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How do I find the minimum value of $a/\sin x+b/\cos x$?

I tried using AM>GM and I am getting a value of $\sqrt{8ab}$ at $x=\pi/4$.

That works for some values of $a$ and $b$, but I observed errors for other values on WolframAlpha.

$a$ and $b$ are constants and $x$ is a variable.

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  • $\begingroup$ Which parameters are constant and which are variable (a,b,x)? $\endgroup$ Commented Nov 5, 2021 at 3:39
  • $\begingroup$ The expression is unbounded both below and above as a function of $x$, unless there are additional constraints that you forgot to mention. $\endgroup$
    – dxiv
    Commented Nov 5, 2021 at 3:40
  • $\begingroup$ $\pi/4$ is a local minima in $(0,\pi/2)$, if that helps. $\endgroup$ Commented Nov 5, 2021 at 3:51
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    $\begingroup$ Set the derivative to 0. $\endgroup$
    – Eric
    Commented Nov 5, 2021 at 3:58
  • $\begingroup$ The solution in the accepted answer does not exist if $b=0$, is never an actual (global) minimum, and is not even a local minimum if $a=b=-1$ for example. If this is the answer you were looking for (since you accepted it), then the question is missing essential information. Voting to close for lack of clarity. $\endgroup$
    – dxiv
    Commented Nov 5, 2021 at 5:25

2 Answers 2

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Suppose for simplicity that $b\gt 0$. When $x$ approaches $\pi/2$ on the left your expression approaches $-\infty$. So the infimum is $-\infty$.

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According to fermat's principle of stationary points, to find a minimum point if any function set it's derivative to $0$. Let, $$f(x)=\frac{a}{\sin(x)}+\frac{b}{\cos(x)}$$ Consider it's derivative, $$f'(x)=\frac{-a\cos(x)}{sin^{2}(x)}+\frac{b\sin(x)}{cos^{2}(x)}$$ Set it equal to $0$, $$f'(x)=\frac{-a\cos(x)}{sin^{2}(x)}+\frac{b\sin(x)}{cos^{2}(x)}=0$$ $$\frac{a\cos(x)}{sin^{2}(x)}=\frac{b\sin(x)}{cos^{2}(x)}$$ Simplification leads to, $$\tan^{3}(x)=\frac{a}{b}$$ $$\tan(x)=\sqrt[3]{\frac{a}{b}}$$ $$x=arctan(\sqrt[3]{\frac{a}{b}})$$ So it's the minimum point. It depends on constants $a,b$. For example if we let, $a=b=1$ then $x=\frac{\pi}{4}$

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  • $\begingroup$ Not by $x$ but by $\ x$ with a space. $\endgroup$
    – markvs
    Commented Nov 5, 2021 at 4:28
  • $\begingroup$ The Fermat principle works for bounded functions. This one is not bounded. $\endgroup$
    – markvs
    Commented Nov 5, 2021 at 4:31
  • $\begingroup$ @markvs, I am not sure about it ,but theorem of stationary points was proved in Wikipedia article without considering the condition you mentioned. $\endgroup$
    – RAHUL
    Commented Nov 5, 2021 at 4:37
  • $\begingroup$ @RAHUL: See my answer here. You are finding local extreme points (which may not be even minimal but maximal) and may be not global which is the case here. $\endgroup$
    – markvs
    Commented Nov 5, 2021 at 4:44
  • $\begingroup$ @RAHUL: In fact what you find may be neither a max nor a min but an inflection point. Like the zero of $x^3$. $\endgroup$
    – markvs
    Commented Nov 5, 2021 at 4:53

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