Question

In $△ABC$, prove that ${(a-b)}^2{\cos }^2\frac{C}{2}+{(a+b)}^2{\sin }^2\frac{C}{2}=c^{2.}$.

Answer 1

L.H.S

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

$=a^2{\sin }^2\frac{C}{2}+b^2{\sin }^2\frac{C}{2}+2ab{\sin }^2\frac{C}{2}+a^2{\cos }^2\frac{C}{2}+b^2{\cos }^2\frac{C}{2}-2ab{\cos }^2\frac{C}{2}$

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

$=a^2+b^2-2ab\cos C$ $\because \cos 2x={\cos }^{2}x-{\sin }^{2}x$

$=a^2+b^2-2ab\cdot \frac{a^2+b^2-c^2}{2ab}$ $\because cosc=\frac{a^2+b^2-c^2}{2ab}$

$=a^2+b^2-a^2-b^2+c^2$

$=c^2$