If A+B+C=180∘. Find sin2A+sin2B+sin2C =
2+2cosAcosBcosC
sin2A+sin2B+sin2C
We know conditional identities but all of them are in the form of degree one.Here, power of sine is 2.
First we should reduce this power to 1 and then simplify it
We know cos2A=1−2sin2A
⇒sin2A=1−cos2A2
So, 1−cos2A2+1−cos2B2+1−cos2C2
= 32−12 cos2A+cos2B+cos2C....(1)
cos2A+cos2B+cos2C
=2cos(2A+2B2)⋅cos(2A−2B2)+cos(2(π−(A+B)))
=2cos(A+B)⋅cos(A−B)+cos(2π−2(A+B))
=2cos(A+B)⋅cos(A−B)+cos2(A+B)
=2cos(A+B)⋅cos(A−B)+2cos2(A+B)−1
=2cos(A+B)[cos(A−B)+cos(A+B)]−1
=2cos(180−c)[2cos(A−B+A+B)2⋅cos(A−B−A−B)2]−1
=−2cosC×[2cosA⋅cos(−B)]−1
(cos(−θ)=cosθ)
∴ given expression becomes 2+2cosAcosBcosC