Trig

[archiver]

Posted by Nick at 1/19/2001 6:07 AM EST


Student on team #240, Mach V, from Jefferson Monroe High School and Visteon.



For all you Pre Calc and above students-

Is there any way to go from arctangent (X/Y) to Sine and Cosine?
I need to use it in programming and the chip doesn't use arctangent, tangent, arcsine, or arccosine!

[archiver]

Posted by Jason Rukes at 1/19/2001 10:51 AM EST


Engineer on team #109, Arial Systems & Libertyville HS, from Libertyville High School and Arial Systems Corp & SEC Design.


In Reply to: Trig
Posted by Nick on 1/19/2001 6:07 AM EST:



I found this in the CRC Standard Math Tables Book.

Let A = Arctan(x), then
sin(A) = x/(sqrt(1+x^2))
cos(A) = 1/(sqrt(1+x^2))

sqrt is the square root function.

-Jason

[archiver]

Posted by Nick at 1/20/2001 6:33 AM EST


Student on team #240, Mach V, from Jefferson Monroe High School and Visteon.


In Reply to: Re: Trig
Posted by Jason Rukes on 1/19/2001 10:51 AM EST:



The problem is I CAN'T Let A = Arctan(x). I cannot use arctan what so ever. The function is not recognized by the BS2 chip.

[archiver]

Posted by Dan at 1/20/2001 1:07 PM EST


Other on team #247, da Bears, from Berkley High and PICO/Wisne Design.


In Reply to: Problem is...
Posted by Nick on 1/20/2001 6:33 AM EST:



You don't need to have the arctan function. It was mentioned just to help understand how the final answer was derived. It wasn't meant to be a line in your program.

I assume you want the sine and cosine of the angle whose tan is x/y. If you know x and y, the sine and cosine of that angle (which you'll never calculate) are:
sine = x/sqrt(x^2+y^2) and cosine = y/sqrt(x^2+y^2).

These are functions of x and y only, no angle.

Hope that helped.

-Dan
#247

[archiver]

Posted by Dan at 1/19/2001 2:32 PM EST


Other on team #247, da Bears, from Berkley High and PICO/Wisne Design.


In Reply to: Trig
Posted by Nick on 1/19/2001 6:07 AM EST:



Pretty much the same as the last reply:

A (angle) = arctan(x/y)
tan A = x/y
Setting up a right triangle with x opposite A and y adjacent to A, hypotenuse becomes sqrt(x^2+y^2) and the rest follows:
sin A = x/sqrt(x^2+y^2) and cos A = y/sqrt(x^2+y^2)

This works fine if the angles you're dealing with are between 0 and 90. Is this true?

-Dan
#247