\operatorname { cos } A + \operatorname { cos } B + \operatorname { co

1.	\operatorname { cos } A + \operatorname { cos } B + \operatorname { cos } C = 1 + 4 \operatorname { sin } \frac { A } { 2 } \operatorname { sin } \frac { B } { 2 } \operatorname { sin } \frac { C } { 2 }
Answer» LHS = ( cos A + cos B ) + cos C = { 2 · cos[ ( A+B) / 2 ] · cos [ ( A-B) / 2 ] } + cos C = { 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) = RHS ..

\operatorname { cos } A + \operatorname { cos } B + \operatorname { cos } C = 1 + 4 \operatorname { sin } \frac { A } { 2 } \operatorname { sin } \frac { B } { 2 } \operatorname { sin } \frac { C } { 2 }

Answer»

LHS

= ( cos A + cos B ) + cos C

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

= { 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)

= RHS ..