Question

If the triangle PQR varies, then the minimum value of $\cos \left( P+Q \right)+\cos \left( Q+R \right)+\cos \left( R+P \right)$ is
(a) $\dfrac{-3}{2}$
(b) $\dfrac{5}{3}$
(c) $\dfrac{3}{2}$
(d) $\dfrac{-5}{3}$

Answer 1

We will first try to simplify the given expression $\cos \left( P+Q \right)+\cos \left( Q+R \right)+\cos \left( R+P \right)$ for a triangle PQR. Then we will try to find the minimum value of the simplified expression by considering the best suitable case according to the expression.

Complete step-by-step answer:
We know that for any triangle PQR, the sum of all angles in a triangle is $\pi$.
That is, $P+Q+R=\pi$

Now, we can write the sum of any two angles of a triangle in the form of a third angle. Then,
$\begin{aligned} & P+Q=\pi -R \\\ & Q+R=\pi -P \\\ & R+P=\pi -Q \\\ \end{aligned}$

We should remember the identity $\cos \left( \pi -\theta \right)=-\cos \theta$
Thus, taking cosines over the above angles, we get

& \cos \left( P+Q \right)=\cos \left( \pi -R \right)=-\cos R \\\ & \cos \left( Q+R \right)=\cos \left( \pi -P \right)=-\cos P \\\ & \cos \left( R+P \right)=\cos \left( \pi -Q \right)=-\cos Q\text{ } \\\ \end{aligned} \right\\}\ldots \left( i \right)$$ Now, the expression given to us is $\cos \left( P+Q \right)+\cos \left( Q+R \right)+\cos \left( R+P \right)$. This can be written from equation (i) as $\begin{aligned} & \cos \left( P+Q \right)+\cos \left( Q+R \right)+\cos \left( R+P \right) \\\ & =-\cos R-\cos P-\cos Q \\\ & =-\left( \cos P+\cos Q+\cos R \right) \\\ \end{aligned}$ Now to find the minimum value of $\cos \left( P+Q \right)+\cos \left( Q+R \right)+\cos \left( R+P \right)$ , We must find the minimum value of $-\left( \cos P+\cos Q+\cos R \right)$, That is, we must find the maximum value of $+\left( \cos P+\cos Q+\cos R \right)$ Now, as $P+Q+R=\pi $, The values of cosines are maximum when $P=Q=R=\dfrac{\pi }{3}$, that is, when triangle PQR is an equilateral triangle. Thus, $\cos P=\cos Q=\cos R=\cos \dfrac{\pi }{3}=\dfrac{1}{2}$ Therefore, the maximum value of $+\left( \cos P+\cos Q+\cos R \right)$ is $\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}$ That is, minimum value of $-\left( \cos P+\cos Q+\cos R \right)$ is $\dfrac{-3}{2}$ Hence, the minimum value of $\cos \left( P+Q \right)+\cos \left( Q+R \right)+\cos \left( R+P \right)$ is $\dfrac{-3}{2}$ **So, the correct answer is “Option A”.** **Note:** While solving such questions, we should have an approach in mind. We should have a rough idea of how to solve the questions in our mind and then we should assess the trigonometric identities that will be required to solve the question. This helps us to form our answer neatly and avoid mistakes. Because it is easy to confuse the intervals in which our identity is valid. For example, the identity $\cos \left( \pi -\theta \right)=-\cos \theta $is valid only for the interval $\left[ \dfrac{\pi }{2},\pi \right]$ which becomes a key factor in solving the questions.