Question

If $A,B,C$are the angles of a triangle then find $\cos A+\cos B+\cos C$.

Answer 1

Find the required value:

$$\begin{gathered} \cos A+\cos B+\cos C \\ =(\cos A+\cos B)+\cos C \\ =\left[2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)\right]+\cos C \end{gathered}$$ ;$\left[\cos \left(A\right)+\cos \left(B\right)=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)\right]$

Given: $A,B,C$are the angles of a triangle.

Thus, $A+B+C=\mathrm{\pi }$

$\begin{array}{rcl}\cos A+\cos B+\cos C& =& \left[2\cos \left\{\left(\frac{\mathrm{\pi }}{2}\right)-\left(\frac{C}{2}\right)\right\}.\cos \left(\frac{A-B}{2}\right)\right]+\cos C\left[\cos \left(\frac{\mathrm{\pi }}{2}-\theta \right)=\sin \left(\theta \right)\right]\\ & =& \left[2\sin \left(\frac{C}{2}\right).\cos \left(\frac{A-B}{2}\right)\right]+\left\{1-2{\sin }^2\left(\frac{C}{2}\right)\right\}\left[\cos \left(2A\right)=1-2{\sin }^2\left(A\right)\right]\\ & =& 1+2\sin \left(\frac{C}{2}\right).\left[\cos \left\{\frac{A-B}{2}\right\}-\sin \left\{\left(\frac{\mathrm{\pi }}{2}\right)-\frac{(A+B)}{2}\right\}\right]\left[\sin \left(\frac{\mathrm{\pi }}{2}-\theta \right)=\cos \left(\theta \right)\right]\\ & =& 1+2\sin \left(\frac{C}{2}\right).\left[\cos \left\{\frac{A-B}{2}\right\}-\cos \left\{\frac{(A+B)}{2}\right\}\right]\left[\cos \left(A\right)-\cos \left(B\right)=2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{B-A}{2}\right)\right]\\ & =& 1+2\sin \left(\frac{C}{2}\right).2\sin \left(\frac{A}{2}\right).\sin \left(\frac{B}{2}\right)\\ & =& 1+4\sin \left(\frac{A}{2}\right).\sin \left(\frac{B}{2}\right).\sin \left(\frac{C}{2}\right)\end{array}$

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