# Finding the maximum and minimum values of $a^2\sin^2\theta + b^2\csc^2\theta$

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?

• Think of it as $rx+sx^{-1}$ where $-1\le x\le1$, and $r,s$ are constants. – Gerry Myerson Aug 2 '19 at 7:17
• @GerryMyerson You mean $0\le x\le1$. – J.G. Aug 2 '19 at 7:36
• @J.G., yes, thanks. – Gerry Myerson Aug 2 '19 at 9:25

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 $$a, then $$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 $$x=0, \pi.$$

• 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. – Aditya Kshitiz Aug 2 '19 at 11:40
• @ Aditya Kshitz I have added more stuff now you can follow it. – Dr Zafar Ahmed DSc Aug 2 '19 at 12:17
• 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) – Aditya Kshitiz Aug 2 '19 at 15:17
• @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. – Dr Zafar Ahmed DSc Aug 3 '19 at 3:00