minimum value of $a/\sin x+b/\cos x$

Question

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.

Answer 1

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 $x=\frac{\pi}{4}$

Answer 2

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