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


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Question Number 67083 by lalitchand last updated on 22/Aug/19

$CosA + CosB + CosC = 1 + 4 Cos (\frac{B + C}{2}) . Cos (\frac{C + A}{2}) . Cos (\frac{A + B}{2}) = 1 + 4 Cos (\frac{Π - A}{4}) . Cos (\frac{Π - B}{4}) . Cos (\frac{Π - C}{4})$ $prove that if A + B + C = Π$
	Answered by Tanmay chaudhury last updated on 23/Aug/19
	$LHS 2cos(((A+B)/2))cos(((A−B)/2))+1−2sin^2 (C/2) look cos2α=cos^2 α−sin^2 α =1−2sin^2 α or 2cos^2 α−1 now ((A+B)/2)=((π−C)/2)=(π/2)−(C/2) so cos(((A+B)/2))=cos((π/2)−(C/2))=sin(C/2) back to problem 2sin(C/2)cos(((A−B)/2))+1−2sin^2 (C/2) 2sin(C/2){cos(((A−B)/2))−sin((C/2))}+1 2cos(((A+B)/2)){cos(((A−B)/2))−cos(((A+B)/2))}+1 =2cos(((A+B)/2)).2sin((A/2))sin((B/2))+1 =4cos(((A+B)/2))cos(((B+C)/2))cos(((A+C)/2))+1$
	$L H S$ $2 c o s (\frac{A + B}{2}) c o s (\frac{A - B}{2}) + 1 - 2 {s i n}^{2} \frac{C}{2}$ $l o o k c o s 2 α = {c o s}^{2} α - {s i n}^{2} α$ $= 1 - 2 {s i n}^{2} α o r 2 {c o s}^{2} α - 1$ $n o w \frac{A + B}{2} = \frac{π - C}{2} = \frac{π}{2} - \frac{C}{2}$ $s o c o s (\frac{A + B}{2}) = c o s (\frac{π}{2} - \frac{C}{2}) = s i n \frac{C}{2}$ $b a c k t o p r o b l e m$ $2 s i n \frac{C}{2} c o s (\frac{A - B}{2}) + 1 - 2 {s i n}^{2} \frac{C}{2}$ $2 s i n \frac{C}{2} {c o s (\frac{A - B}{2}) - s i n (\frac{C}{2})} + 1$ $2 c o s (\frac{A + B}{2}) {c o s (\frac{A - B}{2}) - c o s (\frac{A + B}{2})} + 1$ $= 2 c o s (\frac{A + B}{2}) . 2 s i n (\frac{A}{2}) s i n (\frac{B}{2}) + 1$ $= 4 c o s (\frac{A + B}{2}) c o s (\frac{B + C}{2}) c o s (\frac{A + C}{2}) + 1$
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