Minimum Value of Sum of Cosines of Angle Sums in Triangle

To find the minimum possible value of the expression $S = \cos(P + Q) + \cos(Q + R) + \cos(R + P)$ where $P$, $Q$, and $R$ are the interior angles of a triangle $PQR$, we begin by using the property that the sum of the angles in any triangle is $\pi$: $P + Q + R = \pi$

Using this identity, we can rewrite each term in the sum: $\cos(P + Q) = \cos(\pi - R) = -\cos R$ $\cos(Q + R) = \cos(\pi - P) = -\cos P$ $\cos(R + P) = \cos(\pi - Q) = -\cos Q$

Substituting these back into the expression for $S$: $S = -\cos R - \cos P - \cos Q = -(\cos P + \cos Q + \cos R)$

To find the minimum value of $S$, we need to find the maximum possible value of the sum $\cos P + \cos Q + \cos R$.

Using trigonometric identities: $\cos P + \cos Q + \cos R = 2\cos\left(\frac{P + Q}{2}\right)\cos\left(\frac{P - Q}{2}\right) + 1 - 2\sin^2\left(\frac{R}{2}\right)$

Since $\frac{P + Q}{2} = \frac{\pi - R}{2} = \frac{\pi}{2} - \frac{R}{2}$, we have $\cos\left(\frac{P + Q}{2}\right) = \sin\left(\frac{R}{2}\right)$. Thus: $\cos P + \cos Q + \cos R = -2\sin^2\left(\frac{R}{2}\right) + 2\sin\left(\frac{R}{2}\right)\cos\left(\frac{P - Q}{2}\right) + 1$

Let $x = \sin\left(\frac{R}{2}\right)$. The equation becomes a quadratic in terms of $x$: $f(x) = -2x^2 + 2x\cos\left(\frac{P - Q}{2}\right) + 1$

The maximum value of a quadratic function $ax^2 + bx + c$ with $a < 0$ occurs at $x = -\frac{b}{2a}$, giving the maximum value: $f_{\text{max}} = c - \frac{b^2}{4a} = 1 - \frac{4\cos^2\left(\frac{P - Q}{2}\right)}{4(-2)} = 1 + \frac{1}{2}\cos^2\left(\frac{P - Q}{2}\right)$

Since $\cos^2\left(\frac{P - Q}{2}\right) \le 1$, the maximum value is: $\cos P + \cos Q + \cos R \le 1 + \frac{1}{2} = \frac{3}{2}$

This maximum is achieved when $P = Q = R = \frac{\pi}{3}$ (an equilateral triangle).

Therefore, the minimum value of $S$ is: $S_{\min} = -\left(\cos P + \cos Q + \cos R\right)_{\max} = -\frac{3}{2}$

Thus, the correct option is B.