Question

If $A+B+C={180}^0$, then

(i)$\cos 2A+\cos 2B-\cos 2C=1-4\sin A\sin B\sin C$

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

• A. True
• B. False

Answer 1

The correct option is A True

$\left(i\right)$ $\cos 2A+\cos 2B-\cos 2C=1-4\sin A\sin B\sin C$

$\Rightarrow$ $L.H.S=\cos 2A+\cos 2B-\cos 2C$

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

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

$=\cos 2A-2\sin ({180}^o-A)\sin (B-C)$ [ $B+C={180}^o-A$ ]

$=\cos 2A-2(\sin {180}^o\cos A-\sin A\cos {180}^o)\sin (B-C)$

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

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

$=1-2\sin A[\sin A-\sin (B-C\left)\right]$

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

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

$=1-4\sin A\cos \left(\frac{{180}^o-C-C}{2}\right).\sin \left(\frac{{180}^o-B-B}{2}\right)$ [ $AB={180}^o-C$ and $A+C={180}^o-B$ ]

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

$=1-4\sin A\cos (-C).\sin (-B)$

$=1-4\sin A\sin B\sin C$

$\therefore$ $\cos 2A+\cos 2B-\cos 2C=1-4\sin A\sin B\sin C$ ---- Hence proved

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

$\Rightarrow$ $L.H.S=\sin A-\sin B+\sin C$

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

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

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

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

$=4\sin \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}$

$\therefore$ $\sin A-\sin B+\sin C=4\sin \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}$