MHB Simplification of Trigonometric Expression

[Dustinsfl]

$$
\cos[k_n(L-\varepsilon)]-\cos k_nL
$$
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?

 

[CaptainBlack]

$$
\cos[k_n(L-\varepsilon)]-\cos k_nL
$$
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

 

[Dustinsfl]

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

CB

What trig identity did you use though?

 

[CaptainBlack]

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

 

[Dustinsfl]

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
$$
\cos k_nL\underbrace{\cos k_n\varepsilon}_{\approx 1 \text{ for small }\varepsilon} + \sin k_nL\sin k_n\varepsilon - \cos k_nL = \sin k_nL\sin k_n\varepsilon
$$

 

[MarkFL]

Recall $\displaystyle \sin\theta\approx\theta$ for small $\displaystyle \theta$.