Question

In a $\Delta ABC$ prove that $\mathrm{cosA}+\cos B+\cos C=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

Answer 1

Consider the L.H.S

$=\cos A+\cos B+\cos C$ ……………… (1)

We know that

$\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$

Therefore,

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

Since,

$A+B+C=\pi$

$\cos \theta =1-2{\sin }^2\frac{\theta }{2}$

Therefore,

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

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

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

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

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

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

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

We know that

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

Therefore,

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

$=1+2\sin \frac{C}{2}\left(2\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)\right)$

$=1+4\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)\sin \left(\frac{C}{2}\right)$

Hence, proved.