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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)

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