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?

  • 4
    $\begingroup$ Hint: $\arccot (x)$ is a decreasing function. $\endgroup$ – MisterRiemann May 21 '18 at 14:56
  • $\begingroup$ Oh yes right thank you, do we prove that $\arccot(x)$ is decrasing using its derivative ? $\endgroup$ – Sami Mir May 21 '18 at 15:01
  • $\begingroup$ See the answers below. $\endgroup$ – MisterRiemann May 21 '18 at 15:30
  • $\begingroup$ @SamiMir Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi May 31 '18 at 22:17

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 $$ f'(x)=-\frac{1}{1+x^2}<0 $$



Recall that

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.