I assume you mean that the three angles are in a triangle, so the sum is PI radians.
So if each angle is the same, or PI/3, then cos(PI/3)=.5
so the square is .25, and the sum is NOT 1. I think you have the problem written wrong.
Prove cos^2 α + cos?^2 β + cos^2 γ = 1 .
Thank You
2 answers
You must be studying vectors in 3-D and
cos α, cos β, and cos γ must be the direction cosines of a vector <a, b, c>
you will recall that
cos α = a/√(a^2 + b^2 + c^2)
cos β = b/√(a^2 + b^2 + c^2)
cos γ = c/√(a^2 + b^2 + c^2)
Here is a great youtube which uses a specific example to further explain this if you need further help.
https://www.youtube.com/watch?v=AHxCQKQLpqY
so to prove: cos^2 α + cos?^2 β + cos^2 γ = 1
we have
LS = a^2/√(a^2 + b^2 + c^2)^2 + b^2/√(a^2 + b^2 + c^2)^2 + c^2/√(a^2 + b^2 + c^2)^2
= a^2/(a^2 + b^2 + c^2) + b^2/(a^2 + b^2 + c^2) + c^2/(a^2 + b^2 + c^2)
= (a^2 + b^2 + c^2)/(a^2 + b^2 + c^2)
= 1
= RS
cos α, cos β, and cos γ must be the direction cosines of a vector <a, b, c>
you will recall that
cos α = a/√(a^2 + b^2 + c^2)
cos β = b/√(a^2 + b^2 + c^2)
cos γ = c/√(a^2 + b^2 + c^2)
Here is a great youtube which uses a specific example to further explain this if you need further help.
https://www.youtube.com/watch?v=AHxCQKQLpqY
so to prove: cos^2 α + cos?^2 β + cos^2 γ = 1
we have
LS = a^2/√(a^2 + b^2 + c^2)^2 + b^2/√(a^2 + b^2 + c^2)^2 + c^2/√(a^2 + b^2 + c^2)^2
= a^2/(a^2 + b^2 + c^2) + b^2/(a^2 + b^2 + c^2) + c^2/(a^2 + b^2 + c^2)
= (a^2 + b^2 + c^2)/(a^2 + b^2 + c^2)
= 1
= RS