If A + B + C = π, prove that

Solution:

= sinA + sinB + sinC

Using,

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

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

Since $A+B+C=\pi$

$\rightarrow \mathrm{B}+\mathrm{C}=180-\mathrm{A}$

And,

$\sin \left(\frac{\pi}{2}-\mathrm{A}\right)=\cos \mathrm{A}$

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

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

Using, $\sin 2 A=2 \sin A \cos A$

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

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

$\rightarrow \mathrm{B}+\mathrm{C}=180-\mathrm{A}$

And

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

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

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

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

= R.H.S