Let $A+B+C=\pi$ and $\sin^2(A)+\sin^2(B)-\sin^2(C)=p$, then find range of $p$

Question

Let $A+B+C=\pi$ and $\sin^2(A)+\sin^2(B)-\sin^2(C)=p$, then find range of $p$

Approach: After simplifying above expression I obatained $p=2\sin A \sin B \cos C$.

As I can see maximum value of above expression is $2$ with guess of $A=\dfrac{\pi}{2}, B=\dfrac{\pi}{2}, C=0$.

But I am not able to find its minimum value. Also is there proper method to solve this problem other than gueesing.

Answer 1

We observe that $\sin(A), \sin(B)$ are positive but $\cos(C)$ can be negative, so, the minimum must be less than $0$ for $C\ge \pi/2$. So, we seeks the minimum value under the constraints $A+B+C = \pi$ and $C \ge \pi/2$, we have: $$\sin(A)\sin(B) \le \left(\frac{\sin(A)+\sin(B)}{2}\right)^2=\sin^2\underbrace{\left(\frac{A+B}{2}\right)}_{=\frac{\pi}{2}-\frac{C}{2}}\cos^2\left(\frac{A-B}{2}\right)\le \cos^2\left( \frac{C}{2}\right)$$ $$\implies p\ge2 \cos^2\left( \frac{C}{2}\right)\cos(C)=-2 \cos^2\left( \frac{C}{2}\right) \left(1-2 \cos^2\left( \frac{C}{2}\right)\right)$$ with $\cos\left( \frac{C}{2}\right) \in \left[\frac{1}{\sqrt{2}}, 1\right]$ (as $\pi/2 \le C\le \pi$)

Applying the AM-GM inequality: $$p \ge -\left(\frac{2 \cos^2\left( \frac{C}{2}\right)+\left(1-2 \cos^2\left( \frac{C}{2}\right)\right)}{2}\right)^2=-\frac{1}{4}$$ The equality occurs if and only $(A,B,C) =(\pi/6,\pi/6,2\pi/3)$.

Answer 2

$$p(A,B)=-2\sin A\sin B\cos(A+B).$$ $$\frac{\partial p}{\partial A}=0\iff\cos(2A+B)=0\text{ or }\sin B=0,$$ $$\frac{\partial p}{\partial B}=0\iff\cos(2B+A)=0\text{ or }\sin A=0.$$ If $\sin A$ or $\sin B$ equals $0$ then $p=0,$ which is not an extremum. $$\cos(2A+B)=\cos(2B+A)=0$$ $$\iff2A+B=\frac\pi2+k\pi\text{ and }2B+A=\frac\pi2+\ell\pi\quad(k,\ell\in\Bbb Z)$$ $$\iff A\equiv\frac\pi6\bmod{\frac\pi3}\text{ and }B\equiv A\bmod\pi$$ $$\implies A\equiv\frac\pi6\bmod{\frac\pi3}\text{ and }p=-2\sin^2A\cos(2A)=:q(A)=q(-A)=q(A+\pi).$$

The max and min of $p(A,B)$ are therefore $$q(\pi/2)=2\text{ and }q(\pi/6)=-\frac14.$$

Answer 3

We have \begin{align*} p &= \sin^2 A + \sin^2 B - \sin^2(\pi - A - B)\\ &= \sin^2 A + \sin^2 B - (\sin A\, \cos B + \cos A\, \sin B)^2\\ &= \sin^2 A + \sin^2 B - \sin^2A \, \cos^2 B - \cos^2 A\, \sin^2 B - 2\sin A\, \cos A \, \sin B\, \cos B\\ &\ge \sin^2 A + \sin^2 B - \sin^2A \, \cos^2 B - \cos^2 A\, \sin^2 B - (\sin^2 A\, \cos^2 A + \sin^2 B\, \cos^2 B)\\ &= (u^2 + v^2)^2 - 3(u^2 + v^2) + 2\\ &= (u^2 + v^2 -3/2)^2 - \frac14\\ &\ge - \frac14 \end{align*} with equality if e.g. $A = B = \pi/6$, where $u = \cos A, \, v = \cos B$; we use $2x y \le x^2 + y^2$ in the first $\ge $.