Question

If $A+B+C=\pi$, then $\sin A+\sin B-\sin C=2\sin \frac{A}{2}\sin \frac{B}{2}\cos \frac{C}{2}$.

• A. True
• B. False

Answer 1

The correct option is B False

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

Since, $A+B+C=\pi$

Then, $A+B=\pi -C$

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

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

Also, $\sin C=2\sin \left(\frac{C}{2}\right)\cos \left(\frac{C}{2}\right)$ ----- ( 3 )

Substituting ( 2 ) and ( 3 ) in ( 1 ) we get,

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

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

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

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

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

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

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

$\Rightarrow$ $4\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)\cos \left(\frac{C}{2}\right)$ [ As $\sin (-x)=-sinx$ ]

$\therefore$ We can see given statement is false.