Question

If $A+B+C={180}^{\circ }$, then the value of $\sin 2A-\sin 2B+\sin 2C=$

• A. $2\sin A\cos B\sin C$
• B. $2\cos A\sin B\cos C$
• C. $4\sin A\cos B\sin C$
• D. $4\cos A\sin B\cos C$

Answer 1

The correct option is D $4\cos A\sin B\cos C$

$\sin 2A-\sin 2B+\sin 2C$
$\Rightarrow 2\cos (180-C)\sin (A-B)+\sin 2C$
$=-2\cos C\sin (A-B)+2\sin C\cos C$
$=2\cos C[-\sin (A-B)+\sin (180-A-B)]$
$=2\cos C-\sin (A-B)+\sin (A+B)]$
$=2\cos C\times 2\cos A\sin B$
$=4\cos A\sin B\cos C$
Hence, option ${}^{\prime }{D}^{\prime }$ is correct.