# If A,B,C are the angles of a triangle then cosA+cosB+cosC=?

Nov 2, 2017

$\cos A + \cos B + \cos C = 1 + 4 \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)$

#### Explanation:

As$A , B$ and$C$ are angles of a triangle, we have

$A + B + C = 180$ and $A + B = 180 - C$ or $\frac{A + B}{2} = 90 - \frac{C}{2}$

Hence $\cos A + \cos B + \cos C$

= $2 \cos \left(\frac{A + B}{2}\right) \cos \left(\frac{A - B}{2}\right) + 1 - 2 {\sin}^{2} \left(\frac{C}{2}\right) - 1$

= $2 \cos \left(90 - \frac{C}{2}\right) \cos \left(\frac{A - B}{2}\right) + 1 - 2 {\sin}^{2} \left(\frac{C}{2}\right)$

= $2 \sin \left(\frac{C}{2}\right) \cos \left(\frac{A - B}{2}\right) + 1 - 2 {\sin}^{2} \left(\frac{C}{2}\right)$

= $2 \sin \left(\frac{C}{2}\right) \left(\cos \left(\frac{A - B}{2}\right) - \sin \left(\frac{C}{2}\right)\right) + 1$

= $2 \sin \left(\frac{C}{2}\right) \left(\cos \left(\frac{A - B}{2}\right) - \cos \left(\frac{A + B}{2}\right)\right) + 1$

= $2 \sin \left(\frac{C}{2}\right) \left(2 \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right)\right) + 1$

= $1 + 4 \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)$