Question

In any triangles ABC prove the identities.
$\cos A+\cos B+\cos C=1+4\sin \left(A/2\right)\sin \left(B/2\right)\sin \left(C/2\right)$.

Answer 1

Here we want $+1$ and so we write
$\cos A=1-2{\sin }^2\left(A/2\right)$
L.H.S. $=1-2{\sin }^2\left(A/2\right)+2\cos \left\{\right(B+C\left)/2\right\}\cos \left\{\right(B-C\left)/2\right\}$
$=1-2{\sin }^2\left(A/2\right)+2\sin \left(A/2\right)\cos \left\{\right(B-C\left)/2\right\}$
$=1-2\sin \left(A/2\right)\left[\sin \right(A/2)-\cos \{(B-C)/2\left\}\right]$
$=1-2\sin \left(A/2\right)\left[\cos \right\{(B+C)/2\}-\cos \{(B-C)/2\left\}\right]$
$=1-2\sin \left(A/2\right)\left[2\sin \right(B/2\left)\sin \right(-C/2\left)\right]$
$=1+4\sin \left(A/2\right)\sin \left(B/2\right)\sin \left(C/2\right)$
$\therefore \sin (-\theta )=-\sin \theta$.