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Why do small angle approximations only hold in radians? All the books I have say this is so but don't explain why.

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    $\begingroup$ @EmilioPisanty perhaps you're correct, but anonymous disparagement is hardly the best way to greet a new user. Anyway, user28435, try Googling one-line answers before asking in future (this must have been asked hundreds of times on Math SE). $\endgroup$ – Meow Aug 19 '13 at 21:36
  • $\begingroup$ Don't think of sine as being defined in terms of angles - rather, arcs length of a circle. (The approximations come from the fact that a circle is approximately a straight line, especially near the point of tangency.) $\endgroup$ – Akiva Weinberger Oct 21 '14 at 10:55
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The reason why we use radians for $\sin x$ (or other trigonometric functions) in calculus is explained here and here in Wikipedia.

Having known that, notice that small angle approximation is just the Taylor expansion of $\sin x$ and $\cos x$ around $x=0$:

$$\sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\tag{1}$$ $$ \cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}\tag{2}$$

If you scale $x$ by some constant $\omega$, then you must replace $x$ with $\omega x$ in $(1)$ and $(2)$. So, working in degrees $($ $\omega=\frac{\pi}{180}$ $)$, the approximation will become: $$\sin \theta\approx \frac{\pi}{180}\theta\tag{$\theta$ in degrees}$$

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There are several correct ways to answer this that illuminate different aspects of what is going on (and I wouldn't be surprised if the answer is present on this site somewhere EDIT: indeed the other responses to this question are different ways of looking at it). Here is a geometrical answer.

The basic formula is

\begin{equation} l = r \theta \end{equation}

where $l$ is the arc length of a segment of a circle with radius $r$ subtending an angle $\theta$. For this formula to be true, $\theta$ needs to be in radians. (Just try it out, is the arc length of the whole circle equal to $360 r$ or $2\pi r$?) In fact, that formula is really how you define what you mean by radians.

Now consider a straight line segment connecting the two endpoints of the arc subtended by $\theta$. Call it's length $a$. You can show by fiddling with triangles that \begin{equation} a=2r\sin\left(\frac{\theta}{2}\right) \end{equation}

In the limit that the angle is small (so only a small piece of the circle is subtended), you should be able to convince yourself that $a\approx l$. This is the core of the small angle approximation.

Using the two relationships above we have

\begin{equation} r\theta \approx 2r \sin\left(\frac{\theta}{2}\right) \end{equation} or \begin{equation} \sin\left(\frac{\theta}{2}\right)\approx\frac{\theta}{2} \end{equation} You can see that using radians was crucial here because it allowed us to use $l=r\theta$.

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A 'small angle' is equally small whatever system you use to measure it. Thus if an angle is, say, much smaller than 0.1 rad, it will be much smaller than the equivalent in degrees. More typically, saying 'small angle approximation' typically means $\theta\ll1$, where $\theta$ is in radians; this can be rephrased in degrees as $\theta\ll 57^\circ$.

(Switching uses between radians and degrees becomes much simpler if one formally identifies the degree symbol $^\circ$ with the number $\pi/180$, which is what you get from the equation $180^\circ=\pi$. If you're differentiating with respect to the number in degrees, then, you get an ugly constant, as you should: $\frac{d}{dx}\sin(x^\circ)={}^\circ \cos(x^\circ)$.)

In real life, though, you wouldn't usually say 'this angle should be small' without saying what it should be smaller than. If the latter is in degrees then the former should also be in degrees.

That said, though: always work in radians! Physicists tend to use degrees quite often, but there is always the underlying understanding that the angle itself is a quantity in radians and that degrees are just convenient units. Trigonometric functions, in particular, always take their arguments in radians, so that all the math will work well. Always differentiate in radians, always work analytically in radians. And at the end you can plug in the degrees.

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It's because this relationship $$\lim_{x\rightarrow0}\frac{\sin(x)}{x}=1$$ (i.e. $\sin(x) \approx x$) only holds if $x$ is in radians, as, using L'Hoptial's rule, $$\lim_{x\rightarrow0}\frac{\sin(x)}{x}=\lim_{x\rightarrow0}\frac{\frac{d}{dx}\sin(x)}{\frac{d}{dx}x}=\lim_{x\rightarrow0}\frac{d}{dx}\sin(x)=\begin{cases} 1 & \text{ if } x\text{ is in radians}\\ \frac{\pi}{180} & \text{ if } x \text{ is in degrees} \end{cases}.$$ That is, $\frac{d}{dx}(\sin(x))=\cos(x)$ only if $x$ is in radians. The reason can be divined from this proof without words by Stephen Gubkin:

enter image description here

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    $\begingroup$ Sure but WHY does this only hold in radians. That's the bit I can't understand. Everyone says it holds in radians but no one explains why lol. Thanks anyway. $\endgroup$ – user28435 Aug 19 '13 at 21:03
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    $\begingroup$ @user28435: because in radians the sin of an angle basically equals the angle itself, at small enough angles, so it very much simplifies the algebra if you don't have to keep converting degrees. $\endgroup$ – Mike Dunlavey Aug 20 '13 at 0:12
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You need to clarify what you are really asking: in physics the small angle approx. is typically used to approximate a non-linear differential equation by a linear one - which is much easier to solve - by allowing us to approximate the sin(x) function by x. The resulting equation is only a good one for small angles. I.e., its a physics thing, not a math thing.

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because the sine of an angle is dimensionless [i.e. just a number], the angle has to be also dimensionless [i.e. the radian is dimensionless] ... units on both sides of an equation have to match

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