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This approximation is used oftenly in physics: $$\sin\theta \approx \theta$$ This approximation is valid for small value of $\theta \leq10°$):

  • But $\sin 1°=0.0174524064$
  • $\sin 2°=0.0348994967$
  • $\vdots$
  • $\sin 10°=0.173648178$

So we see that the approximation $\sin \theta\approx \theta$ gives 99% of error. Please explain.

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    $\begingroup$ The endpoint of this question that you'll understand the connection between functions that return to themselves under differentiation, complex numbers, the trig/circle functions, and the Euler identity $e^{i\pi}+1=0$. With a guide who knows your math background this takes about twenty minutes. Finding an online explanation that matches your level of preparedness might take longer than that, but it's worth the effort. $\endgroup$ – rob Jan 1 at 15:27
  • $\begingroup$ I'm voting to close this question as off-topic because it's a pure mathematics question. $\endgroup$ – rob Jan 1 at 15:27
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    $\begingroup$ Duplicate of: math.stackexchange.com/q/56281/649368 $\endgroup$ – Ertxiem - reinstate Monica Jan 1 at 15:37
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Just to complete the other answers, I've kindly asked my computer to give the relative differences of the approximation up to $10^\circ$.

$$ \begin{array}{cccc} \theta^\circ & \theta_{rad} & \sin \theta & \Delta_{rel} \\ \hline 1 & 0.0174533 & 0.0174524 & 0.00508\% \\ 2 & 0.0349066 & 0.0348995 & 0.02031\% \\ 3 & 0.0523599 & 0.0523360 & 0.04571\% \\ 4 & 0.0698132 & 0.0697565 & 0.08128\% \\ 5 & 0.0872665 & 0.0871557 & 0.12704\% \\ 6 & 0.1047196 & 0.1045285 & 0.18300\% \\ 7 & 0.1221730 & 0.1218693 & 0.24920\% \\ 8 & 0.1396263 & 0.1391731 & 0.32567\% \\ 9 & 0.1570796 & 0.1564345 & 0.41242\% \\ 10 & 0.1745329 & 0.1736482 & 0.50951\% \end{array} $$ where $\theta^\circ$ is the angle presented in degrees, $\theta_{rad}$ is the angle presented in radians, $\sin \theta$ is the sine of the angle and $\Delta_{rel} = \frac{\theta - \sin \theta}{\sin \theta} \times 100\%$ is the relative error of the approximation.

Note that the angle $\theta$ in radians overestimates the value of $\sin \theta$ for positive values of $\theta$. More accurate approximations can be obtained, for instance, through a Taylor expansion: $$ \sin \theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} \ldots , \text{ i.e.,} \\ \sin \theta = \sum_{n=0}^{+\infty} (-1)^n \frac{\theta^{2n+1}}{(2n+1)!} $$

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    $\begingroup$ FWIW, $x(12-x^2)/(12+x^2)$ gives a nice approximation for $\sin x$, for smallish $|x|$ $\endgroup$ – PM 2Ring Jan 1 at 15:30
  • $\begingroup$ The relative error term is $\theta^2/6$ (again, for $\theta\ll 1$). $\endgroup$ – rob Jan 1 at 15:31
  • $\begingroup$ Thanks for your answer. I got the point $\endgroup$ – Kartikey Jan 1 at 15:39
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The approximation $\sin(\theta)\approx \theta$ is valid only if $\theta$ is measured in radians. 1 degree $=0.0175$ radians, and $\sin(0.0175)\approx 0.0175$ (the difference is $8.9322\cdot 10^{-7}$, a relative error of $5\cdot 10^{-5}$).

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    $\begingroup$ excellent point, and glaedig jul og godt nytaar! $\endgroup$ – niels nielsen Jan 1 at 21:11
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Use radians for your angles instead of degrees. Then:

1 degree = 0.0174533 radians. sin(0.0174533) = 0.0174524139...

Which is a very good approximation in most circumstances.

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Convert $1°$ into radians, giving $\theta \approx 0.01745$ rad. $\sin(\theta)$ is also approximately this value so the small angle approximation holds. This formula only works for angles measured in radians because the Taylor series is different if it is measured in degrees.

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  • $\begingroup$ You probably mean $0.01745$ rad instead of $0.1745$ rad. $\endgroup$ – Thomas Fritsch Jan 1 at 15:31
  • $\begingroup$ yeah sorry I'll correct it $\endgroup$ – bemjanim Jan 1 at 16:57

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