Why is $\sin \theta$ just $\theta$ for a small $\theta$? [duplicate]

$\begingroup$

When $\theta$ is very small, why is sin $\theta$ taken to be JUST $\theta$?

$\endgroup$ 5

2 Answers

$\begingroup$

It's not just $\theta$.

What you observe is the fact that $\sin \theta$ and $\theta$ approach zero from either side of the number line at a pretty similar rate. This can be best demonstrated with a graph.

You can see that they are about to overlap just at zero. So when $\sin \theta$ is approaching $0$ for some very very small $\theta$ we can approximate it as $\theta$.

Does this make sense?

$\endgroup$ $\begingroup$

=== Geometric ===

.

The red section on the right, d, is the difference between the lengths of the hypotenuse, H, and the adjacent side, A. As is shown, H and A are almost the same length, meaning cos θ is close to 1 and $\theta^2/2$ helps trim the red away. :

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

The opposite leg, O, is approximately equal to the length of the blue arc, s. Gathering facts from geometry, s = A*θ, from trigonometry, sin θ = O/H and tan θ = O/A, and from the picture, $ O \approx s $ and $ H \approx A $ leads to:

:$\sin \theta = {O\over H}\approx{O\over A} = \tan \theta = {O\over A} \approx {s\over A} = {{A*\theta}\over A} = \theta$.

Simplifying leaves,

:$\sin \theta \approx \tan \theta \approx \theta$.

$\endgroup$ 2