Question

In $△ABC$ with the usual notations prove that
$(a-b{)}^2{\cos }^2\left(\frac{C}{2}\right)+(a+b{)}^2{\sin }^2\left(\frac{C}{2}\right)=c^2$

Answer 1

By the projection rule in $△ABC$
$a=c\cos B+b\cos C$
$b=a\cos C+c\cos A$
$a+b=c\cos B+b\cos C+a\cos C+c\cos A$
$(a+b)(1-\cos C)=c(\cos B+\cos A)$
$(a+b)\left(2{\sin }^2\left(\frac{C}{2}\right)\right)=c(\cos B+\cos A)$
$(a+b{)}^2\cdot {\sin }^2\left(\frac{C}{2}\right)=\frac{c}{2}(\cos B+\cos A\left)\right(a+b)$ .... (i)
and $a-b=c\cos B+b\cos C-a\cos C-c\cos A$
$(a-b)(1+\cos C)=c(\cos B-\cos A)$
$(a-b)\left(2{\cos }^2\left(\frac{C}{2}\right)\right)=c(\cos B-\cos A)$
$(a-b{)}^2{\cos }^2\left(\frac{C}{2}\right)=\frac{c}{2}(\cos B-\cos A\left)\right(a-b)$ .... (ii)
Adding equation (i) and (ii), we get
$(a-b{)}^2{\cos }^2\left(\frac{C}{2}\right)+(a+b{)}^2{\sin }^2\left(\frac{C}{2}\right)=\frac{c}{2}[a\cos B+a\cos A+b\cos B+b\cos A+...a\cos B-a\cos A-b\cos B+b\cos A]$
$\therefore (a-b{)}^2{\cos }^2\left(\frac{C}{2}\right)+(a+b{)}^2{\sin }^2\left(\frac{C}{2}\right)=\frac{c}{2}.2(a\cos B+b\cos A)$
$\therefore (a-b{)}^2{\cos }^2\left(\frac{C}{2}\right)+(a+b{)}^2{\sin }^2\left(\frac{C}{2}\right)=c^2[\because c=a\cos B+b\cos A]$