If sin⁡(β+γ)=sin⁡ϵ and cos⁡βcos⁡γ&lt;sin⁡βsin⁡γ then β+γ=π−ϵ I have the following system: sin⁡β,cos⁡γ+cos⁡β,sin⁡γ=sin⁡ϵ cos⁡β,cos⁡γ&lt;sin⁡β,sin⁡γ In other terms: sin⁡(β+γ)=sin⁡ϵ cos⁡β,cos⁡γ&lt;sin⁡β,sin⁡γ Can you explain me why β+γ=π−ϵ please?

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If sin⁡(β+γ)=sin⁡ϵ and cos⁡βcos⁡γ&amp;lt;sin⁡βsin⁡γ then β+γ=π−ϵ  I have the following system:  sin⁡β,cos⁡γ+cos⁡β,sin⁡γ=sin⁡ϵ   cos⁡β,cos⁡γ&amp;lt;sin⁡β,sin⁡γ   In other terms:  sin⁡(β+γ)=sin⁡ϵ    cos⁡β,cos⁡γ&amp;lt;sin⁡β,sin⁡γ  Can you explain me why β+γ=π−ϵ please?

marzembreax · Accepted Answer

The inequality says cos⁡(β+γ)&amp;lt;0 So your problem is that of finding the relations between A and B in the following:  sin⁡A=sin⁡B  cos⁡A&amp;lt;0  From the first equation, we get A=nπ+(−1)nB. In the interval [0,2π) we have A=B or A=π−B.  If we further know something about the angle B (for example, if it is in [0,π2]) then with cos⁡A&amp;lt;0 we can conclude that A=π−B.  So in your problem, depending on what you know about ϵ ,the conclusion can be inferred from that.

If \sin(\beta+\gamma)= \sin\epsilon and \cos\beta\cos\gamma<\sin\beta\sin\gamma then \beta+\gamma=\pi-\epsilon I have the following

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