For small angles, sin(θ) is approximately θ. This post takes a close look at this familiar approximation.

I was confused when I first heard that sin(θ) ≈ θ for small θ. My thought was “Of course they’re approximately equal. All small numbers are approximately equal to each other.” I also was confused by the footnote “The angle θ must be measured in radians.” If the approximation is just saying that small angles have small sines, it doesn’t matter whether you measure in radians or degrees. **I missed the point**, as I imagine most students do. I run into people who have completed math or science degrees without understanding this tidbit from their freshman year.

The point is that the error in the approximation sin(θ) ≈ θ is small, even **relative to the size of θ**. If θ is small, the error in the approximation sin(θ) ≈ θ is really, really small. I use the phrase “really, really” in a precise sense; I’m not just talking like an eight-year-old child. If θ is small, θ^{2} is really small and θ^{3} is really, really small.

For small values of θ, say θ < 1, the difference between sin(θ) and θ is very nearly θ^{3}/6 and so the absolute error is really, really small. The relative error is about θ^{2}/6. If the absolute error is really, really small, the relative error is really small.

The approximation sin(θ) ≈ θ comes from a Taylor series,** but there’s more going on** than that. For example, the approximation log(1 + *x*) ≈ *x* also comes from a Taylor series, but that approximation is not nearly as accurate.

The Taylor series for log(1 + *x*) is

*x* – *x*^{2}/2 + *x*^{3}/3 – …

This means that for small *x*, log(1 + *x*) is approximately *x*, and the error in the approximation is roughly *x*^{2}/2. So if *x* is small, the absolute error in the approximation is really small and the relative error is small. Notice this is one less “really” than the analogous statement about sines.

The reason the sine approximation is one “really” better than the log approximation is that sine is an odd function, so its Taylor series has only odd terms. The error is approximately the first term left out. The series for sin(θ) is

θ − θ^{3} / 3! + θ^{5}/5! – …

so sin(θ) – θ is on the order of θ^{3}, not just θ^{2}.

**We can say even more**. Not only is the error in sin(θ) ≈ θ *approximately* θ^{3}/6, it is in fact *bounded by* θ^{3}/6 for small, positive θ. This comes from the alternating series theorem. When you make an approximation from an *alternating* Taylor series, the error is bounded by the first term you leave out. The argument has to be small enough that the terms of the series decrease in absolute value. For example, if you stick θ = 1 into the series for sine, each term of the series is smaller than the previous one. But if you stick in θ = 2 then the terms get larger before they get smaller. So the precise meaning of “small enough” is small enough that the terms of the alternating series decrease in absolute value.

The series for log(1 + *x*) also alternates, so the error estimate is also an upper bound on the error in that case. Not all series are alternating series. But when we do have an alternating series, we get a bonus as far as being able to control the error in approximations.

It’s also true that for small θ, cos(θ) is approximately 1. At first this may seem analogous to saying that for small θ, sin(θ) is approximately 0. While both are true, the former is much better.

The series for cosine is

1 − θ^{2}/2! + θ^{4}/4! – …

and so the error in approximating cos(θ) with 1 is approximately θ^{2}/2. In fact, since the series alternates, the error is *bound by* θ^{2}/2, if θ is small enough. And in this case the relative error is approximately equal to the absolute error. This says that for small θ, the approximation cos(θ) ≈ 1 is really, really good. By contrast, saying that sin(θ) ≈ 0 for small θ is a bad approximation: the absolute error is approximately θ and the relative error is 100%. Rounding small values of θ down to 0 produces a very good approximation in one context and a poor approximation in another context.

The formula for cosine series seems to omit the factorial operator. I’m only assuming that all denominators in this series are factorials, but since 2!=2, it was omitted. Would the next in the series be [(theta^6)/6!] ? Fascinating stuff.

I am a bit confused on your argument stating that the approximation of sin(θ) ≈ θ is good “for small values of θ, say θ < 1." Are you referring to 1 radian (which is a rather large angle — 57.3 deg), or are you referring to 1 degrees, or for that matter, some other unit (e.g. grad)?

Other than that, your post is very intriguing and enlightening. I have definitely seen the approximation of sin(θ) ≈ θ made numerous times in my physics, math, and engineering courses. In each of those situations, I have generally accepted the approximation and have not considered it in this depth before.

Mark: You are correct that I left out the factorial because 2! = 2. I edited the article to add the factorial since that is clearer. And yes, the next term is -θ

^{6}/ 6!.Eric: Yes, the post assumes angles are measured in radians. I edited the article to point this out.

there is another thing with cos/sin…

Let us say that you want a sin for any angles, BUT you want speed and are ok with low precision (for example for graphing) (for example on a CPU with no FP unit)…

sin(a+b)=cos(a)sin(b)+sin(a)cos(b)

to calcualte sin(c) where c is, for example a fixed point precision

take a decomposition of c as c=a+b where a=ip(a) and b=fp(a).

sin(b)~=b, cos(b)~=1

you get sin(c)=cos(a)*b+sin(a)

calculate the cos/sin(a) with a lookup table, you have 1 multiplication and 1 addition to calculate your sin to a relatively goop precision (that you can tune by changing the decomposition), is fast and does not take a lot of memory as the lookup table can be made as small as you want (within limits)…

cyrille

‘The reason the sine approximation is one “really” better than the log approximation is that sine is an even function, so its Taylor series has only odd terms.’

Should this not call sine an odd function?

Nick: My mistake. I meant to say sine is an odd function. I updated the post. Thanks for catching that.

Up to what value of x we can say that sinx~x

It depends on your application. The error is about x^3 / 6. For how big an x can you ignore x^3 / 6? That’s your answer.

It took me a while to find this post 13 years on, but I did find it, and it was exactly what I needed for work today. Thank you!

Guess I should have said what I worked out, for the next person that comes along. For small values of θ, we have (1 – cos(θ)) / sin(θ) ≈ θ/2.