# Is there an equivalent of cosx=1-(x^2/2) for the sin function

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1. Oct 24, 2014

### Physgeek64

Hi, i was just wondering since cosx=1-(x^2/2) is there a similar formatted formula for sinx??

much appreciated :) :)

2. Oct 24, 2014

### GFauxPas

One of the definitions of cosine is:

$\cos x = 1 - \frac {x^2}{2!} + \frac {x^4}{4!} - \frac {x^6}{6!}+\ldots$

going on forever. If you take only a finite number of terms, then you'll have an approximation.

The corresponding series (infinite sum) for sine is:

$\sin x = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!}+\ldots$

Google "factorial" if you haven't seen "n!" before.

3. Oct 24, 2014

### Physgeek64

ahh okay!! we're getting round to this in maths after had term. does it set up a GP??

so would it end up as x+(x^3/3) as an approximation?? xx

4. Oct 24, 2014

### .Scott

No. Using only the first 2 terms, it would be: $x-{x^3\over6}$.

5. Oct 24, 2014

### Mentallic

Well no. Your first hint as to why it isn't a GP is that if it were, we would most definitely be applying the formula for the GP of an infinite sum. That would then mean that sin(x) could be easily represented as a simple fraction in terms of x and that would change everything in maths.

Your second hint is that if you divide the first by the second term, the second by the third, etc. you won't get the same result each time, so it can't be a GP.

x-x3/3 :)

6. Oct 25, 2014

### Physgeek64

okay. Thank you so much!! I'm doing these physics papers and they're non-calculator and a lot of the involve using sin and cos for non-standard angles. Is there one for tan as well?? xx

7. Oct 25, 2014

### .Scott

I'll take that smiley as an exclamation point!
$x-{x^3 \over 6}$

Last edited: Oct 25, 2014
8. Oct 25, 2014

### .Scott

Aside from using sin/cos, there is this:
$tan(x)=x+{x^3\over 3}+{2x^5\over 15}+...$
See http://en.wikipedia.org/wiki/Taylor_series. About half way down there is a list of trig functions.

9. Oct 25, 2014

### Mentallic

Yes that's exactly what I was aiming for haha

10. Oct 27, 2014

### Physgeek64

Thank you!!!! xx