Question

With the usual notations, prove that : $2\{a{\sin }^2\frac{C}{2}+c{\sin }^2\frac{A}{2}\}=(a+c-b)$.

Answer 1

We know,

${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$

$L.H.S=2\{a{\sin }^2\frac{C}{2}+c{\sin }^2\frac{A}{2}\}$

$=2\{\frac{a(1-\cos C)}{2}+\frac{c(1-\cos A)}{2}\}$

$=a-a\cos C+c-c\cos A$

$=(a+c)-(a\cos C+c\cos A)$

$=a+c-b$