I've seen a number of books take the small angle approximation of $\sin(a - b)=0$, and I'm confused because small angle approximation of $\sin(a)\approx a, \,\cos(a)\approx1$.

Using trigonometric properties, $$\sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b)$$

So why isn't the small angle approximation $\sin(a - b)=(a - b)$?

  • 1
    $\begingroup$ I assume that you mean that the quantity "a-b" is small? In that case then, yes, the small angle approximation of sin(a-b) gives sin(a-b)≈(a-b). But since a-b is small, it's also true that sin(a-b)≈a-b≈0. Whether you want to use the substitution sin(a-b)≈(a-b) or sin(a-b)≈0 depends on the problem. Sometimes substituting sin(a-b)≈0 is going too far and effectively deletes the behavior that you're interested in, in which case you want to use sin(a-b)≈(a-b). $\endgroup$ – Samuel Weir Sep 21 '17 at 0:16
  • $\begingroup$ Okay, thank you for replying. I think as long as I state that I should be fine. I'm just a bit worried that I also have (a-b) independently in another part of my equation that I do want to keep, but I want sin(a-b)≈0 to eliminate my stiffness term. $\endgroup$ – SolBladeMegiddo Sep 21 '17 at 0:40
  • 2
    $\begingroup$ Is it possible that a term containing $\sin(a-b)$ was dropped as insignificant for some other reason? That is, does the exposition explicitly say $\sin(a-b)\approx 0$? That seems rather odd, but without the context it's hard to say. $\endgroup$ – garyp Sep 21 '17 at 2:35
  • $\begingroup$ -1. No research effort. Did you search the internet? eg Using your title the #1 hit is en.wikipedia.org/wiki/Small-angle_approximation. $\endgroup$ – sammy gerbil Sep 21 '17 at 10:06

You are absolutely right, $\sin(a-b)\approx(a-b)$

In some situations, when $a\approx b$, this term may be small compared to other terms, and in that case setting it equal to zero is valid.

However, it is usually a good idea not to "optimize too early" by removing terms - or you risk ending up with "0=0". Sometimes you need to account for first order and even second order terms in order to get to the right approximation.

If you have the exact context in which you saw this approximation I may be able to give you more insight.

  • $\begingroup$ I have a problem that is similar to a double pendulum, where the center of mass is in the middle of each bar. I also have external forces acting on the middle of the second bar, torsional springs at the joints and I am trying to find the equation of motion. I have looked at other double pendulum problems and saw them do that simplification. $\endgroup$ – SolBladeMegiddo Sep 21 '17 at 1:02
  • $\begingroup$ What are a and b in your problem? I am not liking the idea of just making them disappear in this expression... $\endgroup$ – Floris Sep 21 '17 at 1:07
  • $\begingroup$ a is my angle from the vertical direction to the second bar (about the joint connecting the two bars), and b is my angle from the vertical direction to the first bar. The problem I am having is that in my equations of motion I end up having a (theta_dot)^2 term for each theta, when there shouldn't be any dampening. $\endgroup$ – SolBladeMegiddo Sep 21 '17 at 1:58
  • $\begingroup$ That sounds like an energy term - not necessarily damping. $\endgroup$ – Floris Sep 21 '17 at 3:19
  • $\begingroup$ I figured out that my Kinetic Energy equation was messed up. $\endgroup$ – SolBladeMegiddo Sep 21 '17 at 4:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.