What is the importance of small angle approximation

Question

I have seen many approximations like

$$\tan x \approx x$$

$$\sin x \approx x$$

$$\cos x \approx 1-\dfrac{x^2}{2}$$

$$\dfrac{\cos^2(x)}{\sin(x)\tan(x)} \approx \dfrac{x^2}{4} + x^{-2} - 1 $$

These are for smaller values of $x$ or theta, but what is the need to take this approximation into consideration?

Can anyone explain why?

Answer 1

They are useful in limit situations.

Mathematically, they are the Taylor polynomials of order 2 centered at zero. The Taylor polynomial $T_{x_0,k}$ of order k of a function $f \in C^k(a,b)$ and centered at $x_0 \in (a,b)$ is the polynomial of degree at most k that best approximates the function $f$ near the point $x_0$; in particular it satisfies the condition (that also determines its form, try to think about it): $$ T_{x_0,k}^{(j)}(x_0)=f^{(j)}(x_0) \quad j=0,1, \dots k $$ where $g^{(j)}(x_0)$ denotes the j-th derivative of $g$ at $x_0$ (the zero derivative is the function itself). So the explicit form of the Taylor polynomial is: $$ T_{x_0,k}(x)= \sum_{j=0}^k \frac{1}{j!}f^{(j)}(x_0)(x-x_0)^k = f(x_0)+f’(x-x_0)+\frac{1}{2}f’’(x_0)(x-x_0)^2+ \dots + \frac{1}{k!}f^{(k)}(x_0)(x-x_0)^k $$ Try to compute the derivatives of the sine, cosine, tangent functions and you’ll see that those polynomials are indeed what you called approximations for small angles.

It’s important to underline that these polynomials are not equal to a function, but (using the same notation) $f(x)-T_{x_0,k}(x)=o((x-x_0)^k)$ is a little-o of $(x-x_0)^k$, that is: $$ \lim _{x \to x_0}\frac{f(x)-T_{x_0,k}(x)}{(x-x_0)^k}=0 $$

Now you can choose to approximate a function with its Taylor polynomial where you are in a limit situation. A classical example is when you’re studying the small oscillation of a pendulum, but here ‘small’ depends on your misurations.

You can find a lot more about Taylor polynomials here

https://en.wikipedia.org/wiki/Taylor%27s_theorem

Answer 2

For an illustration, suppose that you want to solve for $x$ the equation $$\sin (x)+\cos (x)+\tan (x)=\frac 32$$ Using the tangent half-angle substitution, the problem will be to find the root of $$5 t^4-4 t^2+8 t-1=0$$ which is feasible with radicals but which will be a nightmare.

Using the approximations, the approximate solution is given by the root of $$x^2-4 x+1=0\quad \implies \quad x=2-\sqrt 3=0.267949$$ while the exact solution is $0.265909$ (that is to say a relative error of $0.77$%)