Question

In $\Delta ABC$ prove that $(a-b{)}^2{\cos }^2(c/2)+(a+b{)}^2$ ${\sin }^2\left(c/2\right)=c^2$

Answer 1

$LHS=(a-b{)}^2{\cos }^2(C/2)+(a+b{)}^2{\sin }^2\left(C/2\right)$

$=(a^2+b^2-2ab){\cos }^2\left(C/2\right)+(a^2+b^2+2ab){\sin }^2\left(C/2\right)$

$=(a^2+b^2)\left({\cos }^2\right(C/2)+{\sin }^2(C/2\left)\right)-2ab\left({\cos }^2\right(C/2)-{\sin }^2(C/2\left)\right)$

$=(a^2+b^2)\left(1\right)-2ab\left(\cos 2\right(C/2\left)\right)$

$=a^2+b^2-2ab\cos C=c^2$