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Question

If A+B+C=π2, then show that sin2A+sin2B+sin2C=12sinAsinBsinC and cos2A+cos2B+cos2C=2+2sinAsinBsinC.

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Solution

A+B+C=π2
sin2A+sin2B+sin2C=1cos2A2+1cos2B2+1cos2C2

=12(3(cos2A+cos2B+cos2C))

=12(3(2cos(A+B)cos(AB)+cos(π2(A+B))))

=12(3(2cos(A+B)cos(AB)cos2(A+B)))

=12(32cos(A+B)cos(AB)2cos2(A+B)1)

=12(22cos(A+B)(cos(AB)cos(A+B)))

=12sinAsinBsinC
cos2A+cos2B+cos2C=cos2A+12+cos2B+12+cos2C+12

=12(cos2A+cos2B+cos2C+3)

=12(2cos(A+B)cos(AB)+cos(π2(A+B))+3)

=12(2cos(A+B)cos(AB)cos2(A+B)+3)

=12(2cos(A+B)cos(AB)cos2(A+B)+3)

=12(2cos(A+B)cos(AB)2cos2(A+B)+4)

=12(2cos(A+B)(cos(AB)cos(A+B))+4)
=2+2sinAsinBsinC

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