sina+sinb+sinc=4cosa/2 cosb/2 cosc/2

Với $A,\ B,\ C$ là $3$ góc của một tam giác

Ta có:

$\quad \sin A + \sin B + \sin C$

$= 2\sin\dfrac{A+B}{2}\cdot\cos\dfrac{A-B}{2} + 2\sin\dfrac C2\cdot\cos\dfrac C2$

$= 2\sin\dfrac{\pi - C}{2}\cdot\cos\dfrac{A-B}{2} + 2\cos\dfrac{\pi - C}{2}\cdot\cos\dfrac C2$

$= 2\cos\dfrac C2\cdot\cos\dfrac{A-B}{2} + 2\cos\dfrac{A+B}{2}\cdot\cos\dfrac C2$

$= 2\cos\dfrac C2\left(\cos\dfrac{A-B}{2} + \cos\dfrac{A+B}{2}\right)$

$= 2\cos\dfrac C2\cdot 2\cos\dfrac{A-B+A+B}{4}\cdot\cos\dfrac{A-B-(A+B)}{4}$

$= 4\cos\dfrac C2\cdot\cos\dfrac{A}{2}\cdot\cos\dfrac{-B}{2}$

$= 4\cos\dfrac A2\cdot\cos\dfrac B2\cdot\cos\dfrac C2$