# Maclaurin expansion of sin(cosx)

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#1

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0
7 years ago
#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.
1
7 years ago
#3
(Original post by 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.
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?
1
7 years ago
#4
(Original post by 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.
Isn't f(x) = cos(x) and so not a polynomial?
0
7 years ago
#5
(Original post by 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?
That's what I was thinking. It's horrible.
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#6
(Original post by 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?
It's really awful! And it's FP2 unfortunately

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0
7 years ago
#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:
Spoiler:
Show

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).
1
7 years ago
#8
I suppose we could say

0
7 years ago
#9
(Original post by Indeterminate)
I suppose we could say

That's what Smaug123 said.
1
7 years ago
#10
(Original post by RebeccaLM)
It's really awful! And it's FP2 unfortunately
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.
0
7 years ago
#11
(Original post by RebeccaLM)

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How far are you expected to expand this?

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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7 years ago
#12
Are you sure it's not cos(sin x)? That way it wouldn't be too bad.
0
7 years ago
#13
(Original post by rawsex)
Are you sure it's not cos(sin x)? That way it wouldn't be too bad.
It would be just as bad.
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7 years ago
#14
(Original post by uncapitalised)
It would be just as bad.
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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#15
(Original post by 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).
Differentiating it WAS a *****....I tried.... And yeah I managed to get that thanks though....and interesting username

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#16
(Original post by rawsex)
Are you sure it's not cos(sin x)? That way it wouldn't be too bad.

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#17
(Original post by 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.
Yeah sorry, they asked up to the x^4

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7 years ago
#18
(Original post by RebeccaLM)
Yeah sorry, they asked up to the x^4

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

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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#19
(Original post by nahomyemane778)
up to x^4 is just madness- am i gonna use the product rule within the product rule up to

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

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7 years ago
#20
(Original post by RebeccaLM)
It's OCR FP2 I've just given up....teacher said not to bother with it in the end

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its not hard just laborious- and if try finding the series expansion of sin(x) and substituting the series expansion of cos(x) into it would not be accurate would it? its an approximation of an approximation if you stopped at x^4 for the expansion of cos x
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