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 $x \in (0,\pi)$:

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

Additional note after comment:

Note that according to AM-GM you have equality if and only if

$$\cos^2(x/2) = 3\sin^2(x/2) \Leftrightarrow \tan^2(x/2) = \frac{1}{3} \stackrel{x \in (0,\pi)}{\Leftrightarrow}x = \frac{\pi}{3}$$

  • $\begingroup$ +1, also: this minimum is obtained by observation: plug in $x = \pi/3$ to get an output of $y = \sqrt{3}$. $\endgroup$ – Benjamin Dickman Apr 26 at 3:39
  • 1
    $\begingroup$ @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. $\endgroup$ – trancelocation Apr 26 at 3:45

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