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What is the reason that for $x<0.5$, $\sin(x)\approx x$?

Are there more known properties of these kind for other trigonometry functions?

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  • $\begingroup$ Such approximations follow from $\sin 0=0$, $\sin' 0=1$ $\endgroup$ Apr 14 '15 at 11:55
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To see that $\sin(x) \approx x$ for small $x$ all you have to do (without using the Taylor series) is look at the graph: enter image description here

You can see that $\sin x = x$ when $x = 0$, and since the gradient of the graph is approximately 1 for $-0.5<x<0.5$, $\sin x$ increases approximately at the same rate as $x$ does. This leads to the result that $\sin x \approx x$ for $-0.5<x<0.5$. For other trigonometric properties like this, see these:

$\cos x \approx 1-\frac{x^2}{2}$

$\tan x \approx x$

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You have $\sin x= x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \cdots$ and for small $x$ only the first term is significant.

Similar expressions for small $x$ are $\cos x \approx 1- \dfrac{x^2}{2}$ and $\tan x \approx x$.

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  • $\begingroup$ So it can be easily seen via the taylor series, Thanks! $\endgroup$
    – gbox
    Apr 14 '15 at 12:04
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Think of the geometric interpretation of $\sin\theta$ and $\theta$ (the one using the unit circle). $\sin\theta$ is the straight-line length from $(\cos\theta,\sin\theta)$ to the $x$-axis. $\theta$ is the curved length along the circle from that point to the $x$-axis. When $\theta$ is small, we're considering a small section of the circle, and a very small section of any smooth curve looks like a line.

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  • $\begingroup$ Remember that, in radians, $\theta$ can be thought of as arc length. $\endgroup$ Apr 14 '15 at 14:34

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