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Question

If A+B+C=0o then sin2A+sin2B+sin2C=?

A
sinAcosBcosC
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B
2sinAcosBsinC
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C
4sinAsinBsinC
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D
4sinAsinBsinC
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Solution

The correct option is D 4sinAsinBsinC

Solve:- Given,

A+B+C=0
A+B=C(i)

sin2A+sin2B+sin2C=

[2sin(2A+2B2)cos(2A2B2)]+sin2C

=[2sin(A+B)cos(AB)]+2sinCcosC

sinc+sinD=
2sin(c+D2)cos(cD2)
and sin2x=2sinxcosx

by using equation (i) we get

=[2sin(c)cos(AB)]+2sinccosc

=2sinc[cos(AB)cos(c)]

because cosC=cos((A+B))=cos(A+B)

So,cos(AB)cos(A+B)=2sinAsinB

sin2A+sin2B+sin2c=4sinAsinBsinC


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