Tera_Dragon 260 Report post Posted September 30, 2004 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)? 0 Share this post Link to post Share on other sites
Guest Anonymous Poster Report post Posted September 30, 2004 It's the inverse of cos, ie if y = acos(x) then x = cos(y). 0 Share this post Link to post Share on other sites
Tera_Dragon 260 Report post Posted September 30, 2004 Quote:Original post by Anonymous PosterIt's the inverse of cos, ie if y = acos(x) then x = cos(y).So I was correct then. :pThank you. 0 Share this post Link to post Share on other sites
LilBudyWizer 491 Report post Posted September 30, 2004 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. 0 Share this post Link to post Share on other sites
Agony 3452 Report post Posted September 30, 2004 Quote:Original post by Tera_DragonSo 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. 0 Share this post Link to post Share on other sites
terminate 259 Report post Posted September 30, 2004 acos is not 1/cos. It is a different function as LilBudyWizer said. y = cos(x)acos(y) = x1/cos is the secent function.1/cos = secant1/sin = cosecant1/tan = cotangentThese are not the same as acos, asin, or atan. 0 Share this post Link to post Share on other sites
Tera_Dragon 260 Report post Posted September 30, 2004 Quote:Original post by Agonyacos(x) is often written as cos^{-1}(x), but this is not the same as 1/cos(x).Confused... 0 Share this post Link to post Share on other sites
Fruny 1658 Report post Posted September 30, 2004 Let f(x) = cos(x)Let g(x) = acos(x) = cos^{-1}(x)Let h(x) = 1/cos(c) = cos(x)^{-1}You have f(g(x)) = x but f(x)*h(x) = 1arc cosine is the inverse function of cosine 0 Share this post Link to post Share on other sites
grhodes_at_work 1385 Report post Posted September 30, 2004 Quote:Original post by Tera_DragonQuote:Original post by Agonyacos(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 1acos(x) is the angle whose cos is x, therefore since cos(0) is 1, we can see that acos(1) is 0But, 1/cos(0) = 1/1 = 1, which is VERY different from acos(1)!I hope this helps a bit. 0 Share this post Link to post Share on other sites
Tera_Dragon 260 Report post Posted September 30, 2004 Thank you both. This has cleared things up a bit :D 0 Share this post Link to post Share on other sites
Charles B 863 Report post Posted October 2, 2004 Quote:arc cosine is the inverse function of cosineIn 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 ) 0 Share this post Link to post Share on other sites
Witchcraven 564 Report post Posted October 3, 2004 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. 0 Share this post Link to post Share on other sites
Tera_Dragon 260 Report post Posted October 3, 2004 Quote:Original post by WitchcravenEven 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 0 Share this post Link to post Share on other sites
Witchcraven 564 Report post Posted October 3, 2004 The basic trig functions give you ratios of the sides of right triangles from an angle in the triangle.Sin(angle) = opposite/hypotnueseCos(angle) = adjacent/hypotnueseTan(angle) = opposite/adjacentThe inverted ones:Asin(opposite/hypotnuese) = angleAcos(adjacent/hypotnuese) = angleAtan(opposite/adjacent) = angleFor 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. 0 Share this post Link to post Share on other sites
Tera_Dragon 260 Report post Posted October 3, 2004 I remember doing that a year or so ago now. Thanks all the same. 0 Share this post Link to post Share on other sites