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Question Number 101382 by student work last updated on 02/Jul/20

Answered by Rio Michael last updated on 02/Jul/20

we know or we are to prove that sin A + sinB + sin C = 4 cos(A/2) cos(B/2) cos(C/2) ((1) but A + B + C = π(180°) ⇒ A + B = π−C and ((A +B)/2) = ((π − C)/2) ⇒ ((A + B)/2) = (π/2)−(C/2) so sin (((A + B)/2)) = cos((π/2) −(C/2)) = cos ((C/2)) (2) also sin C = 2 sin ((C/2)) cos ((C/2))......(3) but sin A+ sin B + sinC = 2 cos((C/2))cos (((A − B)/2)) + 2 sin ((C/2)) cos((C/2)) = 2 cos ((C/2))[cos (((A − B)/2)) + sin ((C/2))] = 2 cos ((C/2))[cos (((A−B)/2)) + sin (((π−(A + B))/2))] since A + B + C = π = 2 cos((C/2))[cos (((A−B)/2)) + sin ((π/2)−((A + B)/2))] = 2 cos ((C/2))[cos (((A−B)/2)) + cos (((A + B)/2))] = 2 cos ((C/2))[2cos (((((A −B)/2) + ((A +B)/2))/2)) cos(((((A−B)/2)−((A +B)/2))/2))] = 2 cos ((C/2))[2cos((A/2))cos (−(B/2))] = 4 cos ((A/2))cos ((B/2)) cos ((C/2)) since cos x is an even function. QED

weknoworwearetoprovethatsinA+sinB+sinC=4cosA2cosB2cosC2((1)butA+B+C=π(180°)⇒A+B=π−CandA+B2=π−C2⇒A+B2=π2−C2sosin(A+B2)=cos(π2−C2)=cos(C2)(2)alsosinC=2sin(C2)cos(C2)......(3)butsinA+sinB+sinC=2cos(C2)cos(A−B2)+2sin(C2)cos(C2)=2cos(C2)[cos(A−B2)+sin(C2)]=2cos(C2)[cos(A−B2)+sin(π−(A+B)2)]sinceA+B+C=π=2cos(C2)[cos(A−B2)+sin(π2−A+B2)]=2cos(C2)[cos(A−B2)+cos(A+B2)]=2cos(C2)[2cos(A−B2+A+B22)cos(A−B2−A+B22)]=2cos(C2)[2cos(A2)cos(−B2)]=4cos(A2)cos(B2)cos(C2)sincecosxisanevenfunction.QED

Answered by 1549442205 last updated on 02/Jul/20

sinA+sinB=2sin((A+B)/2)cos((A−B)/2)=2cos(C/2)cos((A−B)/2) sinC=2sin(C/2)cos(C/2)=2cos(C/2)cos((A+B)/2).Hence, sinA+sinB+sinC=2cos(C/2)(cos((A−B)/2)+cos((A+B)/2)) =2cos(C/2).2cos((((A−B)/2)+((A+B)/2))/2)cos((((A−B)/2)−((A+B)/2))/2) =4cos(C/2)cos(A/2)cos((−B)/2)=4cos(A/2)cos(B/2)cos(C/2) (q.e.d)

sinA+sinB=2sinA+B2cosA−B2=2cosC2cosA−B2sinC=2sinC2cosC2=2cosC2cosA+B2.Hence,sinA+sinB+sinC=2cosC2(cosA−B2+cosA+B2)=2cosC2.2cosA−B2+A+B22cosA−B2−A+B22=4cosC2cosA2cos−B2=4cosA2cosB2cosC2(q.e.d)

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