Question

If $\cos A+\cos B=4{\sin }^2\left(\frac{C}{2}\right)$,prove that sides $a,b,c$ of the triangle $ABC$ are in A.P.

Answer 1

$\cos A+\cos B=4{\sin }^2\left(\frac{C}{2}\right)$

$2\cos \frac{A+B}{2}\cos \frac{A-B}{2}=4{\sin }^2\left(\frac{C}{2}\right)$

$A+B+C=\pi \Rightarrow A+B=\pi -C$

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

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

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

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

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

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

$\sin A+\sin B=2\sin C$

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$

$\sin A=ak,\sin B=bk,\sin C=ck$

$ak+bk=2\left(ck\right)$

$a+b=2c$

Therefore the sides of triangle a,b,c are in A.P.