In a triangle ABC, prove that a 3 cos(B-C)+b 3 cos(C-A)+c 3 cos(A-B) = 3abc

In a triangle ABC,  prove that a3cos(B-C)+b3cos(C-A)+c3cos(A-B) = 3abc


1 Answers

Lab Bhattacharjee
121 Points
7 years ago
As a=2R sinA,   sin A=sin{pi-(B+C)}=sin(B+C)
a^3cos(B-C)= a^2 2RsinA\cos(B-C)=a^2R[2sin(B+C)cos(B-C)]
=a^2R[sin2B+sin2C]=a^2[2RsinB cosB+2R sinC cosC ]=a^2[b cosB+c cos C]
=\sum a^3\cos(B-C)=\sum(a^2 b \cos B+ ab^2 \cos A)=\sum ab(a\cos B+b\cos A)=\sum ab(c)

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