# Explanation of small angle approximation?

• ah4p
In summary, the small-angle approximations for sine and cosine involve truncating the Taylor series about zero for these trigonometric functions. When theta is close to zero, these approximations get better with fewer terms. The approximations can also be explained geometrically. Additionally, when x is small, we have sinx ≈ x and cosx ≈ 1, which leads to the approximation tanx ≈ x. These approximations are also exactly true for x = 0 and x = 2pi*n, where n is any integer, due to the periodic nature of the trigonometric functions.

#### 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

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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.

berkeman
Theage said:
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,

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...

Theage said:
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 $\theta = 0$.

View attachment 76762

View attachment 76763

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.
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

ah4p said:
how does this mean sinx = tanx ??
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.

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Fredrik said:
It doesn't, but when x is small, we have ##\cos x=1##, and therefore
$$\tan x=\frac{\sin x}{\cos x}\approx \frac{\sin x}{1}=\sin x.$$
ah thank you very much that makes sense :D

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
$$\cos x\approx\cos 0+\cos'(0)x= 1-\sin 0\cdot x =1.$$

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

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.).

ah4p said:
thank very much :D but why does that result in sin x = x
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.

## 1. What is the small angle approximation?

The small angle approximation is a mathematical technique used to approximate the value of a trigonometric function when the angle involved is very small, typically less than 15 degrees. It involves simplifying the trigonometric function by replacing the sine or tangent of the angle with the angle itself.

## 2. Why is the small angle approximation useful?

The small angle approximation is useful because it simplifies complex trigonometric calculations, making them easier to solve. It is often used in physics and engineering to approximate the behavior of systems at small angles, where the error introduced by the approximation is negligible.

## 3. How accurate is the small angle approximation?

The accuracy of the small angle approximation depends on the size of the angle being approximated. It becomes more accurate as the angle approaches zero, but the error increases as the angle gets larger. Generally, the small angle approximation is considered accurate for angles less than 15 degrees.

## 4. Can the small angle approximation be used for any trigonometric function?

No, the small angle approximation is only applicable to the sine and tangent functions. It cannot be used for the cosine function, as the cosine of a small angle is always close to 1 and does not need to be approximated.

## 5. Are there any limitations to using the small angle approximation?

Yes, there are some limitations to using the small angle approximation. As mentioned before, it is only accurate for angles less than 15 degrees. It also does not take into account higher-order terms in the trigonometric function, so it may not be suitable for highly precise calculations. Additionally, it may not be applicable in situations where the angle is not small enough to be approximated.