    Question

# If A+B+C=180∘. Find sin2A+sin2B+sin2C =

A

14cosAcosBcosC

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B

1+2cosAcosBcosC

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C

2+2cosAcosBcosC

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D

4sinAsinBsinC

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Solution

## The correct option is C 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  Suggest Corrections  0      Similar questions
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