Taylor Series for sine function

I'm trying to code the sine function using the taylor series with x_0 dependent on x, and -10 <= x <= 10; however, I'm kind of stuck on how to do it. I'm trying to code using only basic operations and the modulo function. What I'm mainly having trouble with is coding the trig identities within my function. For example,

sin(x + pi/2) = cos(x)
cos(x + pi/2) = -sin(x)

Anyone have any helpful tips for trying to switch the cosine and sine values at a particular x, without using a transcedental function, or how to go about coding sine in general that may help relieve me of my problem with the trig identities?

[Walter Roberson]

You would not normally use mod() for taylor series.

Are you trying to do "argument reduction" to take an arbitrary angle and move it into a defined range where you can be more accurate?

If you are operating numerically, then for large angles, using mod() can end up being rather inaccurate

[tsmart15]

Yes, I'm trying to do "argument reduction", since I'm not given a specific x_o. The mod() is only used in my function for coding factorials, as of right now.

[John D'Errico]

Actually, mod works very nicely for a trig Taylor series, though it may not go as far as you want. It gets you into a reasonable start point though, and then you can do further range reduction as desired.

Of course, mod does not apply for all Taylor series. But for trig functions, it is quite convenient, since you can use the periodicity of those functions to your gain.

[Walter Roberson]

mod() isn't going to work for numeric arguments beyond about 3.7E16, by which point eps(x) is greater than 2*pi

[John D'Errico]

Yes, that is very true. But the range here was expressly [-10,10], where mod works just fine for range reduction. Anyway, to compute the sin of numbers that large requires some very careful code written.

https://www.mathworks.com/company/newsletters/articles/the-tetragamma-function-and-numerical-craftsmanship.html

Answer 1

[John D'Errico]

(Note: you might decide to read how I coded trig functions in my HPF toolbox. But that would be wild overkill for basic homework.)

So, think. What does this do for you?

Xhat = mod(x,2*pi);

Where does Xhat live, as compared to x? What do you know about sin(Xhat), as compared to sin(x)? If you are not sure, try some values! Play around. THINK! For example...

sin(10)
ans =
     -0.54402

sin(mod(10,2*pi) - pi)
ans =
      0.54402

Can you do better? Of course. The Taylor series for the sine function will converge better if Xhat lives in the range [-pi,pi), instead of [0,2*pi).

You can pretty easily deal with that, since you would have this identity:

sin(x - pi) = -sin(x)

That suggests you can do this:

Xhat = mod(x,2*pi) - pi;

Where does Xhat now live?

Can you do better? Well, yes. You can reduce the range further. You will first need to decide if you can get satisfactory convergence over that interval, with a reasonable number of terms. For example, if you look at the series for sin(x). Those terms are largest when x is large.

pi.^(1:2:25)./factorial(1:2:25)
ans =
       3.1416       5.1677       2.5502      0.59926     0.082146    0.0073704    0.0004663    2.1915e-05   7.9521e-07   2.2948e-08   5.3927e-10   1.0518e-11   1.7302e-13

(pi/2).^(1:2:25)./factorial(1:2:25)
ans =
       1.5708      0.64596     0.079693    0.0046818   0.00016044   3.5988e-06   5.6922e-08    6.688e-10   6.0669e-12   4.3771e-14   2.5714e-16   1.2539e-18   5.1565e-21

You just need to take a few more terms for the larger domain.

In fact, there are some simple tricks to further reduce the domain of interest. For example, if you do range reduction to +/-pi/4, the series will converge much faster yet. But you may choose a larger interval, just accepting a few more terms in the series. The tradeoff will then be between number of terms and complexity in the range reduction.

[tsmart15]

I'm still missing something. When I graph my sine function, it looks more like tangent, and i'm not getting the correct values of sign for numbers that aren't 30,45,60,90, or equivalently, such as pi/12.

My function:

    f(pi/12) 
ans =

   2.287909096016288

The real answer:

    sin(pi/12) + 2
ans =

   2.258819045102521