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Grade: 11
        
In a triangle ABC,  prove that a3cos(B-C)+b3cos(C-A)+c3cos(A-B) = 3abc
3 years ago

Answers : (1)

Lab Bhattacharjee
121 Points
							
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)
 
 
3 years ago
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