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Question

If A+B+C=π and cosA+cosB+cosC=0=sinA+sinB+sinC then cos3A+cos3B+cos3C=1

A
True
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B
False
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Solution

The correct option is B False
Given

sinA+sinB+sinC=0

cosA+cosB+cosC=0

Let Z1=eiA,Z2=eiB,Z3=eiC

Consider

Z1+Z2+Z3

eiA+eiB+eiC

cosA+isinA+cosB+isinB+cosC+isinC

(cosA+cosB+cosC)+i(sinA+sinB+sinC)

0

As Z1+Z2+Z3=0

Z31+Z32+Z333Z1Z2Z3=0

ei3A+ei3B+ei3C=3ei(A+B+C)

cos3A+cos3B+cos3C=3cos(A+B+C)=3cosπ=3

So the given relation is False

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