How to improve the small angle approximations for $\sin x$ and $\cos x$?

Question

If we start with the approximation

$ \sin x \approx x$

we derive, using the trigonometric relation

$ 1 - \cos 2x = 2\sin^2x$

we get that $\cos x \approx 1 - \frac{x^2}{2} $

Is it possible to use this new approximation to improve the approximation for $\sin x$ in order to get more terms of the Taylor series expansion?

Using the same procedure, I found that

$\sin x \approx x - \frac{x^3}{8} $

where the second term of the Taylor's expansion is

$\sin x \approx x - \frac{x^3}{6} $.

How to get the correct next term using elementary trigonometric formulas?

Simply Beautiful Art · Answer 1 · 2016-11-25T01:40:39.040

2

$\large\text{Picard's Iteration Process:}$

Step I:

See that

$$\sin(x)=\int_0^x\cos(t)dt\approx\int_0^x1-\frac{t^2}2dt=x-\frac{x^3}6$$

Step II:

Use step I and see that

$$\cos(x)=1-\int_0^x\sin(t)dt\approx1-\int_0^xt-\frac{t^3}6dt=1-\frac{x^2}2+\frac{x^4}{24}$$

Step III:

Then repeat step I using what you found in step II.

Step III:

Repeat step II using what you find in step III.

Etc.

edited Nov 25 '16 at 01:40

answered Nov 25 '16 at 01:21

Simply Beautiful Art

74,685

Could you post the name of that method you used so that others can search for similar ideas? – AlgorithmsX Nov 25 '16 at 01:24
@AlgorithmsX no idea what you would call it, but it works for repeating derivatives. – Simply Beautiful Art Nov 25 '16 at 01:25
@AlgorithmsX It was from this question of mine. – Simply Beautiful Art Nov 25 '16 at 01:27
It's called a Picard Iteration Process. Here's a post of mine that discusses it for $\tan x$: http://math.stackexchange.com/questions/2020219/how-does-integrating-the-derivative-of-this-approximation-for-tangent-increase-i – AlgorithmsX Nov 25 '16 at 01:30
Here's a link to that. http://www.sosmath.com/diffeq/first/picard/picard.html – AlgorithmsX Nov 25 '16 at 01:31
@AlgorithmsX nice, I didn't know. – Simply Beautiful Art Nov 25 '16 at 01:34
I would put it in your answer, as it can be near impossible to look up more stuff like this without a name or something to search for. – AlgorithmsX Nov 25 '16 at 01:37
@SimpleArt This doesn't answer the question. – Nov 25 '16 at 01:46
1

@Bye_World Yeah, I decided to do that too in the other answer, albeit something I'd rather not do. – Simply Beautiful Art Nov 25 '16 at 01:50
@SimpleArt Thanks for your answer. As I see, this iteration generates a series similar to the Taylor but with greather denominators. I was doubtful about if this series converges to the sine and cosine functions. Is the convergence guaranteed? – Nov 25 '16 at 02:27
@FractionalInquirer I guarantee convergence for both my answer. Notice that at first, there are greater denominators, though eventually those terms will add onto themselves and approach the values you would expect. – Simply Beautiful Art Nov 25 '16 at 02:38

score 1 · Answer 2 · answered Nov 25 '16 at 01:49

Notice that

$$\cos(2x)=2\cos^2(x)-1$$

So, if $\cos(x)\approx1-\frac{x^2}2$, then

$$\cos(2x)\approx2\left(1-\frac{x^2}2\right)^2-1=1-2x^2+\frac{x^4}2$$

Then let $x\to\frac x2$ to get

$$\cos(x)\approx1-\frac{x^2}2+\frac{x^4}{32}$$

So, we repeat:

$$\cos(2x)=2\cos^2(x)-1\approx2\left(1-\frac{x^2}2+\frac{x^4}{32}\right)^2-1$$

etc., you get the idea.

Now that you have the expansion for cosine, we can use the other double angle identity:

$$\sin(2x)=2\sin(x)\cos(x)\approx2x\cos(x)=?$$

This step really depends on how you approximated your cosine, but the general result is the same.

Once you get a new approximation for $\sin(x)$, repeat the above formula over again, and again, etc.

$\large\text{Remark:}$ Notice we started with the so called "small angle approximation" and merely doubled the small angles to make a "double-whatever-the-small-angle-was approximation".

How to improve the small angle approximations for $\sin x$ and $\cos x$?

2 Answers2