# Explanation of small angle approximation?

1. Dec 20, 2014

### ah4p

< Mentor Note -- The OP's question was in the thread title. They have been advised to please make more detailed and clear thread starters >

thanks

Last edited by a moderator: Dec 20, 2014
2. Dec 20, 2014

### Fredrik

Staff Emeritus
3. Dec 20, 2014

### Theage

The most obvious way to explain these approximations is truncating the Taylor series about zero. If you haven't seen these, they're basically infinite series for the trig functions. In the case of sine and cosine, they are
$$\sin\theta = \sum_{n=0}^\infty \frac{(-1)^n \theta^{2n+1}}{(2n+1)!}\text{ and }\cos\theta = \sum_{n=0}^\infty \frac{(-1)^n \theta^{2n}}{(2n)!}$$
When $\theta\approx 0$, these approximations will get better and better for smaller numbers of terms. Where do they come from, you ask? Well, in many cases this is how mathematicians define sine and cosine. Otherwise it takes quite a bit of calculus to come up with these formulas, but you can check that they indeed work. Now, going back to $\theta\approx 0$, we write out the first term of sine and the second term of cosine (basically, we're taking terms with $\theta^2$ or lower exponents. We end up with, lo and behold
$$\sin\theta\approx\theta\text{ and }\cos\theta\approx 1-\frac{\theta^2}{2}.$$
Indeed, if you use a graphing program to graph this you get some very promising results around $\theta = 0$.

The first image plots sine in black and $f(x)=x$ in red, the second image plots cosine in black and its approximation in red. Amazingly, the approximations don't look so bad!

Of course, there are always geometric explanations too, but this is the slightly more sophisticated way to view the approximations.

4. Dec 20, 2014

### Astudious

I wouldn't say so - taking the derivative of sin(x) and cos(x) repeatedly and plugging into Taylor's equation is simple enough for anyone who asks this question!

Rationalising why Taylor's theorem works is slightly harder...

5. Dec 20, 2014

### ah4p

thank you very much for that. I noticed when I tried to find the point of intersection of sinx and tanx graphically they were x = 0, pi, 2pi

how does this mean sinx = tanx ?? I thought it meant sinx only equals tan x when x is 0 pi or 2pi??
tah again

6. Dec 20, 2014

### Fredrik

Staff Emeritus
It doesn't, but when x is small, we have $\cos x\approx1$, and therefore
$$\tan x=\frac{\sin x}{\cos x}\approx \frac{\sin x}{1}=\sin x.$$

Edit: I typed $\cos x=1$ above, but I meant $\cos x\approx 1$. I fixed that after the reply below.

Last edited: Dec 20, 2014
7. Dec 20, 2014

### ah4p

ah thank you very much that makes sense :D

8. Dec 20, 2014

### Fredrik

Staff Emeritus
Recall the definition of the derivative:
$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$ This implies that when $h$ is small, we have
$$f'(x)\approx \frac{f(x+h)-f(x)}{h}$$ for all $x$, and therefore
$$f(x+h)\approx f(x)+f'(x)h$$ for all $x$. This implies that when $h$ is small, we have
$$f(h)\approx f(0)+f'(0)h.$$ This result tells us in particular that when $x$ is small, we have
$$\sin x\approx\sin 0+\sin'(0)x =0+\cos 0\cdot x=x,$$ and
$$\cos x\approx\cos 0+\cos'(0)x= 1-\sin 0\cdot x =1.$$

9. Dec 20, 2014

### ah4p

thank very much :D but why does that result in sin x = x

10. Dec 20, 2014

### Astudious

sin(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! + ...
cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + x^8/8! + ...

For x<1, x^n<x(<1) for all n>1. For very small x, by extension, x^n is even smaller for all n>1; so for x close to 0, x^3 and x^5 and further is extremely close to 0 and can be called 0. So, for x close to 0,

sin(x) = x
cos(x) = 1

approximately.

(So tan(x) = sin(x)/cos(x) = sin(x)/1 (using cos(x)=1) = x/1 (using sin(x)=x) = x approximately, for x close to 0)

These relationships are all exactly true for x=0 (just as they are approximately true for x close to 0) and equivalently for x=2pi*n where n is any integer, because the trigonometric functions are periodic over 2pi (because 2pi is 360 degrees and the functions are defined geometrically, sin(x+2pi*n)=sin(x) etc.).

11. Dec 21, 2014

### Fredrik

Staff Emeritus
It doesn't. I explained why $\sin x\approx x$ for small $x$ in post #8, i.e. the post I wrote just before you asked the question I'm quoting now. It was also explained three times in post #2 (use the link), and twice in post #3. Now it's also explained in post #10.