Binomial series expansion of 1/sqrt(1-b²sin²x)

[Oxymoron]

I have been doing some questions on Binomial Series expansion and have been stuck on this particular question for a long time and desperately need some guidance.

Q) Expand (1/(sqrt(1-b^2(sin^2)x)))), where b = sin(1/2(theta)) as a binomial series.

Here is what I have done so far...

Let x = (b^2(sin^2)x) because I want the expression in binomial form.

So it becomes 1/sqrt(1 - x) with k = -1/2

(1-x)^-1/2 can be written in binomial form... (S is capital sigma)

= S(-1/2 n)(-x)^n
= 1 + (-1/2)(-x) + ((-1/2)(-3/2)/2!)*(-x)^2 + ...
= 1 + 1/2x - 3/8x^2 + ...
= 1 + 1/2k^2sin^2x - 3/8k^4sin^4x + ...

Any help on this question would be excellent!

 

[HallsofIvy]

What kind of help do you want? Do you have any reason to believe that what you have is not correct?

 

[Oxymoron]

Sorry about that. What I meant was that if my working was incorrect could someone correct me or offer a simpler way to do it (if any).

Thanks.