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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=12sin2A

sin2A=1cos2A2

So, 1cos2A2+1cos2B2+1cos2C2

= 3212 cos2A+cos2B+cos2C....(1)

cos2A+cos2B+cos2C
=2cos(2A+2B2)cos(2A2B2)+cos(2(π(A+B)))

=2cos(A+B)cos(AB)+cos(2π2(A+B))
=2cos(A+B)cos(AB)+cos2(A+B)
=2cos(A+B)cos(AB)+2cos2(A+B)1
=2cos(A+B)[cos(AB)+cos(A+B)]1
=2cos(180c)[2cos(AB+A+B)2cos(ABAB)2]1
=2cosC×[2cosAcos(B)]1
(cos(θ)=cosθ)
given expression becomes 2+2cosAcosBcosC


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