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Trigonometry Simplification of Trigonometric Expression

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
\cos[k_n(L-\varepsilon)]-\cos k_nL
$$
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

Well-known member
Jan 26, 2012
890
$$
\cos[k_n(L-\varepsilon)]-\cos k_nL
$$
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
 

dwsmith

Well-known member
Feb 1, 2012
1,673

CaptainBlack

Well-known member
Jan 26, 2012
890
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:

dwsmith

Well-known member
Feb 1, 2012
1,673
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

Administrator
Staff member
Feb 24, 2012
13,775
Recall $\displaystyle \sin\theta\approx\theta$ for small $\displaystyle \theta$.