0
$\begingroup$

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 $\alpha$?

$\endgroup$
1
  • 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

1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

We can clarify this by considering the respective Taylor series:

$\sin{x}=x-\frac{x^3}{6}+...$

So near zero,to first order we have,

$\sin{x} \approx x$.

On the other hand,for

$\cos{x}=1-\frac{x^2}{2}+...$

Therefore for cosine to first order,we have,

$\cos{x} \approx 1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.