# Minimum value of $\frac{2-\cos(x)}{\sin(x)}$ without differentiation

I have to find the minimum value of the expression $$\frac{2 - \cos x}{ \sin x}$$ Also $$x$$ lies between $$0$$ to $$\pi$$. One way is to find the minima using differentiation. But it is not taught in my grade so my teacher asked me to do it without differentiation.

Here's what I did

Let$$\frac{2 - \cos x }{ \sin x} = y~,$$ so that $$(2-y \sin x)^2 = 1 - \sin^2(x) \\ \implies \sin^2(x) \cdot (y^2 +1) - 4 y \sin x +3=0$$

Now I am struck. I tried using Discriminant $$\ge{0}$$ but no use as our variable $$\sin x$$ lies between $$0$$and $$1$$. Please help.

Let $$k=\frac{2-\cos x}{\sin x}\Rightarrow 2-\cos x=k\sin x$$

So we have $$k\sin x+\cos x=2$$

Using $$|a\sin x+b\cos x|\leq \sqrt{a^2+b^2}$$

So we have $$|k\sin x+\cos x|\leq \sqrt{k^2+1}$$

$$2\leq \sqrt{k^2+1}\Rightarrow k^2+1\geq 4\Rightarrow k\geq \sqrt{3}.$$

You can use AM-GM as follows for $$x \in (0,\pi)$$:

$$\frac{2-\cos(x)}{\sin(x)}= \frac{2-(\cos^2(x/2) - \sin^2(x/2))}{2\sin(x/2)\cos(x/2)}$$ $$= \frac{\cos^2(x/2) +3\sin^2(x/2)}{2\sin(x/2)\cos(x/2)}\stackrel{AM-GM}{\geq} \frac{\sqrt{\cos^2(x/2) \cdot 3\sin^2(x/2)}}{\sin(x/2)\cos(x/2)} = \sqrt{3}$$

$$\cos^2(x/2) = 3\sin^2(x/2) \Leftrightarrow \tan^2(x/2) = \frac{1}{3} \stackrel{x \in (0,\pi)}{\Leftrightarrow}x = \frac{\pi}{3}$$
• +1, also: this minimum is obtained by observation: plug in $x = \pi/3$ to get an output of $y = \sqrt{3}$. – Benjamin Dickman Apr 26 at 3:39
• @BenjaminDickman : I have added a note on this, as the specific value for $x$ also follows directly from the equality condition of AM-GM. Thanks for mentioning it. – trancelocation Apr 26 at 3:45