# Why is $\sin(d\Phi) = d\Phi$ where $d\Phi$ is very small?

I haven't touched Physics and Math (especially continuous Math) for a long time, so please bear with me.

In essence, I'm going over a few Physics lectures, one which tries to calculate the Force exerted by uniform magnetic field on a semi circular current carrying wire.

The mathematics that puzzles me is this, that:

$$\sin(d \phi) \thickapprox d\phi$$

where $d\phi$ is very small. Link to video.

• The glib answer: because there's not much difference between the lengths of a chord and an arc length when the angle of a circular arc is small. – Semiclassical Jul 17 '14 at 17:13
• Do you understand what "$\sin(x)\sim x$ when $x$ is small" means? – Git Gud Jul 17 '14 at 17:16
• It might help to know that radians are defined for the sake of making $\sin(x) \approx x$ true for small $x$. – DanielV Jul 17 '14 at 20:45
• The video in the link talks about a charged particular in a uniform magnetic field and does not use the approximation in your question so far as I saw. Since this is calculus, $d\phi$ is a differential. You can't perform an operation on a differential like taking the sine of it. Either this is some kind of typographical error or of abuse of notation. (If the latter, the real understanding is being glossed right over.) (Can't say I liked that video, anyway. It seems to encourage memorizing and guesswork over a more rigid approach; I always found that was far too error prone for me.) – jpmc26 Jul 18 '14 at 2:07
• @jpmc26 Physicists have been taking functions of differentials longer than pure mathematicans have known about differential calculus ^_^ – DanielV Jul 18 '14 at 13:26

Just draw the diagram!

What does $\sin x$ mean? it's the ratio of the opposite side to the hypotenuse in a triangle.

Now, let's draw a triangle with a small angle $x$ inside the unit circle:

$\quad\quad\quad$ Now clearly, when the angle becomes really small, the opposite side is approximately the arc length. In radians, the arc length in a unit circle is exactly the angle $x$, and so we have for small angles:

$$\sin x = \frac{\text{opposite}}{\text{hypot}} = \frac{\text{opposite}}{1}\approx \frac{x}{1} = x$$

• Thanks perfectly explained :) – Games Brainiac Jul 17 '14 at 17:31
• This is really a bad explanation, because it is misleading and ends up not justifying precisely what is in question. It is easy to draw diagrams that "clearly" show all sorts of false relationships. See here, for instance. – Andrés E. Caicedo Jul 17 '14 at 17:44
• @AndresCaicedo - you are of course correct. This wasn't meant as a "proof", but rather as a good way to gain intuition. – nbubis Jul 17 '14 at 17:50
• You can make this proof more rigorous by considering areas instead of arclengths. Specifically, the area of sector $BAC$ is $\frac{1}{2}\theta$ while the area of triangle $BAC$ is $\frac{1}{2}\sin \theta$. – JimmyK4542 Jul 17 '14 at 19:42
• I like this answer because it also tells us that $\cos x\approx1$ for small values of $x$. – user1729 Jul 18 '14 at 8:24

If you are familiar with Taylor series you know that the series of $\sin(x)$ expanded at $0$ is:

$$\sin{(x)} = x - \frac{x^3}{6} + \frac{x^5}{120} + \cdots$$

Then, if $x$ is very small you can neglect all term of order greater than one getting:

$$\sin{(x)} \approx x$$

You can also show this result using basic trigonometry but this approach seems easier to me.

• Yes, but this is an overkill, just the Taylor approximation of order one (or order two) suffices: There is a constant $C$ such that for all $x$ sufficiently small, $|\sin(x)-x|\le C|x|^3$. Note that for $x$ small, $x^2$ is much smaller, so (even if $C$ is "big", which is not the case here), $C|x|^3$ is orders of magnitude smaller than $|x|$. This is more precise than simply saying $\sin x\approx x$, and conceptually easier than establishing the validity of the Taylor series. – Andrés E. Caicedo Jul 17 '14 at 17:26
• This is also very interesting. Thanks for this, its a new perspective on things. – Games Brainiac Jul 17 '14 at 17:31

You can give a linear approximation for $\sin$ near $0$ based on this formula: $$f(x)\approx f(x_0)+(x-x_0)f'(x_0),$$ and using the fact that: $\sin^\prime=\cos$, you get: \eqalign{\sin x&\approx \sin0+(x-0)\cos(0)= x.} So when $x$ is very small, you have that $\sin x\sim x.$

What this intuitively means, is that when you observe closely the graph of the curve $\color{darkmagenta}{\sin x}$ near $0$, it starts to resemble a line, and this line is described by $y=\color{darkblue}x$.

$\phantom{XXX}$ • Can you perhaps give a reference/name for your formula? – user1729 Jul 18 '14 at 8:22
• @user1729: that's just the first-order Taylor expansion, with the remainder "hidden" in the $\approx$ symbol. A slightly more rigorous form would be $f(x) = f(x_0) + (x-x_0) f'(x_0) + o(x-x_0)$ (using the Peano form for the remainder). – Matteo Italia Jul 18 '14 at 10:12
• @user1729 It is also known as Newton's approximation (because from it one can easily derive Newton's method although it is misnamed since it's Raphson's version of the method, rather than Newton's, that is to be found in most textbooks.) – Hakim Jul 28 '14 at 0:57

I think one way to think of it is $\displaystyle\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$. Which means that as $x$ becomes very small, the ratio goes to one, i.e, $\sin x$ can be approximated by $x$.

• That a bit of a circular argument.. – nbubis Jul 17 '14 at 17:23
• @nbubis I don't see how the argument is circular; could you explain? – Jam Jul 17 '14 at 17:26
• @nbubis Most trig arguments are circular ;) – dlev Jul 17 '14 at 17:32
• @dlev I had to edit my answer to avoid that pun. Also, ALL trig arguments are circular (excuse my hyperbole!). – Jam Jul 17 '14 at 17:34
• @Hakim - Saying that the limit of a ratio is $1$ is just re- stating the original statement in more formal terms. It does not explain why this is so. Arguably, my answer is not a "proof", just an intuitive explanation, but neither is this one :) – nbubis Jul 17 '14 at 17:36

Substituting $x$ for $d \Phi$...

I would say $\sin x \approx x$ when $x \approx 0$ because...

1. $\sin x$ is a smoothly varying function with no discontinuities.
2. $\sin x = 0$ when $x = 0$
3. The gradient of $\sin x$ is equal to the gradient of $x$ when $x = 0$
4. The second order derivative of $\sin x$ is $-\sin x$, which is $0$ when $x=0$

On point 3, the derivative of $\sin x$ is $\cos x$ which evaluates to $1$ when $x=0$, and the derivative of $x$ is $1$ (at all points).

This is closely related to the Taylor series argument.

As already was said Taylor's theorem answer this question in a precise way. $\sin(d\Phi)=d\Phi+O((d\Phi)^3).$ I don't see another better explanation than this one.