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.

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?

share|improve this question
    
What is "adj", what is "hyp"...?! –  DonAntonio May 31 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 at 18:06
1  
Presumably, OP means the adjacent side and hypotenuse (of a right triangle). –  Cadenza May 31 at 18:08
    
updated question –  Starkers May 31 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 at 18:16

2 Answers 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 $0 \leq \theta \leq \pi$.

share|improve this answer

$\arccos$ 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.

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.