Why is the approximation of small angles valid for cosine but not for sine?

[mathfilip]

Hi! I have a question about approximation of functions with small angles. I was looking through some notes from my teacher and didnt understand why the following approximation is valid. We have a system which is at equlibrium at an angle, say a. Now we wanted to se what happens with the equilibrim if we deviate from a, with a small angle say, n. All the calculations isn't necessary here so i cut them out.

He chooses to write cos(a+n) = cosa * cos n - sina * sin n, just by expanding.

However for sin(a + n) he just writes sin(a+n) = sin(a)

Can you please validate this and tell me why this is ok to do? I mean why we don't expand sin(a+n) in the same manner?

 

[Simon43254]

If you check it and expand it yourself you get:

sin(a+n)=sin(a)cos(n)+sin(n)cos(a)

if N is small, Sin(n) is insignificant. let's just say close enough to 0. and using the same principle, if N is small, Cos(n) is close enough to 1. Think of the sin and cos curves for small values. This leaves you with;

sin(a+n)=sin(a)1+0cos(a) which therefore approximates to Sin(a) as stated.

All he has done is missed out the calculation because he knows something about the values of N for sin and cos when they are small.

Btw this should be in the maths section not the physics.

Hope this helped.

Si

 

[betel]

I'm not sure this expansion is correct. If you want to expand for small n you have to keep all term that are linear in n. In the case of the sine this would be
\sin(a+n)=\sin(a)\cos(n)+\sin(n)\cos(a)\approx \sin(a) + n \cos(a)+ \mathcal{O}(n^2)

for the cosine it would be
\cos(a+n)=\cos(a)\cos(n)-\sin(a)\sin(n) \approx \cos(a) - n \sin(a)+ \mathcal{O}(n^2)

It is inconsistent to keep the linear term for the epxansion of the cosine but not the sine.