax

Expert

The inequality says $\mathrm{cos}\left(\beta +\gamma \right)<0$ So your problem is that of finding the relations between A and B in the following:
$\mathrm{sin}A=\mathrm{sin}B$
$\mathrm{cos}A<0$
From the first equation, we get $A=n\pi +{\left(-1\right)}^{n}B$. In the interval $\left[0,2\pi \right)$ we have A=B or $A=\pi -B$.
If we further know something about the angle B (for example, if it is in $\left[0,\frac{\pi }{2}\right]$) then with $\mathrm{cos}A<0$ we can conclude that $A=\pi -B$.
So in your problem, depending on what you know about $ϵ$ ,the conclusion can be inferred from that.

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