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How do I find the maximum and minimum values of the following?

$$a^2\sin^2\theta + b^2\csc^2\theta$$ Is the max value $\infty$?

I tried to find the minimum value by using A.M$\geq$G.M. inequality (is there any other way?) and got it as 2|a||b| which seem to work for all the positive values when a>b (I checked it on Desmos by putting different values of a and b) but when a$\le$b it just don't match in the graph. So, what am I doing wrong here? And what should be the correct answer?

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    $\begingroup$ Think of it as $rx+sx^{-1}$ where $-1\le x\le1$, and $r,s$ are constants. $\endgroup$ Aug 2 '19 at 7:17
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    $\begingroup$ @GerryMyerson You mean $0\le x\le1$. $\endgroup$
    – J.G.
    Aug 2 '19 at 7:36
  • $\begingroup$ @J.G., yes, thanks. $\endgroup$ Aug 2 '19 at 9:25
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The function is periodic with period $\pi$ so let us consider $ 0\le x \le \pi.$

Case-1: When $a>b$ in $$f(x)=a^2\sin^2 x+ b^2 \csc^2 x~~~(1)$$ then $$f'(x)=2a^2 \sin x \cos x-2b^2 \cot x \csc^2 x.= 2 \cos x(a^2 \sin^4 -b^2)/\sin^2 x ~~~(2)$$ $f'(x)=0,$ gives $\cos x=0$ or $\sin^4 x = b^2/a^2 \Rightarrow \sin^2 x= b/a \Rightarrow \sin x = \pm \sqrt{b/a} \Rightarrow x_1 = \pi/2, ~ \mbox{or} ~ x_{2,3}=\pm \sin^{-1} \sqrt{b/a}$. Next $$f''(x)= b^2(4 \cot^2x \csc^2 x + 2 \csc^2 x)+2a^2(\cos^2 x- \sin^2 x)~~~~~~(3).$$ Get $f''(x_1)=-2(a^2-b^2)< 0,$ so local max. at $x=\pi/2$. Use $\sin^2 x=b/a, \cos^2 x=(a-b)/a$ etc. in (3) to get $f''(x_{2,3})=8a(a-b)>0,$ so two local minima at $x=x_{2,3}.$

Hence when $a>b$, the local max. and min. are given as $$f_{max}=f(\pi/2)=a^2+b^2, f_{min}f(x_{2,3})=2ab.$$

Case-2: When $a<b$, then $x_{2,3}$ do not exist and $f''(\pi/2)=-2(a^2-b^2)>0,$ so there exists only one local min. and $f_{min}=f(\pi/2)=a^2+b^2$

In both the cases the function $f(x)$ is positive and unbounded which can becomes $\infty$ at $x=0, \pi.$

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  • $\begingroup$ Thanks, I guessed the correct answer myself which matches with yours but could you give a much easier proof of it or explain it a bit? I don't seem to get it after the "cos x = 0" line. $\endgroup$
    – asks281
    Aug 2 '19 at 11:40
  • $\begingroup$ @ Aditya Kshitz I have added more stuff now you can follow it. $\endgroup$
    – Z Ahmed
    Aug 2 '19 at 12:17
  • $\begingroup$ Okay now it makes sense but shouldn't f max be infinity other than a^2 +b^2 when a>b? And why doesn't A.M≥G.M. inequality only give the result with a>b and not b>a. (i.e. only 2ab not a^2 + b^2) $\endgroup$
    – asks281
    Aug 2 '19 at 15:17
  • $\begingroup$ @Aditya Kshitz Please note that $infunuty$ can never be answer usually. Your function has local max and two local minima. These minima are also global . In second case there is no local max. Mentioning that the function can become infinitely large is correct rather than saying max is $\infty$Like in the equation $1/x=0$ one does not sat that $x=\infty$ is a root. $\endgroup$
    – Z Ahmed
    Aug 3 '19 at 3:00
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Let the max/min values of $a^2\sin^2\theta + b^2\csc^2\theta$ be $k$. Thus:

$$a^2\sin^2\theta + b^2\csc^2\theta = k \implies a^2 \sin^4 \theta-k\sin^2 \theta+b^2=0$$ $$\Delta = 0: (-k)^2 - 4a^2b^2 = 0 \implies k^2 = 4a^2b^2, k = 2|a||b|$$

and there is no maximum value because the range of $\csc^2 \theta$ is $[1, \infty)$.

Restricting the domain to $[0, \pi)$, there are two local minimums which match the absolute minimum value, but there is also another local maximum at $x = \frac{\pi}{2}$ which cannot be found using this method.

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