# Maclaurin expansion of sin(cosx)

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#2

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) = , so sin(cos(x)) = .

But you know f(x) is a polynomial, so you can expand it in powers of 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) = , so sin(cos(x)) = .

But you know f(x) is a polynomial, so you can expand it in powers of x.

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#3

(Original post by

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) = , so sin(cos(x)) = .

But you know f(x) is a polynomial, so you can expand it in powers of x.

**Smaug123**)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) = , so sin(cos(x)) = .

But you know f(x) is a polynomial, so you can expand it in powers of x.

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

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#4

**Smaug123**)

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) = , so sin(cos(x)) = .

But you know f(x) is a polynomial, so you can expand it in powers of x.

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#5

(Original post by

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

@OP

**rawsex**)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 i**t?
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(Original post by

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?

**rawsex**)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?

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#7

But cos(x) has its own polynomial expansion: 1-x^2/2+…

So substituting that into the sin(x) expansion gives the required result. The point of Maclaurin expansions is that they go on forever but you can choose to stop calculating them at some point; so if I were to choose to stop at the x^3 term, I would know the answer was ((1-x^2/2)-(1-x^2/2)^3/3!) but truncated at the x^3 term, so I know the answer must be 1-x^2/2 - 1/6 + 3x^2/4 = 5/6-x^2/4.

Alternatively stated, taking a Maclaurin expansion "infects" all other functions: if I take an expansion of a single function in an expression, then I'm free to take the expansion of all other functions in the expression, including cos.

The method I have given gives an expansion to ten terms of:

which is accurate to three d.p. at x=1, and accurate to 7 d.p. at x=0.1.

The general term of the expansion is indeed not nice, but the method for finding the expansion up to n terms is not particularly complicated (though you end up with lots of disgusting number-crunching to do).

So substituting that into the sin(x) expansion gives the required result. The point of Maclaurin expansions is that they go on forever but you can choose to stop calculating them at some point; so if I were to choose to stop at the x^3 term, I would know the answer was ((1-x^2/2)-(1-x^2/2)^3/3!) but truncated at the x^3 term, so I know the answer must be 1-x^2/2 - 1/6 + 3x^2/4 = 5/6-x^2/4.

Alternatively stated, taking a Maclaurin expansion "infects" all other functions: if I take an expansion of a single function in an expression, then I'm free to take the expansion of all other functions in the expression, including cos.

The method I have given gives an expansion to ten terms of:

which is accurate to three d.p. at x=1, and accurate to 7 d.p. at x=0.1.

The general term of the expansion is indeed not nice, but the method for finding the expansion up to n terms is not particularly complicated (though you end up with lots of disgusting number-crunching to do).

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#10

(Original post by

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

**RebeccaLM**)It's really awful! And it's FP2 unfortunately

*general*term.

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#11

(Original post by

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

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

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You can go straight from the definition of a Maclaurin series with

f(x) = f(0) + xf'(0) + x^2f''(x)/2! + ...

Note that sin(cosx) is an even function, so you shouldn't expect to get any odd powers of x

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#13

(Original post by

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

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

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#14

(Original post by

It would be just as bad.

**uncapitalised**)It would be just as bad.

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(Original post by

Differentiating it would be a *****, 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).

**rawsex**)Differentiating it would be a *****, 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).

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(Original post by

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

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

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(Original post by

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

**Lord of the Flies**)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.Posted from TSR Mobile

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#18

iv already got:

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?

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(Original post by

up to x^4 is just madness- am i gonna use the product rule within the product rule up to

iv already got:

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?

**nahomyemane778**)up to x^4 is just madness- am i gonna use the product rule within the product rule up to

iv already got:

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?

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#20

(Original post by

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

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

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