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In physics, particularly in waves, we make use of the fact that for small angles (less than $\pi/12$-ish), the sine function value of an angle is pretty close to the value of the angle itself (in radians of course). Can anyone give a mathematical explanation for why this is?

Dmoreno
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Just_a_fool
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  • If you draw a triangle, and limit the angle $\theta$ to zero, i.e., make it smaller and smaller, you see that the opposite side is very small compared to the hypotenuse. Therefore, $\sin \theta\to0$ as $\theta\to 0$. –  Feb 09 '14 at 16:26
  • You may want to see this question on the corresponding question for cosine: http://math.stackexchange.com/questions/113416/how-to-justify-small-angle-approximation-for-cosine – Batman Feb 09 '14 at 16:27

4 Answers4

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Hint

Any mathematical function can be, at least locally, approximated by so called Taylor or Mc Laurin expansions.

To make it as simple as possible, the tangent to a curve is, at the point where it is defined, a local approximation of the curve.

So, write the equation of the tangent to the curve $y=sin(x)$ at $x=0$ and you will obtain, for the tangent line, $y = 0 + (x-0) = x$. So, close to $x=0$, $sin(x)$ is close to $x$.

Uisng the same approach, you could show by yourself that, close to $x=0$, $e^x$ is close to $1+x$, that $log(1+x)$ is close to $x$ and so on. For sure, the approximations can be made better and better at the price of more terms.

Claude Leibovici
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Hint

Consider the Taylor expansion of $\sin{x}$ about $x = 0$.

Cheers!

Dmoreno
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  • Of course! But now I wonder, is there a way to angle this useful behavior without such series? – Just_a_fool Feb 09 '14 at 16:18
  • I was about to show you that the tangent curve of $\sin{x}$ about $x = 0$ is precisely $y(x) = x$, so $\sin{x} \sim x$ in the neighbourhood of $x \to 0$, but... @Claude Leibovici was quicker than me, haha! – Dmoreno Feb 09 '14 at 16:28
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There is a very simple geometrical observation you can make. The arch in a circle with radius 1 equals the angle x (in radians), and for small x the arch is close to sin x. Draw this and you will see!

Peter Svahn
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Alternative solution, if you do not want to deal with series expansion, you could calculate

$$\lim_{x\to0} \frac{\sin x}{x} = 1\quad\text{and/or}\quad \lim_{x\to0}\frac{x}{\sin x}=1$$

Thus $\sin x \sim x$ for $x$ close to $0$.

user127.0.0.1
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