Math, asked by aswin1128, 10 months ago

can anyone solve this sum

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Answered by yuvrajsinhz588

1

A+B+C=π

A+B=π−C

cos3A+cos3B+cos3C=1⟶

2cos3A+3B2cos3A−3B2+cos3C=1

2cos3A+3B2cos3A−3B2+cos(3π2−3A−3B2)=1 [since A−B=π−C]

2cos3A+3B2cos3A−3B2−sin3A+3B2=1

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