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Question Number 32240 by mondodotto@gmail.com last updated on 22/Mar/18
Answered by MJS last updated on 22/Mar/18
![((cos α)/(sin β×sin γ))+((cos β)/(sin α×sin γ))+((cos γ)/(sin α×sin β))= =((cos α×sin α+cos β×sin β+cos γ×sin γ)/(sin α×sin β×sin γ)) I. cos α×sin α+cos β×sin β+cos γ×sin γ= [using cos θ×sin θ=((sin 2θ)/2)] =((sin 2α+sin 2β+sin 2γ)/2) with γ=π−(α+β) and sin 2(π−(α+β))=sin(2π−(2α+2β))= =−sin(2α+2β) this becomes ((sin 2α+sin 2β−sin(2α+2β))/2)=Z/2 II. sin α×sin β×sin γ with γ=π−(α+β) and sin(π−(α+β))=sin(α+β) this becomes sin α×sin β×sin(α+β)= =sin α×sin β×(cos α×sin β+sin α×cos β)= =sin α×cos α×sin^2 β+sin β×cos β×sin^2 α= [using sin^2 θ=((1−cos 2θ)/2)] =((sin 2α)/2)×((1−cos 2β)/2)+((sin 2β)/2)×((1−cos 2α)/2)= =((sin 2α−sin 2α×cos 2β+sin 2β−sin 2β×cos 2α)/4)= [using sin θ×cos φ=((sin(θ−φ)+sin(θ+φ))/2)] =((sin 2α−((sin(2α−2β)+sin(2α+2β))/2)+sin 2β−((sin(2β−2α)+sin(2β+2α))/2))/4)= =((sin 2α+sin 2β−((sin(2α−2β)+sin(2α+2β)−sin(2α−2β)+sin(2α+2β))/2))/4)= =((sin 2α+sin 2β−sin(2α+2β))/4) =((sin 2α+sin 2β−sin(2α+2β))/4)=Z/4 ((Z/2)/(Z/4))=2](Q32310.png)
Commented by mondodotto@gmail.com last updated on 23/Mar/18
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Question and Answers Forum | ||
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Question Number 32240 by mondodotto@gmail.com last updated on 22/Mar/18 | ||
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Answered by MJS last updated on 22/Mar/18 | ||
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Commented by mondodotto@gmail.com last updated on 23/Mar/18 | ||
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