In a triangle ABC, prove that a3cos(B-C)+b3cos(C-A)+c3cos(A-B) = 3abc
In a triangle ABC, prove that a3cos(B-C)+b3cos(C-A)+c3cos(A-B) = 3abc
Satviki Pathak , 10 Years ago
Grade 11
Trigonometry> In a triangle ABC, prove that a 3 cos(B-C...
1 Answers
Lab Bhattacharjee
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]
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