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CosA-CosB-CosC-1-4Cos-B-C-2-Cos-C-A-2-Cos-A-B-2-1-4Cos-A-4-Cos-B-4-Cos-C-4-prove-that-if-A-B-C-

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Question Number 67083 by lalitchand last updated on 22/Aug/19
CosA+CosB+CosC=1+4Cos(((B+C)/2)).Cos(((C+A)/2)).Cos(((A+B)/2))=1+4Cos(((Π−A)/4)).Cos(((Π−B)/4)).Cos(((Π−C)/4)) prove that if A+B+C=Π
CosA+CosB+CosC=1+4Cos(B+C2).Cos(C+A2).Cos(A+B2)=1+4Cos(Π−A4).Cos(Π−B4).Cos(Π−C4)provethatifA+B+C=Π
Answered by Tanmay chaudhury last updated on 23/Aug/19
LHS 2cos(((A+B)/2))cos(((A−B)/2))+1−2sin^2 (C/2) look cos2α=cos^2 α−sin^2 α =1−2sin^2 α or 2cos^2 α−1 now ((A+B)/2)=((π−C)/2)=(π/2)−(C/2) so cos(((A+B)/2))=cos((π/2)−(C/2))=sin(C/2) back to problem 2sin(C/2)cos(((A−B)/2))+1−2sin^2 (C/2) 2sin(C/2){cos(((A−B)/2))−sin((C/2))}+1 2cos(((A+B)/2)){cos(((A−B)/2))−cos(((A+B)/2))}+1 =2cos(((A+B)/2)).2sin((A/2))sin((B/2))+1 =4cos(((A+B)/2))cos(((B+C)/2))cos(((A+C)/2))+1
LHS2cos(A+B2)cos(A−B2)+1−2sin2C2lookcos2α=cos2α−sin2α=1−2sin2αor2cos2α−1nowA+B2=π−C2=π2−C2socos(A+B2)=cos(π2−C2)=sinC2backtoproblem2sinC2cos(A−B2)+1−2sin2C22sinC2{cos(A−B2)−sin(C2)}+12cos(A+B2){cos(A−B2)−cos(A+B2)}+1=2cos(A+B2).2sin(A2)sin(B2)+1=4cos(A+B2)cos(B+C2)cos(A+C2)+1

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