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cosA+cosB+cosC=sinA+sinB+sinC=0 THEN PROVE (1)cos3A+cos3B+cos3C=3cos(A+B+C), (2)sin3A+sin3B+sin3C=3SIN(A+B+C)

GAURAV AGRAWAL , 13 Years ago

Grade 11

1 Answers

Arun Kumar

Hi
$\\A^3+B^3+C^3-3ABC = (A+B+C)(A^2+B^2+C^2-AB-AC-BC) \\$Assume $ \\A \to e^{ia} \\B \to e^{ib} \\C \to e^{ic} \\$fitting them into the equation$ \\e^{i3a}+e^{i3b}+e^{i3c}-3e^{i3a}e^{i3b}e^{i3c}=0$ since $\sum sina=0,\sum cosa=0$ so $\sum e^{ia}=0 \\e^{i3a}+e^{i3b}+e^{i3c}=3e^{i3a}e^{i3b}e^{i3c} \\$equate real and imaginary part we get$ \\\sum sin3a=3 \prod sina \\\sum cos3a=3 \prod cosa$
Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty

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