Question for Trig Gurus

[Gerson W. Barbosa]

(12-02-2014 04:57 PM)Namir Wrote:  

(12-02-2014 10:14 AM)Didier Lachieze Wrote:  http://lmgtfy.com/?q=casio+fx10+manual

Thanks Didier. As I suspected, the FX-10 has NO inverse trigonometric functions. The manual casually mentions how to calculate inverse hyperbolic sine and cosine, but that's not what I am looking for.

As I said in my last post, it looks like Newton's method is the best way, if I can live with the simple iterations.

:-(

Namir

This should be good enough for practical purposes:

asin(x) ~ (1/2*e^(-x)*(e^(2*x)-1)+e^(-e^(2.700492094-1.821362659*x)))*180/pi

or

asin(x) ~ 90*(2*e^(-14.88705575*e^(-1.821362659*x)) + e^(x) - e^(-x))/pi


http://www.wolframalpha.com/input/?i=plo...x%3D0..0.5

http://www.wolframalpha.com/input/?i=plo...x%3D0..0.5

Examples:

asin(0.5) ~ 30.000 degrees

asin(0.4) ~ 23.578 degrees

asin(0.2) ~ 11,538 degrees


-------------------------------

Idea:

Let's start with the expression for asinh(x):

y = ln(x + sqrt(x^2 + 1))

A plotting shows this function and y = asin(x) match at y = 0, but the curves soon bend themselves to opposite directions. But visually it appear that the inverse function of y = ln(x + sqrt(x^2 +1)), equivalent to y = asinh(x), would follow y = asin(x) a bit further.

Then we invert the function so it gets closer to y = asin(x), at least in the 0..1/2 range:

solve y == ln(x+sqrt(x^2+1)) for x

--> x = 1/2*e^(-y)*(e^(2*y) - 1)

--> y = 1/2*e^(-x)*(e^(2*x) - 1)


By comparing the latter with y = asin(x), we find the greatest difference to be 0.00250347010455165, at x = 0.5. Then some curve fitting to improve the approximation (the second part of the first equation). For the range 1/2..1, we use simmetry. Or you can use the first part as a first guess for the solve, if you want full accuracy.