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Question

Say true or false:
If A+B+C=π, then cos2A+cos2B+cos2C+4cosAcosBcosC is positive.

A
True
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B
False
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Solution

The correct option is B False
As A+B+C=π

cos2A+cos2B+cos2C

=2cos(A+B)cos(AB)+2cos2C1

=2cos(πC)cos(AB)+2cos2C1 since cosC+cosD=2cosC+D2cosCD2,cos2x=2cos2x1

=2cosC(cos(AB)+cos(A+B))1

=2cosC(cos(AB)+cos(A+B))1

=4cosAcosBcosC1

cos2A+cos2B+cos2C=4cosAcosBcosC1

Hence cos2A+cos2B+cos2C+4cosAcosBcosC=1<0.

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