If sin⁡(β+γ)=sin⁡ϵ and cos⁡βcos⁡γ<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<sin⁡β,sin⁡γ In other...

Alvin Pugh

Alvin Pugh

Answered

2022-01-29

If sin(β+γ)=sinϵ and cosβcosγ<sinβsinγ then β+γ=πϵ
I have the following system:
sinβ,cosγ+cosβ,sinγ=sinϵ
cosβ,cosγ<sinβ,sinγ
In other terms:
sin(β+γ)=sinϵ
cosβ,cosγ<sinβ,sinγ
Can you explain me why β+γ=πϵ please?

Answer & Explanation

marzembreax

marzembreax

Expert

2022-01-30Added 13 answers

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

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