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When $\theta$ is very small why $\sin \theta$ is similar to $\theta$ and $\cos\theta$ similar to $1$? Is it related to limits or we can prove it simply by using diagrams?


marked as duplicate by uranix, graydad, colormegone, 6005, Inactive - avoiding CoC Aug 26 '15 at 16:54

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  • $\begingroup$ maybe you would like to prove $\lim_{x\to0}(\sin x - x) =0$ :-) $\endgroup$ – Math-fun Aug 26 '15 at 6:19
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    $\begingroup$ @Math-fun How would that help? One might also prove $\lim_{x\to 0}(\sin x - x^2) = 0$ alas that does not imply $\sin x \approx x^2$,,,, $\endgroup$ – CiaPan Aug 26 '15 at 6:56
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    $\begingroup$ @CiaPan ups you are right, thanks! one could look at the ratio $\frac{\sin x}{x}$ then turn to math.stackexchange.com/questions/75130/… $\endgroup$ – Math-fun Aug 26 '15 at 7:05
  • $\begingroup$ Yes, you can visualize it using a diagram. I have uploaded a diagram and explanation. However, this can't be called a 'proof', because a proof needs to be rigorous. $\endgroup$ – voyager Aug 26 '15 at 7:25
  • $\begingroup$ @CiaPan: Yes it does. $\endgroup$ – user14972 Aug 26 '15 at 12:55

On the unit circle, $\theta$ is the length of the arc (as well as the angle extended by that arc). (Thus, perimeter of the unit circle is $2\pi$). Whereas, $\cos\theta$ is the length of the $X$ intercept, and $\sin\theta$ is the length of the $Y$ intercept.

Look at the following diagram: enter image description here

You can now easily visualize that when Point P approaches closer to $(1,0)$, then $\theta \rightarrow \ 0$. At this time, the arc in question will become almost a vertical line, and the $Y$ intercept of the arc is almost the same length as the arc.

Hence as $\theta \rightarrow \ 0$ then $\sin\theta \rightarrow \theta$

And, at that time, the length of the $X$ intercept will get closer and closer to $1$.

Hence as $\theta \rightarrow \ 0$ then $\cos\theta \rightarrow 1$

Also, from this figure, you can easily visualize that when Point P approaches $(0,1)$, the $Y$ intercept will approach $1$ and the $X$ intercept will have same length as the length of the remaining part of the arc (from point P to point $(0,1)$) which is $(\frac{\pi}{2} - \theta)$. (Remember that total length of the arc from $(1,0)$ to $(0,1)$ is $\frac{\pi}{2}$).

Thus, we have:

$\theta \rightarrow \frac{\pi}{2}$ then $\sin\theta \rightarrow 1$, and

$\theta \rightarrow \frac{\pi}{2}$ then $\cos\theta \rightarrow (\frac{\pi}{2}-\theta)$

  • $\begingroup$ Why do you consider the length of arc to be $\theta$? $\endgroup$ – Aneek Aug 26 '15 at 7:27
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    $\begingroup$ I am not considering the length of the arc to be $\theta$, actually it is a fact. The length of the arc of the unit circle IS taken as the angle extended by that arc, this is called Radian angle. Please refer to this: en.wikipedia.org/wiki/Radian. Thus, $90\textdegree$=$\frac{\pi}{2}$, and $180\degree$=$\pi$, and $270\degree$=$\frac{3\pi}{2}$, and $360\degree$=$2\pi$, and so on... $\endgroup$ – voyager Aug 26 '15 at 7:33
  • $\begingroup$ Let me guess, is an unit circle is a circle of radius 1 unit? $\endgroup$ – Aneek Aug 26 '15 at 7:34
  • $\begingroup$ Yes, dear friend! A unit circle is a circle of radius 1, that's why we call it Unit Circle. That's why it intersects $X$ axis at $(1,0)$, and $(-1,0)$, and it intersects $Y$ axis at $(0,1)$, and $(0,-1)$. $\endgroup$ – voyager Aug 26 '15 at 7:41
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    $\begingroup$ @voyager Apparently there is no \degree or \textdegree command in the TeX implementation on MSE. You migh use superscript circle instead — 2^\circ renders as $2^\circ$ $\endgroup$ – CiaPan Aug 26 '15 at 8:03

Using the Taylor series expansion, we can express $\sin x$ and $\cos x$ as

$$\displaystyle\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$$

$$\displaystyle\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$$

Note that when $x \ll 1$, then $x^2 \ll x$ and $x^3 \ll x^2$ and so on. Any higher powers of $x$ will be very small. Then the above functions can be approximated by their first terms.

$$\displaystyle\sin(x) \approx x $$

$$\displaystyle\cos(x) \approx 1 $$

  • $\begingroup$ I get it sir. Thanks $\endgroup$ – Aneek Aug 26 '15 at 6:25

If you use circular function approach to define $\cos \theta$ then for very small but positive $\theta$, we have $\cos \theta = x$ where $x$ is the $x$ coordinate of the point on the unit circle which lies on the terminal side of the angle $\theta$. From the unit circle, you can see that as $\theta$ gets smaller and smaller, $x$ gets bigger and bigger and tends toward the point $(1,0)$ on the $x$ axis. This works like a proof of your claim.


See Wikipedia Small-angle approximation — Geometric:

in a right triangle the side opposite to the small angle is almost same length as the corresponding circular arc, which in turn is equal to the angle (see the definition of radian), hence sine approximately equal the angle. At the same time the long leg is almost equal to hypotenuse, making the cosine approximately equal $1$.


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