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

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Discussion Overview

The discussion revolves around finding a formula for the sine function that parallels the approximation for cosine, specifically the expression cos(x) = 1 - (x^2/2). Participants explore the Taylor series for sine and its relationship to cosine, as well as the concept of geometric progressions (GP) in this context.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant inquires if there is a similar formula for sin(x) as there is for cos(x).
  • Another participant provides the Taylor series for sine, noting it as an infinite sum: sin(x) = x - (x^3/3!) + (x^5/5!) - ...
  • There is a discussion about whether the approximation for sine could be expressed as a geometric progression (GP), with some participants arguing that it cannot be represented as such.
  • One participant suggests an approximation of sin(x) as x - (x^3/3) based on the first two terms of the series.
  • A later reply confirms the approximation and expresses interest in whether a similar series exists for tan(x), leading to the mention of tan(x) = x + (x^3/3) + (2x^5/15) + ...

Areas of Agreement / Disagreement

Participants generally agree on the Taylor series for sine and its approximation, but there is disagreement regarding the classification of the series as a geometric progression. The discussion remains unresolved on the implications of this classification.

Contextual Notes

Limitations include the assumption that participants are familiar with Taylor series and factorial notation, as well as the lack of resolution on the GP classification.

Physgeek64
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Hi, i was just wondering since cosx=1-(x^2/2) is there a similar formatted formula for sinx??

much appreciated :) :)
 
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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.
 
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
 
No. Using only the first 2 terms, it would be: x-{x^3\over6}.
 
Physgeek64 said:
ahh okay! we're getting round to this in maths after had term. does it set up a GP??
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.

Physgeek64 said:
so would it end up as x+(x^3/3) as an approximation?? xx
x-x3/3 :)
 
Mentallic said:
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 :)
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
 
Mentallic said:
x-x3/3 :)
I'll take that smiley as an exclamation point!
x-{x^3 \over 6}
 
Last edited:
Physgeek64 said:
Is there one for tan as well??
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.
 
.Scott said:
I'll take that smiley as an exclamation point!
x-{x^3 \over 6}
Yes that's exactly what I was aiming for haha
 
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Thank you! xx
 
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