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Tera_Dragon

acos

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Tera_Dragon    260
I was just reading through some vetor math, and came across the following formula: O = acos( (a.b)/(||a|| * ||b||) ) Where O is the angle between the two vectors. I was just wondering what acos is. Is it 1/cos (cos to the -1)?

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LilBudyWizer    491
It's inverse cosine. With numbers the inverse of x is the number y such that y*x=1. With functions the inverse of f is the function g such that g(f(a))=a. So more or less acos(cos(t))=t. More or less because cos(t)=cos(t+2*n*pi) where n is an integer and cos(t)=cos(2*pi-t). So t could be any real number but you are going to get back a number between 0 and pi. So only if 0<=t<=pi does acos(cos(t))=t.

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Agony    3452
Quote:
Original post by Tera_Dragon
So I was correct then.
Not to be rude, but just so that you are corrected, no, you weren't correct. acos(x) does not equal (cos(x))-1. acos(x) is often written as cos-1(x), but this is not the same as 1/cos(x).

Basically, if cos(x) takes an angle and returns a ratio of side-adjacent to hypotenuse, then acos(y) takes a ratio of side-adjacent to hypotenuse, and returns an angle.

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terminate    259
acos is not 1/cos. It is a different function as LilBudyWizer said.

y = cos(x)
acos(y) = x

1/cos is the secent function.


1/cos = secant
1/sin = cosecant
1/tan = cotangent

These are not the same as acos, asin, or atan.

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grhodes_at_work    1385
Quote:
Original post by Tera_Dragon
Quote:
Original post by Agony
acos(x) is often written as cos-1(x), but this is not the same as 1/cos(x).

Confused...


It is confusing notation. I would just take it for granted that when you see cos-1(x) that what it means is acos(x) instead of 1/cos(x). This is simply an oddity of math notation. There are other such oddities out there! You do have to be careful of the context of the equation, since in some cases people will write in a way that is inconsistent with this oddity. Being able to spot the inconsistency is EXACTLY why it is really, really good to understand geometry and trig!

You can see that acos is not the same as 1/cos by looking at this example:

cos(0) is equal to 1
acos(x) is the angle whose cos is x, therefore since cos(0) is 1, we can see that acos(1) is 0

But, 1/cos(0) = 1/1 = 1, which is VERY different from acos(1)!

I hope this helps a bit.

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Charles B    863
Quote:

arc cosine is the inverse function of cosine


In the appropriate definition domains ;) For instance : [0, PI] for cos and [-1,1] for acos.

(I lost so many points in analysis as a student for not precising the restrictions of the functions on a definition domain f -> fd )

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Witchcraven    564
Even though your your question is probably answered:

I learned about acos when looking at a cosine table. I knew cos would take the angle and give me the ratio, but I had the ratio and needed the angle. My teacher informed me what I wanted was acos. Hopefully I didnt get that backwards.

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Tera_Dragon    260
Quote:
Original post by Witchcraven
Even though your your question is probably answered:

I learned about acos when looking at a cosine table. I knew cos would take the angle and give me the ratio, but I had the ratio and needed the angle. My teacher informed me what I wanted was acos. Hopefully I didnt get that backwards.


Ah, thank you for pointig this out. I undstood what acos did (or at least the button for it on my calc ;)), but as I don't really know what cos does it doesn't mean much. You say cos gives you the ratio, the ratio of what?
Tera_Dragon

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Witchcraven    564
The basic trig functions give you ratios of the sides of right triangles from an angle in the triangle.

Sin(angle) = opposite/hypotnuese
Cos(angle) = adjacent/hypotnuese
Tan(angle) = opposite/adjacent

The inverted ones:

Asin(opposite/hypotnuese) = angle
Acos(adjacent/hypotnuese) = angle
Atan(opposite/adjacent) = angle

For example, tan(45 degrees) = 1.0 (I think). It's 1.0 because the opp/adjacent sides are the same length when that triangle is using a 45 degree angle.

The functions can be used to scale triangle and a whole bunch of useful crap, like rotating stuff an trajectories.

Hopefully I didnt explain a bunch you already know. Even more hopefully, I got all that right.

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