Question

Prove that: $co{s}^2\frac{A}{2}+co{s}^2\frac{B}{2}+co{s}^2\frac{C}{2}=2+\frac{r}{2R}$

Answer 1

We know for any triangle $\Delta$ ABC, $r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$.

Now, LHS will be $2+$ $2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$.

Now, RHS

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

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

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

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

$=2+\left(\cos \frac{\pi -C}{2}\cos \frac{A-B}{2}\right)-{\sin }^2\frac{C}{2}$ [Since, $A+B+C=\pi$]

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

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

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

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

$=2+2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$.

So, RHS$=$ LHS.