Linear approximations are an application of tangent lines to a function. Not exactly a derivative application, but you'll notice that in order to get the tangent line we must take derivatives. So here is how it works. We have a function $f(x)$, its tangent is defined at $x = a$. So, the equation of the tangent line, known as $L(x)$, is $L(x) = f(a)+f'(a)(x-a)$. The point of the tangent line would be located at $(a, f(a))$. So, at $x=a$, the tangent line is the linear approximation to our function $f(x)$.

Let's create an example. Our function is $f(\theta) = \cos{\theta}$ at a point $\theta = 0$, meaning $f(0)=1$. So, this means $f'(\theta)=-\sin{\theta}$ and f'(0)=0. Our linear approximation is $L(\theta)=f(0)+f'(0)(\theta-a)\\= 1 + (0)(\theta-0)\\= 1$
This tells us that $\cos{\theta}$ is approximately 1. Something to note is that it's important that $\theta$ stays small, because as it moves away from our point $a$, in this case, 0, the approximation becomes less and less accurate.

We can take this a step further. Obviously. Now that we have our local approximation at $\theta=0$ which came to be 1, we can use this to find approximations at other points. For example, if I wanted to know the approximation for $f(\theta)=\frac{\pi}{6}$, we would see first what the derivative of our function is at our original point 0. Since the derivative is $f'(\theta) = \sin{\theta}$, that means at 0, the slope is 0. So, this tells us that the tangent line must be a flat line, and is therefore 1. However, we can tell using a unit circle or our brains that the value of $\cos{\frac{\pi}{6}}$ is $\frac{\sqrt{3}}{2}$, which is about 0.866...not quite 1. That's why it's called an approximation right? You're supposed to take it with a grain of salt. If we were to choose values larger than $\frac{\pi}{6}$, that margin of error would grow larger and larger.