My lab manual gives this:

$B$ is a function of $A$, Greek are uncertainties...

$$B + \beta = \sin(A + \alpha) = \sin(A)\cdot\cos(\alpha) + \sin(\alpha)\cdot\cos(A)$$

--> because $\alpha$ is taken to be (at least relatively) small, $\cos(\alpha) \to 1$, and $\sin(\alpha) \to \alpha$, measured in radians.

I see that $\cos(\alpha) \to 1$, but I would have expected that $\sin(\alpha)$ for a small would, by the same logic, go to 0.

Has it got something to do with the rate of change of either function near zero? Why is $\cos(\alpha)$ of small $\alpha$ not also proportional or written by relation to $\alpha$?

  • 1
    $\begingroup$ Have a look at Taylor series. Normally you have to keep at least the linear term to have consistent results. $\endgroup$
    – DarioP
    Jan 17, 2014 at 16:33

3 Answers 3


The easiest way is to plot $x$, $\sin(x)$, and $\cos(x)$. Fortunately, Wikipedia has done that for us:

enter image description here enter image description here

From the first graph, when $x\lesssim0.2$ rad, $\sin(x)\simeq x$. From the second graph, the approximation that $\cos(x)\simeq1$ really only holds when $x\lesssim0.1$ rad; normally one writes it as $\cos(x)\approx1-x^2/2$.

The reason for these approximations come from their series expansion: $$\sin(x)=x-\frac16x^3+\frac{1}{120}x^5+\cdots\\ \cos(x)=1-\frac12x^2+\frac{1}{24}x^4+\cdots $$ when $x$ is small, $x^3\approx0$, and all other higher terms are also zero, thus we eliminate them from the expansion.

  • $\begingroup$ As an aside, $0.2\,{\rm rad}\approx11^\circ$ is when the deviation between the $x$ and $\sin(x)$ is about 0.001. $\endgroup$
    – Kyle Kanos
    Jan 17, 2014 at 16:48

Yes, it has to do with the rate of change of these function near zero. For small $\alpha$ you are able to approximate most functions (including $\cos$ and $\sin$) in a power series of $\alpha^n$, called Taylor series: $$f(0+x)=f(0)+f'(0)\cdot x+f''(0)\cdot \frac{x^2}{2!}+f'''(0)\cdot \frac{x^3}{3!}+\dots$$

Note that the relvance of each further term decreases as $|x|<1$ becomes smaller due to the higher power $n$ of $x^n$.

If you calculated these series for $\cos$ and $\sin$ you get:

$$\cos(\alpha)=1-\frac{\alpha^2}{2}+\frac{\alpha^4}{24}+\dots$$ $$\sin(\alpha)=0+\alpha-\frac{\alpha^3}{6}+\dots$$

Let us now consider that $|\alpha|$ is such small, that you can neglect any power of $\alpha^2$ or higher your approximation would be

$$\cos(\alpha)\approx1$$ $$\sin(\alpha)\approx\alpha$$

So in both approximation you consider a small change of the $\alpha$ around $0$, but as the slope of $\cos$ is zero at that point (in contrast to $\sin$) the value of $\cos(\alpha)$ does approximately not change if you vary $\alpha$ slightly.


We can clarify this by considering the respective Taylor series:


So near zero,to first order we have,

$\sin{x} \approx x$.

On the other hand,for


Therefore for cosine to first order,we have,

$\cos{x} \approx 1$.


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