Trigonometric inequalities $\sqrt 3 \gt {3\over 2} \implies {\pi\over 6} \lt \operatorname{arccot}{3 \over 2}) $

Question

Can someone explain me why this implication is true:

$$\sqrt3 \gt {3\over 2} \implies {\pi\over 6} \lt \operatorname{arccot}({3 \over 2}) $$

where $\operatorname{arccot}$ is defined as $\operatorname{arccot}: \mathbb R \to ]0,\pi[ $ such that, $\operatorname{arccot}(\cot(x))=x$ for $x \in ]0,\pi[ $

What I have done is I know that \begin{align*} \sqrt 3 &\gt {3\over 2} \\ \cot({\pi\over 6})&= \sqrt 3 \\ \operatorname{arccot} \cot {3\over 2} &={ 3\over 2 } \end{align*}

so since $\operatorname{arccot}$ is always positive I have: \begin{align*}\sqrt 3 \gt {\dfrac 32} &\iff \operatorname{arccot}(\sqrt 3)\gt \operatorname{arccot}({3\over 2}) \\ &\iff \operatorname{arccot}(\cot({\pi\over 6})) \gt \operatorname{arccot}({3\over 2}) \\ &\iff {\pi \over 6} \gt \operatorname{arccot}({3\over 2}) \end{align*}

Can someone explain me what's wrong and how to prove this implication?

Answer 1

The arccotangent function is decreasing on the interval $(0,\pi)$. Suppose $0<x<y<\pi$. Then $$ \cot x-\cot y= \frac{\cos x}{\sin x}-\frac{\cos y}{\sin y}= \frac{\cos x\sin y-\sin x\cos y}{\sin x\sin y}= \frac{\sin(y-x)}{\sin x\sin y} $$ From $0<x<y<\pi$ we deduce that $0<y-x<\pi$, so $$ \sin(y-x)>0,\qquad \sin x>0,\qquad \sin y>0 $$ and therefore $$ \cot x>\cot y $$

The inverse function of a decreasing function is decreasing as well.

If you know derivatives, you can simplify the computation: if $f(x)=\operatorname{arccot}x$, then $$ f'(x)=-\frac{1}{1+x^2}<0 $$

Answer 2

HINT

Recall that

• for an increasing function $x\ge y \iff f(x)\ge f(y)$

• for an decreasing function $x\ge y \iff f(x)\le f(y)$