Computer Science Canada sine arc function? |
Author: | midnite13 [ Fri Jan 06, 2006 12:04 am ] |
Post subject: | sine arc function? |
Hello... I was wondering if turing has a predefined sine arc function (I need it in degrees) the sind function only returns the ratio, but it's the angle that I want. I'm not sure my wording is the best, so I'll provide an example sind (30) would return a value of 0.5, however, I"m wondering if there's an arcsind that woudl allow for arcsind (0.5) which would return 30. Thanks for all your help |
Author: | MysticVegeta [ Fri Jan 06, 2006 1:04 am ] |
Post subject: | Re: sine arc function? |
midnite13 wrote: Hello...
I was wondering if turing has a predefined sine arc function (I need it in degrees) the sind function only returns the ratio, but it's the angle that I want. I'm not sure my wording is the best, so I'll provide an example sind (30) would return a value of 0.5, however, I"m wondering if there's an arcsind that woudl allow for arcsind (0.5) which would return 30. Thanks for all your help you know arcsind does work. |
Author: | Saad85 [ Fri Jan 06, 2006 1:17 am ] |
Post subject: | |
if not that then you could go arcsin*(180/pi) to turn radians into degrees |
Author: | MysticVegeta [ Sun Jan 08, 2006 1:45 pm ] |
Post subject: | |
What I forgot to mention is that, arcsind works for Turing 4.05 only. |
Author: | Phonon [ Mon Apr 03, 2006 10:22 pm ] |
Post subject: | older versions |
I have an older version of Turing. I plan on getting the newest one but I need to get this done. The version I have does include an arctan function. Is there any way to manipulate trigonometric functions so that I can use a combination of arctan and other functions to get arcsin? I can't see a way of doing this with Taylor series expansions. Maybe I should be thinking about using exponentials and complex variables (but leaving 'i' out of it)? |
Author: | Delos [ Tue Apr 04, 2006 9:53 am ] | ||
Post subject: | |||
Disclaimer: I am not a Math major. The furthest math I took was 1st year intro calculus. I don't really like math all that much - but I am quite capable of it. Take this proof with a bit of salt, and perhaps oregano for taste. I'll going to work in radians, because I want to and because all the cool people do so anyway. You've already seen a way to convert from radians to degrees above. Conversions from arctan to arcsin are possible, but you'll just have to limit yourself to (-1<= x <= 1). That's not too difficult. You can always manipulate your answer later on to suit your needs. Either way:
Try it. It should work. |
Author: | NikG [ Tue Apr 04, 2006 10:40 am ] |
Post subject: | |
I'm just wondering... what exactly is arcsin even necessary for anyways? I mean if you're just looking for the angle in between 2 lines, arctand is more than sufficient. |
Author: | Andy [ Tue Apr 04, 2006 11:02 am ] | ||
Post subject: | |||
wow delos.. you're trying wayyyyyyyyyyy too hard.... consider the following: what is sine? draw a unit circle with the centre as the origin, and start from (1,0) then walk B units counter clock wise, now let's call that point (a,b) sin(B) in this case would be b. insimpler terms, sin(theta) yeilds the y co ordinate on a unit circle to find arcsin(A): let arcsin(A) = B => sin(B) = A this means that A is the y co ordinate of the arc length B from the (1,0) along the unit circle. so let's find both co ordinates of this point. take the equation of the circle: x^2 + y^2 = 1 and the equation of the line y = A. turns out, the co ordinate is (A, sqrt(1-A^2)) now to find how long is that arc between (A, sqrt(1-A^2)) and (1,0) along the unit circle: simply take the arctan(A/sqrt(1-A^2))) and you're done. note, this method only works for the first quadrant, but making the rest work shouldnt be too hard.
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Author: | Delos [ Tue Apr 04, 2006 11:34 am ] |
Post subject: | |
Yes dodge, yes I am. . Ah well, I enjoyed the proof anyway, so it was worth it. But yes, do what dodge says, he knows more maths than I do. |
Author: | Phonon [ Tue Apr 04, 2006 6:20 pm ] |
Post subject: | |
NikG wrote: I'm just wondering... what exactly is arcsin even necessary for anyways?
I mean if you're just looking for the angle in between 2 lines, arctand is more than sufficient. I am writing a program for x-ray crystallography. In order to find theta from bragg's law, I need to use arcsin. There's simply no other way I can do it, as far as I see, because: sin (theta) = (n * radiation wavelength)/2*d(hkl). Thus, knowing n, the radiation wavelength and d(hkl), I am using that information to find theta. |
Author: | Phonon [ Wed Apr 05, 2006 12:09 am ] |
Post subject: | works |
Anyway, I got it to work. Thanks very much for the help! |
Author: | Andy [ Wed Apr 05, 2006 12:56 pm ] |
Post subject: | |
of course you got it to work.. i gave you the exact code |