Trigonometric Limit: Solve the Expression

[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.

Please, help.

 

[tiny-tim]

Hi intervoxel!

(have a pi: π and an epsilon:ε )

Learn your trigonometric identities …

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

 

[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.

 

[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. ​

 

[intervoxel]

Thank you for your help.