rjr@mgweed.UUCP (Bob Roehrig) (01/31/84)
Perhaps some math whiz can help me out with a problem. I have a computer program that gives me an answer that is the sin of x. How can I get the value of x without an arc-sin function? My computer just has sin, cos, tan and arc-tan. (I also have the formula ACS(x)=-ATN(x/sqr(-x*x+1))+pi/2 which gives me the arc-cosine function). Tnx Bob
ags@pucc-i (Seaman) (02/01/84)
> Perhaps some math whiz can help me out with a problem. > I have a computer program that gives me an answer that is > the sin of x. How can I get the value of x without an > arc-sin function? My computer just has sin, cos, tan > and arc-tan. (I also have the formula ACS(x)=-ATN(x/sqr(-x*x+1))+pi/2 > which gives me the arc-cosine function). > > Tnx Bob If y = sin(x) and -PI/2 < x < PI/2 then cos(x) = sqrt(1-sin^2(x)) = sqrt(1-y^2) and tan(x) = sin(x) / cos(x) = y / sqrt(1-y^2) so that arcsin(y) = x = arctan(y/sqrt(1-y^2)) -- Dave Seaman ..!pur-ee!pucc-k:ags "Against people who give vent to their loquacity by extraneous bombastic circumlocution."
leichter@yale-com.UUCP (Jerry Leichter) (02/03/84)
The request was for a way to compute arcsin(x), given sin(x). This is actually quite easy. Given x, you want y = arcsin(x); or, inverting the equation, you want to solve for y, given sin(y) = x. A Newton-Raphson approximation should be quite effective. Recall that to solve f(y)=x for y, you will need to be able to compute f and its first derivative. Since f=sin, you have that; and f'=cos, which you also have since cos(x)=sin(90-x) (in degrees). -- Jerry
rpw3@fortune.UUCP (02/05/84)
#R:mgweed:-668600:fortune:6200004:000:747
fortune!rpw3 Feb 5 01:13:00 1984
You have a formula for arc-cosine, and you have sin(x), so if you
could go from sin(x) ==> cos(x) you could arc-cos(cos(x)) and be home, eh?
Well, look at a unit circle and note that (sin(x))^2 + (cos(x))^2 = 1^2 = 1,
since the sine and cosine are the opposite and adjacent of a right triangle
of hypoteneuse (sp?) 1 (i.e., radius of the unit circle), by the Pythagorean
Theorem.
So cos(x) = sqrt(1 - (sin(x)^2).
All details of guessing which quadrant "x" was originally from are left
to the student. No way to know, from the info given. I assume here 1st
quadrant. Don't.
Rob Warnock
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