# Trigometric functions - Inverse/Recriprocal

1. Oct 2, 2007

### AbedeuS

Hey, I think I am pretty sussed out on all of this, but it's best to be 100% sure right?

Inverse: Does the opposite operation of a function

$$Sin(\theta) = \frac{a}{b}$$

so:

$$Arcsin(\frac{a}{b})=\theta$$

My own example

$$f(x)=x^{2}$$

so:

$$f^{-1}(x)=\sqrt{x}$$

So Arctan, Arcsin and Arccos are all INVERSE functions

Reciprocal notation of those functions:

$$csc(\theta)=\frac{1}{cos(\theta)}$$

This isnt inverse, right? when people use these they genuinely mean reciporical, and not inverse?

Also, just so I know my maths is ok with working with these:

Example usage

$$Cos(\frac{\pi}{2}) = \frac{30}{x}$$

So:

$$x = 30*Csc(\frac{\pi}{2})$$

Ive tried to lay this out as easily to read as possible, I am pretty sure of my abilities with maths (I do chemistry at uni) but I sometimes get confused, and I like to be fully certain of stuff I do rather then waffleing to a maths lecturer "Oh yeah Csc is a inverse function" for him to go "No...its a reciprocal" rather pedantically.

2. Oct 2, 2007

### arildno

We occasionally talk about reciprocals as multiplicatively inverses, that is, 1/a is the multiplicative inverse of a.
In a similar vein, (-a) is occasionally called the additive inverse of a.

This is the cause of the confusion.

In both of these cases, we have that performing two operations on the neutral element of the operation brings us back to the neutral element:
$$0\to{0+a}\to{a+(-a)}=0$$
$$1\to{1}*a\to{a}*\frac{1}{a}=1$$

In an analogous sense, for functional inverses in which we could call the argument the "neutral" element, we have:
$$x\to{f}(x)\to{f^{-1}}(f(x))=x$$

A slight note, sec(x)=1/cos(x), csc(x)=1/sin(x)

3. Oct 2, 2007

### AbedeuS

Ach damn crappy, scs sec and cot notation >,<