Simplification of Trigonometric Expression

Click For Summary

Discussion Overview

The discussion revolves around the simplification of the trigonometric expression $$\cos[k_n(L-\varepsilon)]-\cos k_nL$$, particularly focusing on the implications of small epsilon in the context of trigonometric identities and Taylor expansions. The scope includes mathematical reasoning and technical explanations related to trigonometric functions.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • Some participants propose that the expression can be simplified to $$2\sin k_nL$$ using a trigonometric identity for small epsilon.
  • Others argue that this simplification is incorrect and suggest it approximates to $$\sim k_n \varepsilon\, \sin(k_n L)$$ instead.
  • A participant inquires about the specific trigonometric identity used in the proposed simplification.
  • Another participant mentions the cosine of a sum formula and notes that a Taylor expansion of the first term yields the same result.
  • There is a suggestion that using the cosine sum formula leads to $$\sin k_nL\sin k_n\varepsilon$$, assuming small epsilon.
  • A participant recalls that $$\sin\theta\approx\theta$$ for small theta, which may relate to the approximations being discussed.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus, as there are competing views on the correct simplification of the expression and the appropriate mathematical approach to take.

Contextual Notes

The discussion includes assumptions about the behavior of trigonometric functions for small angles, and the validity of different mathematical identities and expansions is not resolved.

Dustinsfl
Messages
2,217
Reaction score
5
$$
\cos[k_n(L-\varepsilon)]-\cos k_nL
$$
I just read this for small epsilon this result can be further simpli ed by using the trigonometric identity for $\cos[k_n(L-\varepsilon)]$. Doing so result is
$$
2\sin k_nL
$$
I don't see this? Can someone explain?
 
Mathematics news on Phys.org
dwsmith said:
$$
\cos[k_n(L-\varepsilon)]-\cos k_nL
$$
I just read this for small epsilon this result can be further simplied by using the trigonometric identity for $\cos[k_n(L-\varepsilon)]$. Doing so result is
$$
2\sin k_nL
$$
I don't see this? Can someone explain?

It's not true, it is \(\sim k_n \varepsilon\, \sin(k_n L)\)

CB
 
CaptainBlack said:
It's not true, it is \(\sim k_n \varepsilon\, \sin(k_n L)\)

CB

What trig identity did you use though?
 
dwsmith said:
What trig identity did you use though?

You can use the cosine of a sum formula, but you get the same result by taking a Taylor expansion of the first term.

CB
 
Last edited:
CaptainBlack said:
You can use the cosine of a sum formula, but you get the same result by taking a Mclaurin expansion of the first term.

CB

If we use cosine sum, shouldn't it be
$$
\cos k_nL\underbrace{\cos k_n\varepsilon}_{\approx 1 \text{ for small }\varepsilon} + \sin k_nL\sin k_n\varepsilon - \cos k_nL = \sin k_nL\sin k_n\varepsilon
$$
 
Recall $\displaystyle \sin\theta\approx\theta$ for small $\displaystyle \theta$.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K