1. Analyze the given equations:
We are given:
cosα+cosβ=0⟹cosα=−cosβ
sinα+sinβ=0⟹sinα=−sinβ
2. Relate the angles:
Squaring both equations and adding them:
cos2α=(−cosβ)2⟹cos2α=cos2β
sin2α=(−sinβ)2⟹sin2α=sin2β
Adding these results:
cos2α+sin2α=cos2β+sin2β
1=1
(This confirms consistency but does not provide α and β directly).
However, consider the complex representation or the sum-to-product identities:
cosα+cosβ=2cos(2α+β)cos(2α−β)=0
sinα+sinβ=2sin(2α+β)cos(2α−β)=0
Since α=β, cos(2α−β)=0. Therefore, we must have:
cos(2α+β)=0andsin(2α+β)=0
This is impossible for real angles because sin2θ+cos2θ cannot be 0. Let's re-evaluate using the complex form:
eiα+eiβ=0⟹eiα=−eiβ=eiπ⋅eiβ=ei(β+π)
Thus, α=β+π+2nπ, or α−β=π (since α=β).
3. Substitute into the expression:
We want to find E=cos2α+cos2β+2cos(α+β).
Since α=β+π:
α+β=β+π+β=2β+π
Substitute α=β+π:
cos2α=cos2(β+π)=cos(2β+2π)=cos2β
cos(α+β)=cos(2β+π)=−cos2β
Now substitute these into the expression:
E=cos2β+cos2β+2(−cos2β)
E=2cos2β−2cos2β=0
Correct Option: (a) 0
