If cosα=12x+1x and cosβ=12y+1y,(xy>0) ;x,y,α,β,∈R then
sin(α+β+γ)=sinγ∀γ∈R
cosαcosβ=1∀α,β∈R
(cosα+cosβ)2=4∀α,β∈R
sin(α+β+γ)=sinα+sinβ+sinγ∀a,b,γ∈R
cosα=12x+1x and cosβ=12y+1y
since xy > 0, we havex+1x≥2 or ≤−2 and y+1y≥2 or ≤−2
⇒cosα=1,cosβ=1----1or cosα=−1,cosβ=−1----2cosαcosβ=1⇒α+β is an even multiple of π(cosα+cosβ)2=4⇒sin(α+β+γ)=sin(2nπ+γ)=sinγAlso ,sinα=sinβ=0.