# Trigonometric limit

1. Feb 1, 2010

### intervoxel

I met the following expression in a QM book:

$\frac{sin[(n+1/2)\pi+\epsilon]}{cos[(n+1/2)\pi+\epsilon]}=\frac{(-1)^n\cos(\epsilon)}{(-1)^{n+1}\sin(\epsilon)}$

where $\epsilon << 1$

No matter how hard I try (sine of sum, etc.), I can't see the intermediate steps to this result.

2. Feb 1, 2010

### tiny-tim

Hi intervoxel!

(have a pi: π and an epsilon:ε )

sin(A+B) = sinAcosB + cosAsinB

cos(A+B) = cosAcosB - sinAsinB

sin(n + 1/2)π = (-1)n

cos(n + 1/2)π = (-1)n+1

3. Feb 1, 2010

### intervoxel

Oh, come on, tiny-tim, I'm stuck in this problem.

I arrive at the (wrong) answer: -(cos(ε)-sin(ε)) / (cos(ε)+sin(ε))=-(1-ε)/(1+ε) and not -1/ε, which is correct.

Last edited: Feb 1, 2010
4. Feb 2, 2010

### tiny-tim

oops!

oops!

What was I thinking?

(lesson: check what people tell you, not only to see why it works but sometimes to see whether it works, or you'll never learn anything!)

Try it again … this time with cos(n + 1/2)π = 0.

5. Feb 2, 2010