Question

If the triangle $PQR$ varies, then the minimum value of $\cos (P+Q)+\cos (Q+R)+\cos (R+P)$ is

• A. $-\frac{3}{2}$
• B. $\frac{5}{3}$
• C. $\frac{3}{2}$
• D. $-\frac{5}{3}$

Answer 1

The correct option is A $-\frac{3}{2}$

Given $PQR$ is a triangle.

Hence, $P+Q+R=\pi$

$\therefore P+Q=\pi -R,Q+R=\pi -P,P+R=\pi -Q$

$\therefore E=\cos (P+Q)+\cos (Q+R)+\cos (P+R)$
$=\cos (\pi -R)+\cos (\pi -P)+\cos (\pi -Q)=-\cos R-\cos P-\cos Q=-(\cos P+\cos Q+\cos R)$

To minimize the given expression (which is under negative sign), we need to maximize $\cos P+\cos Q+\cos R$.

We know that $\cos P+\cos Q+\cos R$ will be maximum when $\cos P=\cos Q=\cos R$

$\Rightarrow P=Q=R=\frac{\pi }{3}$ (i.e, $PQR$ is an equilateral triangle)

$\therefore E_{min}=-3\cos \left(\frac{\pi }{3}\right)=-\frac{3}{2}$

Hence, option $A$ is correct.