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cosA+cosB+cosC=sinA+sinB+sinC=0 THEN PROVE cos3A+cos3B+cos3C=3cos(A+B+C) sin3A+sin3B+sin3C=3SIN(A+B+C)

cosA+cosB+cosC=sinA+sinB+sinC=0
THEN PROVE
cos3A+cos3B+cos3C=3cos(A+B+C)
sin3A+sin3B+sin3C=3SIN(A+B+C)

Grade:11

2 Answers

mycroft holmes
272 Points
9 years ago

Kaushik Kumar
32 Points
9 years ago

cos A + cos B + cos C=0 ; i*(sin A + sin B + sin C)=0

cos A + i sin A + cos B + i sin B + cos C + i sin C=0

Z1 + Z2 + Z3 = 0

now use this identity

a^3 + b^3 + c^3 - 3abc = (a+b+c) (a^2 + b^2 + c^2 - ab - bc - ac)

Z1^3 + Z2^3 + Z3^3 - a Z1 Z2 Z3 = 0

cos 3A + i sin 3A + cos 3B + i sin 3B + cos 3C + i sin 3C = 3 }cos (A+B+C) + i sin (A+B+C)}

now compare real and imaginary parts.


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