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Question Number 86540 by jagoll last updated on 29/Mar/20
prove that cos ((A/2))+cos ((B/2))+cos ((C/2)) = 4 cos (((π+A)/4))cos (((π+B)/4))cos (((π−C)/4)) where A+B+C = π
provethatcos(A2)+cos(B2)+cos(C2)=4cos(π+A4)cos(π+B4)cos(π−C4)whereA+B+C=π
Commented by jagoll last updated on 29/Mar/20
RHS 2 { 2cos (((π+A)/4))cos (((π+B)/4))} cos (((π−C)/4)) 2 {cos (((2π+A+B)/4))+cos (((A−B)/4))} cos (((π−C)/4)) 2 {−sin (((A+B)/4)) + cos (((A−B)/4))}sin (((π+C)/4)) 2{cos (((A−B)/4))sin (((π+C)/4))−sin (((A+B)/4))sin (((π+C)/4))}
RHS2{2cos(π+A4)cos(π+B4)}cos(π−C4)2{cos(2π+A+B4)+cos(A−B4)}cos(π−C4)2{−sin(A+B4)+cos(A−B4)}sin(π+C4)2{cos(A−B4)sin(π+C4)−sin(A+B4)sin(π+C4)}
Commented by john santu last updated on 29/Mar/20
A+B+C = π⇒B−C = π−(A+C) LHS A = π−(B+C) (A/2) = (π/2)−(((B+C)/2)) cos ((A/2)) = sin (((B+C)/2)) ⇒cos ((B/2))+cos ((C/2)) = 2cos (((B+C)/4))cos (((B−C)/4)) sin (((B+C)/2)) = 2sin (((B+C)/4))cos (((B+C)/4)) ⇒2cos (((B+C)/4)) {sin (((B+C)/4))+cos (((B−C)/4))} 2 cos (((π−A)/4)) {sin (((π−A)/4))+cos (((π−(A+C))/4))}
A+B+C=π⇒B−C=π−(A+C)LHSA=π−(B+C)A2=π2−(B+C2)cos(A2)=sin(B+C2)⇒cos(B2)+cos(C2)=2cos(B+C4)cos(B−C4)sin(B+C2)=2sin(B+C4)cos(B+C4)⇒2cos(B+C4){sin(B+C4)+cos(B−C4)}2cos(π−A4){sin(π−A4)+cos(π−(A+C)4)}
Commented by jagoll last updated on 29/Mar/20
sorry sir A+B+C = π
sorrysirA+B+C=π

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