Question for Trig Gurus

12012014, 07:49 PM
Post: #1




Question for Trig Gurus
I recently bought a vintage Casio fx10 calculator. The machine has trigonometric functions but not their inverse counterpart. Other than trial and error (like Newton's method), is there a secret formula to calculate or estimate the inverse trig functions from one of the trig functions??
Namir 

12012014, 08:08 PM
Post: #2




RE: Question for Trig Gurus
(12012014 07:49 PM)Namir Wrote: I recently bought a vintage Casio fx10 calculator. The machine has trigonometric functions but not their inverse counterpart. Other than trial and error (like Newton's method), is there a secret formula to calculate or estimate the inverse trig functions from one of the trig functions?? The FX10 has Arcsine, Arccosine, Arctangent built in. Which functions are you looking for? Tom L Tom L Cui bono? 

12012014, 08:46 PM
Post: #3




RE: Question for Trig Gurus  
12012014, 10:54 PM
(This post was last modified: 12012014 11:00 PM by Namir.)
Post: #4




RE: Question for Trig Gurus
(12012014 08:46 PM)PANAMATIK Wrote:(12012014 08:08 PM)toml_12953 Wrote: The FX10 has Arcsine, Arccosine, Arctangent built in. How does one invoke the inverse trig functions? I don't see a shift key. This review confirms the absence of inverse trig functions. Namir 

12012014, 11:00 PM
(This post was last modified: 12022014 02:01 AM by Mark Hardman.)
Post: #5




RE: Question for Trig Gurus
(12012014 07:49 PM)Namir Wrote: I recently bought a vintage Casio fx10 calculator. The machine has trigonometric functions but not their inverse counterpart. Other than trial and error (like Newton's method), is there a secret formula to calculate or estimate the inverse trig functions from one of the trig functions?? This is an odd one. It has trig functions but not their inverse. Casio fx10 Interesting how it displays zeros as well. Ceci n'est pas une signature. 

12022014, 12:09 AM
(This post was last modified: 12022014 12:14 AM by Thomas Klemm.)
Post: #6




RE: Question for Trig Gurus
(12012014 07:49 PM)Namir Wrote: is there a secret formula to calculate or estimate the inverse trig functions from one of the trig functions?? You could use these trigonometric identities to calculate \(\tan \frac{x}{4}\) from \(\cos x\): \[ \begin{align} \cos \frac{x}{2} &= \sqrt{\frac{1+\cos x}{2}} \\ \tan \frac{x}{2} &= \sqrt{\frac{1\cos x}{1+\cos x}} \\ \end{align} \] From this you could use the Taylor series of \(\arctan x \approx x\frac{x^3}{3}+\frac{x^5}{5}+\dots\) to estimate \(x\). Thus we can calculate \(\arccos x\). Example: \(\cos x = 0.5\) \(\cos \frac{x}{2} = \sqrt{\frac{1.5}{2}} = \sqrt{0.75} \approx 0.866025\) \(\tan \frac{x}{4} = \sqrt{\frac{10.866025}{1+0.866025}} = \sqrt{\frac{0.133975}{1.866025}} \approx 0.267949\) We slightly rewrite the Taylor series using Horner's method: \(((\frac{x^2}{5}\frac{1}{3}) x^2+1) x\). Just keep in mind that we already calculated \(x^2 = \frac{0.133975}{1.866025} \approx 0.071797\). This is how you could calculate it: 0.071797 ÷ 5 − 3 [1/x] × 0.071797 + 1 × 0.267949 = 0.261813 The correct value is: 0.261799 We can use these two identities to calculate \(\arcsin x\) or \(\arctan x\): \[ \begin{align} \cos x &= \sqrt{1\sin^2 x} \\ \cos x &= \frac{1}{\sqrt{1+\tan^2 x}} \\ \end{align} \] Example: \(\tan x = \frac{1}{\sqrt{3}}\) \(\cos x = \frac{1}{\sqrt{1+\frac{1}{3}}} \approx 0.866025\) \(\cos \frac{x}{2} = \sqrt{\frac{1.866025}{2}} \approx 0.965926\) \(\tan \frac{x}{4} = \sqrt{\frac{10.965926}{1+0.965926}} = \sqrt{\frac{0.034074}{1.965926}} \approx 0.131652\) Here we end up with: 0.523598 The correct value is: 0.523599 Kind regards Thomas PS: From John Wolff's Web Museum: Quote:Thus we're not too bad with our results. 

12022014, 12:16 AM
Post: #7




RE: Question for Trig Gurus
Thanks Thomas for the trigonometric identities. I am looking for some identities between the inverse trigonometric functions and the regular trigonometric functions, since the calculator has sin, cos, and tan functions but not their inverses. So for example what is f() and g() that define for example the arcsine as a function of the sine and a function of the angle:
arcsin(x) = f(sin(g(x))) Namir 

12022014, 01:12 AM
Post: #8




RE: Question for Trig Gurus
(12022014 12:16 AM)Namir Wrote: So for example what is f() and g() that define for example the arcsine as a function of the sine and a function of the angle: f(x)=arcsin(g^{1}(arcsin(x))) But that's probably not what you are looking for. I assume that f and g should be "simple" functions that are available on this calculator or at least a simple combination of them. However I doubt that there are any. You could try and set up f and g as power series and then try to figure out their coefficients. But the composition of functions is used twice. Good luck! Cheers Thomas 

12022014, 01:30 AM
(This post was last modified: 12022014 01:32 AM by Gerson W. Barbosa.)
Post: #9




RE: Question for Trig Gurus
(12022014 01:12 AM)Thomas Klemm Wrote:(12022014 12:16 AM)Namir Wrote: So for example what is f() and g() that define for example the arcsine as a function of the sine and a function of the angle: WA finds relationships in the complex domain, but they're useless for his purposes. http://www.wolframalpha.com/input/?i=asi...+relations Cheers, Gerson. 

12022014, 01:49 AM
Post: #10




RE: Question for Trig Gurus
(12022014 01:12 AM)Thomas Klemm Wrote:(12022014 12:16 AM)Namir Wrote: So for example what is f() and g() that define for example the arcsine as a function of the sine and a function of the angle: Thomas, The calculator has no inverse trig functions. I am trying to see if there is a math trick to calculate the inverse trig functions using the trig functions (without using an iterative method like Newton's method). What I am looking for is hard to find if it exists at all. I am looking for a math trick to do the job. Namir 

12022014, 01:50 AM
Post: #11




RE: Question for Trig Gurus
(12022014 01:30 AM)Gerson W. Barbosa Wrote:(12022014 01:12 AM)Thomas Klemm Wrote: f(x)=arcsin(g^{1}(arcsin(x))) Gerson, I think you are right. However for a simple calculator like the Casio fx10 imaginary numbers are not supported :(. Namir 

12022014, 02:30 AM
(This post was last modified: 12022014 02:32 AM by toml_12953.)
Post: #12




RE: Question for Trig Gurus
(12012014 10:54 PM)Namir Wrote: [quote='PANAMATIK' pid='22183' dateline='1417466789'] The manual I found was for both the FX10 and FX50. It doesn't say there's a difference between the keyboards so I figured both had all the functions. Here's a BASIC subroutine to calculate ATN(X): 30000 REM ARCTANGENT: A9=ATN(X9) 30300 REM VARIABLES USED: A9,B9,C9,T9,X9 30310 T9=SGN(X9): X9=ABS(X9): C9=0 : IF X9>1 THEN C9=1 : X9=1/X9 30330 A9=X9*X9 : B9=((2.86623E3*A91.61657E2)*A9+4.29096E2)*A9 30330 B9=((((B97.5289E2)* A9+.106563)* A9.142089)*A9+.199936)*A9 30340 A9=((B9.333332)*A9+1)*X9 : IF C9=1 THEN A9=1.5708A9 30350 A9=T9*A9 : RETURN Tom L Cui bono? 

12022014, 09:21 AM
Post: #13




RE: Question for Trig Gurus
(12022014 02:30 AM)toml_12953 Wrote:(12012014 10:54 PM)Namir Wrote: [quote='PANAMATIK' pid='22183' dateline='1417466789'] Do you have a link to the fx10 manual? 

12022014, 10:14 AM
Post: #14




RE: Question for Trig Gurus
(12022014 09:21 AM)Namir Wrote: Do you have a link to the fx10 manual?http://lmgtfy.com/?q=casio+fx10+manual 

12022014, 04:57 PM
Post: #15




RE: Question for Trig Gurus
(12022014 10:14 AM)Didier Lachieze Wrote:(12022014 09:21 AM)Namir Wrote: Do you have a link to the fx10 manual?http://lmgtfy.com/?q=casio+fx10+manual Thanks Didier. As I suspected, the FX10 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 

12022014, 07:34 PM
Post: #16




RE: Question for Trig Gurus
(12022014 04:57 PM)Namir Wrote:(12022014 10:14 AM)Didier Lachieze Wrote: http://lmgtfy.com/?q=casio+fx10+manual This should be good enough for practical purposes: asin(x) ~ (1/2*e^(x)*(e^(2*x)1)+e^(e^(2.7004920941.821362659*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. 

12042014, 02:36 AM
(This post was last modified: 12052014 01:07 AM by Gerson W. Barbosa.)
Post: #17




RE: Question for Trig Gurus
(12022014 07:34 PM)Gerson W. Barbosa Wrote: For the range 1/2..1, we use simmetry. This is completely wrong (including spelling), sorry! Desirable accuracy should be granted in the range [0..sqrt(2)/2], then for the range [sqrt(2)/2..1] we use the formula asin(x) = 90°  asin(sqrt(1  x^2)) So the previous approximation has been adjusted for a wider range. It's somewhat less accurate, but still good for practical purposes: asin(x) ~ 90*(2*e^(16.01031715*e^(1.950569752*x)) + e^(x)  e^(x))/pi Examples: x asin(x) 0.00 00.00 0.05 02.87 0.10 05.74 0.15 08.63 0.20 11.54 0.25 14.48 0.30 17.46 0.35 20.48 0.40 23.57 0.45 26.74 0.50 29.99 (30.00) 0.55 33.37 0.60 36.88 0.65 40.55 0.70 44.43 0.71 45.23 http://www.wolframalpha.com/input/?i=plo...282%29%2F2  Somewhat better: asin(x) ~ 180/pi*(0.000005592*(x + 1)^15.096 + sinh(x)) or asin(x) ~ 90/pi*e^(x)*(0.000011184*e^x*(x + 1)^15.096 + e^(2*x)  1) Examples: x asin(x) 0.00 00.000 0.05 02.867 (02.866) 0.10 05.740 (05.739) 0.15 08.629 (08.627) 0.20 11.541 (11.537) 0.25 14.483 (14.478) 0.30 17.465 (17.458) 0.35 20.495 (20.487) 0.40 23.586 (23.578) 0.45 26.750 (26.744) 0.50 30.002 (30.000) 0.55 33.365 (33.367) 0.60 36.864 (36.870) 0.65 40.536 (40.542) 0.70 44.429 (44.427) 0.71 45.239 (45.235) http://www.wolframalpha.com/input/?i=plo...282%29%2F2  Less accurate, but more simple: asin(x) ~ sinh(x)*(2  (1  x^6)^(1/6))*180/pi 

12052014, 02:48 AM
Post: #18




RE: Question for Trig Gurus
Thank you for all these approximations and the work you put in showing the tables of results and errors!
Namir 

12052014, 02:44 PM
(This post was last modified: 12052014 06:57 PM by Gerson W. Barbosa.)
Post: #19




RE: Question for Trig Gurus
(12052014 02:48 AM)Namir Wrote: Thank you for all these approximations and the work you put in showing the tables of results and errors! Not so much work, just copied & pasted from Excel :) I had never heard of a calculator which offered sin, cos and tan but not their inverses. Thanks for sharing this information with us! If that were my only calculator, I'd use the simplest approximation as a basis for the others (perhaps there are even more simple approximations for acos or atan, but I haven't investigated yet). As we can see in the plot, it's good enough for practical purposes. No cumbersome constants to enter (no numerical curve fitting used to obtain it, only three or four plottings). If slightly more accure results are required, just replace the 6's in formula with 5.8: 180/pi*(sinh(x)*(2  (1  x^5.8)^(5/29)) x asin(x) 0.00 00.00 0.05 02.87 0.10 05.74 0.15 08.63 0.20 11.54 0.25 14.47 (11.48) 0.30 17.45 (17.46) 0.35 20.47 (20.49) 0.40 23.55 (23.58) 0.45 26.71 (26.74) 0.50 29.95 (30.00) 0.55 33.31 (33.37) 0.60 36.81 (33.87) 0.65 40.51 (23.58) 0.70 44.46 (44.43) √2/2 45.05 (45.00) max abs error: ~ 6/100 degrees Gerson.  P.S.: The maximum absolute error of the latter in the range [0..√2/2] compares to that in asin(x) ~ (sinh(x) + (689*x^9 + 1008*x^7 + 1512*x^5)/22680)*180/pi, which can be obtained on the HP49G/50G this way: 'ASIN(X)' 'X' 9 TAYLR 'SINH(X)' 'X' 9 TAYLR  EXPAND 

12092014, 02:19 PM
(This post was last modified: 12102014 07:41 PM by Namir.)
Post: #20




RE: Question for Trig Gurus
Based on Gerson's fine comments and math work, I did some curve fitting for asin(x), sinh(x), and the error between the two for 0.5<= x <= 0.9 and I found:
asin(x) = sinh(x) for 0 <= x <= 0.5 The first error correction formula (based on log10(error) vs x: asin(x) = sinh(x) + 10^(4.48918286080606 + 3.8564664209301*x) And a somewhat less accurate formula (based on log10(error) vs log10(x)): asin(x) = sinh(x) + 10^(0.770263767622744 + 6.27135730622513*log10(x)) I guess you can use the first approximation even up to x=1. Also, acos(x) = pi/2  asin(x) and, atan(x) = tanh(x) for 0 <= x <= 1 Since the Casio FX10 has the e^x function, I can calculate hyperbolic functions with some reasonable ease. :) Namir 

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