Question

What is the Taylor series expansion for the function f(x)=[1-cos(x)]/sin(x) ?

Answer 1

`(1-cosx)/sinx = 2sum_(n=1)^oo ((−1)^n(2^(2n)-1)B_(2n) )/((2n)!)x^(2n−1)`

Explanation:

Use the trigonometric identities:

`2sin^2 alpha = 1-cos 2alpha`

`2sin alpha cos alpha = sin 2alpha`

Then:

`(1-cosx)/sinx = (2sin^2(x/2))/(2sin(x/2)cos(x/2)) = tan(x/2)`

The McLaurin expansion of the tangent is very difficult to obtain, but you can it from a manual:

`tan(x/2) = sum_(n=1)^oo ((−1)^n2^(2n)(2^(2n)-1)B_(2n) )/((2n)!)(x/2)^(2n−1)`

`tan(x/2) = sum_(n=1)^oo ((−1)^n color(blue)(2^(2n))(2^(2n)-1)B_(2n) )/((2n)! color(blue)( 2^(2n-1)))x^(2n−1)`

`tan(x/2) = 2sum_(n=1)^oo ((−1)^n(2^(2n)-1)B_(2n) )/((2n)!)x^(2n−1)`