Trigonometry Reference Angle algorithm?

Forum Attachments

Trigonometry Reference Angle algorithm?

Silverwings (guest) 15 Sep 2014 23:46

Hi,

I'd just like to know what an optimal algorithm would be to find the reference angle of a given value.

ht*tp://ww*w.mathopenref.c*om/reference-angle.ht*ml

(please get rid of the *)

For example,
reference angle of 153 degrees = 27 degrees
reference angle of -58 degrees = 58 degrees
reference angle of -220 degrees = 40 degrees

Thanks for any help. I know enough TIBasic to know how to ask for inputs and display the answer, I just need some help for the actual algorithm.

Re: Trigonometry Reference Angle algorithm?

MasterChief2552 16 Sep 2014 18:16

Given an degree θ, the following code will return the reference angle in Ans:

:Prompt θ
:{θ,180-θ,θ-180,360-θ
:Ans(1+int(θ/90
:Disp Ans

I used the list data type to compress all the quadrants into a single line. Of course you also could use If statements, but it would be a bit longer.
I hope this helps.

To enable negative degrees as well, use this code with an extra line instead:

:Prompt θ
:θ+360(θ<0→θ
:{θ,180-θ,θ-180,360-θ
:Ans(1+int(θ/90
:Disp Ans

Edit: If the program should also be able to handle degrees greater than 360 or less then -360, you can take advantage of the modulo function, which only exists on the newer OSs as the remainder( function. If you do not have the latest OS, you can achieve A mod B (remainder(A,B)) by B x fPart(A/B)
I do not have my TI 84 here at the moment, so I cannot look, how remainder( handles negative numbers, and therefore I am not able to optimize the code that much, but it should look similar to this:

:remainder(θ,360→θ

If nothing else works for negative numbers less than -360, you maybe want to use this:

:1-2(θ<0)remainder(abs(θ),360→θ

So:

:Prompt θ
:1-2(θ<0)remainder(abs(θ),360→θ
:θ+360(θ<0→θ
:{θ,180-θ,θ-180,360-θ
:Ans(1+int(θ/90
:Disp Ans

---

Wake me, when you need me.

Last edited on 16 Sep 2014 18:34 by MasterChief2552 Show more