How is arccos derived?

Question

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?

Answer 1

$\arccos$ 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.

Answer 2

This is a common point of confusion: Cosine has a multiplicative inverse and an inverse function.

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 $0 \leq \theta \leq \pi$.