# How is arccos derived?

Stupid question but I need to understand:

If $\cos = \dfrac{\text{adjacent}}{\text{hypotenuse}}$

What is arcosine? $\text{adjacent} \cdot \text{hypotenuse}$? Is this the same for arcsine and arctangent?

-
What is "adj", what is "hyp"...?! – DonAntonio May 31 '14 at 18:05
You must make yourself a bit more clear - arc-osine? You mean arccos(x)? What is adj or hyp? – Nicky Hekster May 31 '14 at 18:06
Presumably, OP means the adjacent side and hypotenuse (of a right triangle). – White Shirt May 31 '14 at 18:08
updated question – Starkers May 31 '14 at 18:11
You seem to be uptaking the basic, first geometric definition of cosine using right angled triangle and stuff. Perhaps you should wait until the basic trigonometric functions's definition is extended to the whole real line.. – DonAntonio May 31 '14 at 18:16

The multiplicative inverse is a function that, when you multiply $\cos \theta$ by it, you get $1$ (assuming $\cos \theta \neq 0$). Cosine's multiplicative inverse is $\sec \theta$, which you've probably seen written as $\frac{\text{hypotenuse}}{\text{adjacent}}$.
The inverse function is a function that, when composed with $\cos \theta$, returns the original input, at least for a certain set of inputs. This is the function you're asking about. If $y = \cos \theta$, then $\arccos y = \arccos \left(\cos \theta\right) = \theta$, assuming that $-1 \leq y \leq 1$ and $0 \leq \theta \leq \pi$.
$\arccos$ returns the angle, not a ratio or multiplication of the sides of a triangle. Thus, $$\arccos\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right) = \theta$$ Where $\theta$ is the angle.