# Thread: [SOLVED] Prove the trigonometric inequality?

1. ## [SOLVED] Prove the trigonometric inequality?

Prove that cot x/2 - cot x ≥ 1 for x between 0 and pi.

2. Originally Posted by fardeen_gen
Prove that cot x/2 - cot x ≥ 1 for x between 0 and pi.
Note that $\frac{\cos x}{\sin x} = \frac{\cos^2 \left( \frac{x}{2} \right) - \sin^2 \left( \frac{x}{2}\right) }{2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)} = \frac{2 \cos^2 \left( \frac{x}{2} \right) - 1 }{2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)}$.

Therefore the left hand side becomes

$\frac{\cos \left( \frac{x}{2} \right) }{\sin \left( \frac{x}{2} \right) } - \frac{2 \cos^2 \left( \frac{x}{2} \right) - 1 }{2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)} = \frac{2\cos^2 \left( \frac{x}{2} \right) }{2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)} - \frac{2 \cos^2 \left( \frac{x}{2} \right) - 1 }{2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)}$

$= \frac{1}{2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)} = \frac{1}{\sin x}$.

Now note that $0 \leq \sin x \leq 1$ over $0 \leq x \leq \pi$.

3. Hello, fardeen_gen!

Another approach . . .

Identity: . $\cot\frac{x}{2} \;=\;\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}} \;=\;\frac{\sqrt{\frac{1+\cos x}{2}}}{\sqrt{\frac{1-\cos x}{2}}} \;=\;\sqrt{\frac{1 + \cos x}{1-\cos x}}$

Prove that: . $\cot\frac{x}{2} - \cot x \:\geq \:1\quad\text{ for }0 \leq x \leq \pi$

We have: . $\cot\frac{x}{2} - \cot x \;\;=\;\;
\sqrt{\frac{1+ \cos x}{1-\cos x}} - \frac{\cos x}{\sin x}$

Rationalize: . $\sqrt{\frac{1+\cos x}{1-\cos x}\cdot{\color{blue}\frac{1+\cos x}{1+\cos x}}} -\frac{\cos x}{\sin x}\;\;=\;\;\sqrt{\frac{(1+\cos x)^2}{1-\cos^2x}} - \frac{\cos x}{\sin x}$

. . $= \;\sqrt{\frac{(1+\cos x)^2}{\sin^2\!x}} - \frac{\cos x}{\sin x} \;\;=\;\;\frac{1+\cos x}{\sin x} - \frac{\cos x}{\sin x} \;\;\frac{1}{\sin x} \;\;=\;\;\csc x$

And: . $\csc x \:\geq \:1\;\text{ for }0 \:\leq x\;\leq \pi\quad\hdots\quad\text{Q.E.D.}$