There is something I don't understand with arccos and inequalities.

Suppose I have this inequality

$cos(x) ≤ \frac{1}{2}$

Having $x = 90$, satisfies this since $cos(90) = 0$.

Then since arccos is defined on [-1,1], I should be able to arccos both sides, which gives me

$x ≤ 60$

But then when $x = 90$, the inequality isn't satisfied.

Why does this happen, even though I kept everything in degrees format?


1 Answer 1


The reason is that $\arccos$ (as it is usually defined) is a decreasing function. So after you apply it to both sides, you need to switch the direction of the inequality. In fact, that is precisely what it means for the function to be decreasing: $f$ is decreasing on the interval $(x,y)$ if for all $a, b \in (x,y)$, $a \leq b$, implies $f(a) \geq f(b)$.

Here are two examples of decreasing functions besides $\arccos$:

  • $f(x) = -x$ [This is decreasing on $(-\infty,\infty)$.]
  • $g(x) = 1/x$ [This function is decreasing on the interval $(-\infty,0)$ and also on $(0, \infty)$.]

Of course, this begs the question, "Why is $\arccos$ a decreasing function?"

  • 1
    $\begingroup$ Of course!! Thank you, I've been struggling on a problem for one hour, and this answer completely clarified the issue! $\endgroup$
    – Tropilio
    Apr 4, 2018 at 17:38

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