Limits of cosine and sine 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?
 A: 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:

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 $\cos\theta \rightarrow (\frac{\pi}{2}-\theta)$
A: 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. 
$$\displaystyle\sin(x) \approx x $$
$$\displaystyle\cos(x) \approx 1 $$
A: 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.
A: 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 $1$.
