Question

$A+B+C={180}^0$ find the value of $sin2A-sin2B+sin2C$

Answer 1

We have,

$A+B+C={180}^0.......\left(1\right)$

Then, the value of

$\sin 2A-\sin 2B+\sin 2C=?$

Solve that:-

$\sin 2A-\sin 2B+\sin 2C$

$=\sin 2A+\sin 2C-\sin 2B$

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

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

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

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

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

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

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

$=2\sin B\left[2\sin A\sin C\right]$

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

Hence, this is the answer.