Assume: α+β+γ=π (Say, angles of a triangle)
Prove: sinα+sinβ+sinγ=4cosα2cosβ2cosγ2
There is already a solution on Math-SE, however I want to avoid using the sum-to-product identity because technically the book I go by hasn't covered it yet.
So, is there a way to prove it with identities only as advanced as sinα2?
Edit: Just giving a hint will probably be adequate (i.e. what identity I should manipulate).
Answer
You may go the other way around:
cosγ2=cosπ−α−β2=sinα+β2=sinα2cosβ2+cosα2sinβ2
so the right hand side becomes
4cosα2cosβ2sinα2cosβ2+4cosα2cosβ2cosα2sinβ2
Recalling the duplication formula for the sine we get
2sinαcos2β2+2sinβcos2α2
and we can recall
2cos2δ2=1+cosδ
to get
sinα+sinαcosβ+sinβ+sinβcosα=sinα+sinβ+sin(α+β)=sinα+sinβ+sinγ
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