Maclaurin expansion of sin(cosx)

You know the Maclaurin expansion of cos(x) and you know the expansion of sin(x).
In order to avoid doing the entire thing for you, I'll pretend the expansion of cos(x) is f(x).
Then sin(x) = $x-x^3/3!+x^5/5!…$, so sin(cos(x)) = $f(x) - f(x)^3/3! + f(x)^5/5!…$.
But you know f(x) is a polynomial, so you can expand it in powers of x.

You know the Maclaurin expansion of cos(x) and you know the expansion of sin(x).
In order to avoid doing the entire thing for you, I'll pretend the expansion of cos(x) is f(x).
Then sin(x) = $x-x^3/3!+x^5/5!…$, so sin(cos(x)) = $f(x) - f(x)^3/3! + f(x)^5/5!…$.
But you know f(x) is a polynomial, so you can expand it in powers of x.

That's an awful way to do it. f(x) goes on forever, how on earth would you simplify this?

@OP the maclaurin series is awful for this expression. Surely it's not for A Level FM is it?

That's an awful way to do it. f(x) goes on forever, how on earth would you simplify this?

@OP the maclaurin series is awful for this expression. Surely it's not for A Level FM is it?

I suppose we could say


$\displaystyle \sin(\cos(x)) =\cos{x} - \frac{\cos^3(x)}{3!} + \frac{\cos^5(x)}{5!} - ...$

It's really awful! And it's FP2 unfortunately

Can anyone take me through how to go about this question?


Posted from TSR Mobile

Are you sure it's not cos(sin x)? That way it wouldn't be too bad.

It would be just as bad.

Differentiating it would be a bitch, but with sin(cos x) you end up with cos(1) and sin(1) terms in your expansion; whereas, you don't with cos(sin x).

Are you sure it's not cos(sin x)? That way it wouldn't be too bad.

Then surely the question includes something like "up to the x^3 term"? I find it hard to believe that you are being asked for the general term.

Yeah sorry, they asked up to the x^4


Posted from TSR Mobile

up to x^4 is just madness- am i gonna use the product rule within the product rule up to $f^4(x)$
iv already got:

$f^3(x)= \sin(x)cos(cos(x))+sin^3(x)cos(cos(x)) - \frac{3}{2}sin(2x)sin(cos(x))$

Besides you would need to find sin (1) and cos (1) to evaluate the series (1 radian btw) which will be messy

where did you get this question and what syllabus?

P.S Lol can you imagine trying to integrate the above derivative?

It's OCR FP2 I've just given up....teacher said not to bother with it in the end


Posted from TSR Mobile