Using the concept of finite difference

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Discussion Overview

The discussion revolves around finding the expression for the finite difference of the cosine function, specifically ▲cos x, and how to express this in terms of sine. Participants explore various mathematical identities and approaches related to finite calculus and trigonometric functions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that ▲cos x can be expressed as cos(x+1) - cos(x) and seeks to rewrite this in terms of sine.
  • Another participant provides a breakdown of cos(x+1) using trigonometric identities, questioning the feasibility of expressing the difference solely in terms of sine.
  • A different participant mentions a standard form for finite differences involving sine and cosine, suggesting that the expression can be simplified but acknowledges it results in a multiplicative form.
  • One participant proposes an expression involving sine, but another critiques it as being "a little off" and suggests an alternative form using cofunction identities.
  • Several participants discuss the derivation of the expression -2sin(1/2)sin(x+1/2), referencing a known trigonometric identity for the difference of cosines.
  • One participant seeks clarification on the derivation of the expression -2sin(1/2)sin(x+1/2) and references a previous participant's work.

Areas of Agreement / Disagreement

Participants express differing views on the best way to express ▲cos x in terms of sine, with no consensus on a single preferred form. There are competing approaches and identities discussed, leading to some unresolved questions about the derivation and utility of the expressions presented.

Contextual Notes

The discussion includes various mathematical identities and assumptions that may not be universally accepted or understood, leading to potential limitations in the clarity of the derivations presented.

irony of truth
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I am trying to find the ▲cos x. By using its definition.

It simply turned out to be cos(x+1) - cos(x). How do I express this in
terms of sine? (and as only one term)?
 
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cos(x+1)= cos(x)cos(1)- sin(x) sin(1) so cos(x+1)- cos(x)= cos(x)(cos(1)-1)- sin(x)sin(1). What makes you think you can write it in terms of sine only?
 
irony of truth said:
I am trying to find the ▲cos x. By using its definition.

It simply turned out to be cos(x+1) - cos(x). How do I express this in
terms of sine? (and as only one term)?
That is a completely correct equation for the difference. It is also not a particularly useful form. The standard form for such things in finite calculus is
[tex]{\Delta}^n\cos(x+a)=(2\sin(\frac{h}{2}))^n\cos(x+a+\frac{h}{2}n+\frac{\pi}{2}n)[/tex]
in this specific case where a=0 n=1 h=1
[tex]\Delta\cos(x)=-2\sin(\frac{1}{2})\sin(x+\frac{1}{2})[/tex]
This is one term additive, but two terms multiplicative.
 
Last edited:
irony of truth said:
I am trying to find the ▲cos x. By using its definition.

It simply turned out to be cos(x+1) - cos(x). How do I express this in
terms of sine?

[tex]sin(x+1 - \frac{\pi}{4}) - sin(x-\frac{\pi}{4})[/tex]
 
Antiphon said:
[tex]sin(x+1 - \frac{\pi}{4}) - sin(x-\frac{\pi}{4})[/tex]
little off
you may have been attempting to use the cofunctional identity
[tex]\cos(x)=\sin(\frac{\pi}{2}-x)[/tex]
getting
[tex]\sin(\frac{\pi}{2}-x-1) - \sin(\frac{\pi}{2}-x)[/tex]
or equivalently
[tex]\sin(x-\frac{pi}{2})-\sin(x+1-\frac{\pi}{2})[/tex]
but this form is not perfered for most purposes that would be
[tex]-2\sin(\frac{1}{2})\sin(x+\frac{1}{2})[/tex]
 
Last edited:
Excuse me, how was [tex]-2\sin(\frac{1}{2})\sin(x+\frac{1}{2})[/tex]
derived? My derivation just ended with that of HallsofIvy.
 
irony of truth said:
Excuse me, how was [tex]-2\sin(\frac{1}{2})\sin(x+\frac{1}{2})[/tex]
derived? My derivation just ended with that of HallsofIvy.
A well known identity in trigonometry is
[tex]\cos(a)-\cos(b)=-2\sin(\frac{a-b}{2})\sin(\frac{a+b}{2})[/tex]
The result is mode clear by taking a=x+1; b=x
This identity can be derived by writing
a=(a+b)/2+(a-b)/2
b=(a+b)/2-(a-b)/2
Then using the identity
cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
and then adding the like terms
 
Thank you very much.
 

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