Question

a cos A+bcos B+c cos C/a+b+c

Answer 1

=> acosA+bcosB+ccosC/a+b+c

=>a+b+c(cosA+cosB+cosC)/a+b+c

=>cosA+cosB+cosC

=> ( cos A + cos B ) + cos C

=> { 2 · cos[ ( A+B) / 2 ] · cos [ ( A-B) / 2 ] } + cosC

=> { 2 · cos [ (π/2) - (C/2) ] · cos [ (A-B) / 2 ] } + cos C

=> { 2 · sin( C/2 ) · cos [ (A-B) / 2 ] } + { 1 - 2 · sin² ( C/2 ) }

=>1 + 2 sin ( C/2 )· { cos [ (A -B) / 2 ] - sin ( C/2 ) }

= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - sin [ (π/2) - ( (A+B)/2 ) ] }

= 1 + 2 sin ( C/2 )· { cos [ (A-B) / 2 ] - cos [ (A+B)/ 2 ] } 

= 1 + 2 sin ( C/2 )· 2 sin ( A/2 )· sin( B/2 ) ... ... ... (2)

= 1 + 4 sin(A/2) sin(B/2) sin(C/2)