If the lines xsin^2a + are concurrent where a b c are angles of a tria

1.	If the lines xsin^2a + are concurrent where a b c are angles of a triangle
Answer» Answer: Step-by-step EXPLANATION: Let α, β and γ be the ANGLES of a triangle; then it holds sin2α+sin2β+sin2γ=2+2cosαcosβcosγ If this equals 2, we conclude cosαcosβcosγ=0 so ONE of the angles is a right angle. Proof of the claim Let's use that γ=π−α−β, so cosγ=−cos(α+β). Then cos2α+cos2β+cos2γ=cos2α+cos2β+cos2(α+β)=cos2α+cos2β+cos2αcos2β+sin2αsin2β−2sinαsinβcosαcosβ=cos2α+cos2β+1−cos2α−cos2β+2cos2αcos2β−2sinαsinβcosαcosβ=1+2cosαcosβ(cosαcosβ−sinαsinβ)=1−2cosαcosβcosγ giving the FINAL relation cos2α+cos2β+cos2γ=1−2cosαcosβcosγ Now sin2α+sin2β+sin2γ=3−cos2α−cos2β−cos2γ=2+2cosαcosβcosγ as claimed at the beginning.

If the lines xsin^2a + are concurrent where a b c are angles of a triangle

Answer»

Answer:

Step-by-step EXPLANATION:

Let α, β and γ be the ANGLES of a triangle; then it holds

sin2α+sin2β+sin2γ=2+2cosαcosβcosγ

If this equals 2, we conclude

cosαcosβcosγ=0

so ONE of the angles is a right angle.

Proof of the claim

Let's use that γ=π−α−β, so cosγ=−cos(α+β). Then

cos2α+cos2β+cos2γ=cos2α+cos2β+cos2(α+β)=cos2α+cos2β+cos2αcos2β+sin2αsin2β−2sinαsinβcosαcosβ=cos2α+cos2β+1−cos2α−cos2β+2cos2αcos2β−2sinαsinβcosαcosβ=1+2cosαcosβ(cosαcosβ−sinαsinβ)=1−2cosαcosβcosγ

giving the FINAL relation

cos2α+cos2β+cos2γ=1−2cosαcosβcosγ

Now

sin2α+sin2β+sin2γ=3−cos2α−cos2β−cos2γ=2+2cosαcosβcosγ

as claimed at the beginning.

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