Question

If A + B + C = π, then prove that $cos^{2}\frac{A}{2} + cos^{2}\frac{B}{2} + cos^{2}\frac{C}{2} = 2(1 + sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2})$.

Answer 1

Answer:

Step-by-step explanation:

if A + B + C = π

then to prove

$cos^{2}\frac{A}{2} + cos^{2}\frac{B}{2}+ cos^{2}\frac{C}{2} = 2 (1+sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2})$

let us take LHS

$=cos^{2}\frac{A}{2} + cos^{2}\frac{B}{2} +cos^{2}\frac{C}{2}$

as we know from half angle formula

$cos^{2}\frac{A}{2} =\frac{1}{2} (1+cos\:A)$

so put all these values

$=\frac{1}{2} (1+cos\:A)+\frac{1}{2} (1+cos\:B)+\frac{1}{2} (1+cos\:C)\\\\=\frac{3}{2}+\frac{1}{2}(cosA+cosB+cosC)\\\\=\frac{3}{2}+\frac{1}{2}[(2cos\frac{A+B}{2}cos\frac{A-B}{2}+cos\:C]\\\\\\=\frac{3}{2}+(cos\frac{\pi-C}{2}cos\frac{A-B}{2}+\frac{1}{2}(1-2sin^{2}\frac{C}{2}]$

∵[(A+B)/2=π/2-C/2 and cosC=1-2sin²C/2]

$=2+sin\frac{C}{2} cos\frac{A-B}{2} -sin^{2}\frac{C}{2}\\ \\\\=2+sin\frac{C}{2} [cos\frac{A-B}{2} -sin\frac{C}{2}]\\ \\\\=2+sin\frac{C}{2} [cos\frac{A-B}{2} -sin(\frac{\pi}{2}-\frac{A+B}{2}]$

$=2+sin\frac{C}{2} [cos\frac{A-B}{2} -cos(\frac{A+B}{2})]\\\\\\=2+sin\frac{C}{2}(2 sin\frac{A}{2} sin\frac{B}{2})\\\\=2(1+sin\frac{C}{2}sin\frac{A}{2} sin\frac{B}{2})\\\\\\=RHS$