show that sin^2 theta + sin^2 ( theta + 60° ) + sin^2 ( theta - 60° ) = 3/2
Answers
Answered by
0
Answer:
Answer
Open in answr app
Correct option is
A
23
sin2θ+sin2(60∘+θ)+sin2(60∘−θ)=sin2θ+21(1−cos(120∘+2θ))+21(1−cos(120∘−2θ))
=sin2θ+1−21(cos(120∘+2θ)+cos(120∘−2θ))
=1+sin2θ−21(2cos120∘cos2θ)
=1+sin2θ+21(1−2sin2θ)
=1+21=23
Similar questions