Question

IF $\cos A+\cos B+\cos C=1+p\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)\sin \left(\frac{C}{2}\right),$then p =.

Answer 1

Given $A+B+C=\pi$ (angles of $△$)

$\cos A+\cos B+\cos C$

$=2\cos \frac{(A+B)}{2}.\cos \frac{(A-B)}{2}+\cos C$ (standard formula)

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

$=2\sin \frac{C}{2}\cos \frac{(A-B)}{2}+1-2{\sin }^2\left(\frac{C}{2}\right)$ $[\because \cos (\frac{\pi }{2}-x)=\sin xstandardformula]$

$=1+2\sin \frac{C}{2}[\cos \frac{(A-B)}{2}-\sin \frac{(\pi -(A+B\left)\right)}{2}]$

$=1+2\sin \frac{C}{2}[\cos \frac{(A-B)}{2}-\cos \frac{(A+B)}{2}]$ [standard formula]

$=1+2\sin \frac{C}{2}.2\sin \frac{A}{2}\sin \frac{B}{2}$

$=1+4\sin \frac{C}{2}.\sin \frac{A}{2}.\sin \frac{B}{2}$

$\Rightarrow p=4$