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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

$$\mathrm{CosA}+\mathrm{CosB}+\mathrm{CosC}=\mathrm{1}+\mathrm{4Cos}\left(\frac{\mathrm{B}+\mathrm{C}}{\mathrm{2}}\right).\mathrm{Cos}\left(\frac{\mathrm{C}+\mathrm{A}}{\mathrm{2}}\right).\mathrm{Cos}\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)=\mathrm{1}+\mathrm{4Cos}\left(\frac{\Pi−\mathrm{A}}{\mathrm{4}}\right).\mathrm{Cos}\left(\frac{\Pi−\mathrm{B}}{\mathrm{4}}\right).\mathrm{Cos}\left(\frac{\Pi−\mathrm{C}}{\mathrm{4}}\right) \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{if}\:\mathrm{A}+\mathrm{B}+\mathrm{C}=\Pi \\ $$

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

$${LHS} \\ $$$$\mathrm{2}{cos}\left(\frac{{A}+{B}}{\mathrm{2}}\right){cos}\left(\frac{{A}−{B}}{\mathrm{2}}\right)+\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \frac{{C}}{\mathrm{2}} \\ $$$${look}\:\:{cos}\mathrm{2}\alpha={cos}^{\mathrm{2}} \alpha−{sin}^{\mathrm{2}} \alpha \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \alpha\:\:\:{or}\:\:\mathrm{2}{cos}^{\mathrm{2}} \alpha−\mathrm{1} \\ $$$${now}\:\frac{{A}+{B}}{\mathrm{2}}=\frac{\pi−{C}}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}−\frac{{C}}{\mathrm{2}} \\ $$$${so}\:{cos}\left(\frac{{A}+{B}}{\mathrm{2}}\right)={cos}\left(\frac{\pi}{\mathrm{2}}−\frac{{C}}{\mathrm{2}}\right)={sin}\frac{{C}}{\mathrm{2}} \\ $$$${back}\:{to}\:{problem} \\ $$$$\mathrm{2}{sin}\frac{{C}}{\mathrm{2}}{cos}\left(\frac{{A}−{B}}{\mathrm{2}}\right)+\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \frac{{C}}{\mathrm{2}} \\ $$$$\mathrm{2}{sin}\frac{{C}}{\mathrm{2}}\left\{{cos}\left(\frac{{A}−{B}}{\mathrm{2}}\right)−{sin}\left(\frac{{C}}{\mathrm{2}}\right)\right\}+\mathrm{1} \\ $$$$\mathrm{2}{cos}\left(\frac{{A}+{B}}{\mathrm{2}}\right)\left\{{cos}\left(\frac{{A}−{B}}{\mathrm{2}}\right)−{cos}\left(\frac{{A}+{B}}{\mathrm{2}}\right)\right\}+\mathrm{1} \\ $$$$=\mathrm{2}{cos}\left(\frac{{A}+{B}}{\mathrm{2}}\right).\mathrm{2}{sin}\left(\frac{{A}}{\mathrm{2}}\right){sin}\left(\frac{{B}}{\mathrm{2}}\right)+\mathrm{1} \\ $$$$=\mathrm{4}{cos}\left(\frac{{A}+{B}}{\mathrm{2}}\right){cos}\left(\frac{{B}+{C}}{\mathrm{2}}\right){cos}\left(\frac{{A}+{C}}{\mathrm{2}}\right)+\mathrm{1} \\ $$

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