Với A, B, C là 3 góc trong 1 tam giác, chứng minh sin A + sin B + sin C = $4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$ .

sin A + sin B + sin C

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

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

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

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

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

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

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