What is the minimum value of $\csc x - \sin x$?

Question

What is the minimum value of $\csc x - \sin x$? Differentiating and setting it to zero yields nothing meaningful. How can I find the minimum value?

Answer 1

There is no (absolute) minimum if we do not put restrictions on the domain. This is obvious since $\csc x$ is large negative when $x$ is negative and close to $0$. No need of derivatives.

If we restrict the domain to $(0,\pi)$, then differentiation is perhaps warranted. I would set $f(x)=\frac{1}{\sin x}+\sin x$ and differentiate. We get $-\frac{\cos x}{\sin^2 x}+\cos x$, that is, $\frac{\cos x(\sin^2 x-1)}{\sin^2 x}$. This is $0$ at $x=\frac{\pi}{2}$. The top is negative in the interval $(0,\pi/2)$ and positive in $(\pi/2,\pi)$, so we have a minimum at $x=\pi/2$.

There are also non-calculus ways to find the minimum in the interval $(0,\pi)$.

Answer 2

f(x) = Cosec x - sin x = (1/sin x)-sin x

=(1-sin^2(x))/sin x

= cos^2(x)/sinx=1/(sec x tan x). Will that help?

Note: f(x) is minimised when (sec xtan x) is maximized.