Question

if cos(αlpha-β)+cos(β+Gama) +cos(gama-alpha) =-3/2 then

what will be the correct option
a) sigma cos alpha = 0
b) sigma sin alpha = 0
c) sigma(cos alpha +sin alpha) = 0
d) sigma cos alpha + sin alpha =0​

Answer 1

Answer:

Step-by-step explanation:

cos(α−β)+cos(β−γ)+cos(γ−α)=−32  

⇒2(cosαcosβ+sinαsinβ)+2(cosβcosγ+sinβsinγ)+2(cosαcosγ+sinαsinγ)=−3

⇒2(cosαcosβ+cosβcosγ+cosαcosγ)+2(sinαsinβ+sinβsinγ+sinαsinγ)+3=0

⇒2(cosαcosβ+cosβcosγ+cosαcosγ)+2(sinαsinβ+sinβsinγ+sinαsinγ)+(cos2α+sin2α)+(cos2β+sin2β)+(cos2γ+sin2γ)=0

⇒(cos2α+cos2β+cos2γ+2(cosαcosβ+cosβcosγ+cosαcosγ))+(sin2α+sin2β+sin2γ+2(sinαsinβ+sinβsinγ+sinαsinγ))=0

⇒(cosα+cosβ+cosγ)2+(sinα+sinβ+sinγ)2=0

a2+b2=0 is possible onle when a=b=0.

∴cosα+cosβ+cosγ=sinα+sinβ+sinγ=0