Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 0 down vote accepted

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?"

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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