# FP2 sin(cosx)

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Find the Maclaurin series of sin(cosx) up to degree 4.

So the Maclaurin series of sinx and cosx give

Sinx= x-(1/6)x^3+(1/120)x^5-(1/5040)x^7+...

Cosx = 1-(1/2)x^2+(1/24)x^4-(1/720)x^6+...

Unlike the Maclaurin expansion of cos(sinx) in which you can say:

1-(1/2)(x-(1/6)x^3+...)^2+(1/24)(x-(1/6)x^3+...)^4...

You cannot do the same for sin(cosx) since the 1's within the cosx Maclaurin expansion will continue to add to the expansion for an infinite number of terms

So how do I expand this therefore..?

So the Maclaurin series of sinx and cosx give

Sinx= x-(1/6)x^3+(1/120)x^5-(1/5040)x^7+...

Cosx = 1-(1/2)x^2+(1/24)x^4-(1/720)x^6+...

Unlike the Maclaurin expansion of cos(sinx) in which you can say:

1-(1/2)(x-(1/6)x^3+...)^2+(1/24)(x-(1/6)x^3+...)^4...

You cannot do the same for sin(cosx) since the 1's within the cosx Maclaurin expansion will continue to add to the expansion for an infinite number of terms

So how do I expand this therefore..?

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#2

(Original post by

Find the Maclaurin series of sin(cosx) up to degree 4.

So the Maclaurin series of sinx and cosx give

Sinx= x-(1/6)x^3+(1/120)x^5-(1/5040)x^7+...

Cosx = 1-(1/2)x^2+(1/24)x^4-(1/720)x^6+...

Unlike the Maclaurin expansion of cos(sinx) in which you can say:

1-(1/2)(x-(1/6)x^3+...)^2+(1/24)(x-(1/6)x^3+...)^4...

You cannot do the same for sin(cosx) since the 1's within the cosx Maclaurin expansion will continue to add to the expansion for an infinite number of terms

So how do I expand this therefore..?

**Mathematicus65**)Find the Maclaurin series of sin(cosx) up to degree 4.

So the Maclaurin series of sinx and cosx give

Sinx= x-(1/6)x^3+(1/120)x^5-(1/5040)x^7+...

Cosx = 1-(1/2)x^2+(1/24)x^4-(1/720)x^6+...

Unlike the Maclaurin expansion of cos(sinx) in which you can say:

1-(1/2)(x-(1/6)x^3+...)^2+(1/24)(x-(1/6)x^3+...)^4...

You cannot do the same for sin(cosx) since the 1's within the cosx Maclaurin expansion will continue to add to the expansion for an infinite number of terms

So how do I expand this therefore..?

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#4

http://mathworld.wolfram.com/MaclaurinSeries.html

I think trying to manipulate the sinx and cosx expansion is incorrect and unwieldy. Try differentiating a few times and then substituting zero for x to find the coefficients.

I think trying to manipulate the sinx and cosx expansion is incorrect and unwieldy. Try differentiating a few times and then substituting zero for x to find the coefficients.

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#5

**Mathematicus65**)

Find the Maclaurin series of sin(cosx) up to degree 4.

So the Maclaurin series of sinx and cosx give

Sinx= x-(1/6)x^3+(1/120)x^5-(1/5040)x^7+...

Cosx = 1-(1/2)x^2+(1/24)x^4-(1/720)x^6+...

Unlike the Maclaurin expansion of cos(sinx) in which you can say:

1-(1/2)(x-(1/6)x^3+...)^2+(1/24)(x-(1/6)x^3+...)^4...

You cannot do the same for sin(cosx) since the 1's within the cosx Maclaurin expansion will continue to add to the expansion for an infinite number of terms

So how do I expand this therefore..?

Trying to do it any other way, I think you're in for a fairly significant world of pain. The coefficients involve terms like sin(1) and if you want to find those by summing the series you're going to need to consider an infinite number of terms and then spot things like

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#6

(Original post by

I do not understand why sin(cosx) this is different from cos(sinx)

**TeeEm**)I do not understand why sin(cosx) this is different from cos(sinx)

But for sin(cos(x)) the same approach has you considering terms of the form (1-x^2/2 + x^4/24)^n and because of the constant term you can't discard any of them if you want to get the right answer.

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#7

(Original post by

You can find cos(sin(x)) to O(x^4) by substituting x-x^3/6 for sin(x) and then using the known expansion of cos x. Because there's no constant term, you don't need to worry about (x-x^3/6)^5 and higher terms.

But for sin(cos(x)) the same approach has you considering terms of the form (1-x^2/2 + x^4/24)^n and because of the constant term you can't discard any of them if you want to get the right answer.

**DFranklin**)You can find cos(sin(x)) to O(x^4) by substituting x-x^3/6 for sin(x) and then using the known expansion of cos x. Because there's no constant term, you don't need to worry about (x-x^3/6)^5 and higher terms.

But for sin(cos(x)) the same approach has you considering terms of the form (1-x^2/2 + x^4/24)^n and because of the constant term you can't discard any of them if you want to get the right answer.

maybe I had to do them a few times and I remember the pain

Standard differentiation (with a few tricks such as bringing y and y' back in) is the way forward for those

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#8

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I would never substitute in these type of expansions ...

maybe I had to do them a few times and I remember the pain

**TeeEm**)I would never substitute in these type of expansions ...

maybe I had to do them a few times and I remember the pain

For context, he's made a couple of other posts where he's said he needs to do it by substitution/manipulation rather than differentiation, so I was pretty sure that was the direction his brain was running in, but I can't believe you were supposed to not use differentiation here.

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(Original post by

I think it's reasonable for the cos(sin) case, but it fails horribly for the other (as we have been telling the OP).

For context, he's made a couple of other posts where he's said he needs to do it by substitution/manipulation rather than differentiation, so I was pretty sure that was the direction his brain was running in, but I can't believe you were supposed to not use differentiation here.

**DFranklin**)I think it's reasonable for the cos(sin) case, but it fails horribly for the other (as we have been telling the OP).

For context, he's made a couple of other posts where he's said he needs to do it by substitution/manipulation rather than differentiation, so I was pretty sure that was the direction his brain was running in, but I can't believe you were supposed to not use differentiation here.

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#10

(Original post by

the question specifically asks for me to substitute

**Mathematicus65**)the question specifically asks for me to substitute

http://www.wolframalpha.com/input/?i...8cos%28x%29%29

You're getting sin(1) and cos(1), both of which aren't rational numbers.

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Also does anyone have any ideas on how to find the Maclaurin series of ln(1+e^x) by manipulation/substitution. I did it by differentiation and got the correct solution but took far too long and my working looked horrific so is there a better way to do this?

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#12

(Original post by

Also does anyone have any ideas on how to find the Maclaurin series of ln(1+e^x) by manipulation/substitution. I did it by differentiation and got the correct solution but took far too long and my working looked horrific so is there a better way to do this?

**Mathematicus65**)Also does anyone have any ideas on how to find the Maclaurin series of ln(1+e^x) by manipulation/substitution. I did it by differentiation and got the correct solution but took far too long and my working looked horrific so is there a better way to do this?

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#13

(Original post by

I don't think it's wise or possible.

You're getting sin(1) and cos(1), both of which aren't rational numbers.

**Alex:**)I don't think it's wise or possible.

You're getting sin(1) and cos(1), both of which aren't rational numbers.

*possible*- but it requires some fairly inspired recognition to convert various infinite series into sin 1 and cos 1.

Wise? Definitely not...

EdIt: Thought of a better approach (if we absolutely insist on doing this by substitution). It's the fact that cos(0) = 1 that kills us, so we need to finesse that.

So rewrite sin(cos(x)) as sin(1+(cos(x) - 1) = sin(1)cos(cos(x)-1) + cos(1) sin(cos(x) - 1).

Then cos(x) - 1 = (x^4/24 - x^2/2) + O(x^6). So our problem reduces to finding sin(1) cos(x^4/24 - x^2/2) + cos(1) sin(x^4/24-x^2/2) up to terms in O(x^4), which is not TOO bad and doesn't involve infinite sums.

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(Original post by

Okay, you've got to be using differentiation now. Leave manipulation and using standard series behind unless it is trivially easy. Differentiating ln(1+e^x) isn't too difficult.

**Alex:**)Okay, you've got to be using differentiation now. Leave manipulation and using standard series behind unless it is trivially easy. Differentiating ln(1+e^x) isn't too difficult.

Thank you for your help!

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#15

**Mathematicus65**)

Also does anyone have any ideas on how to find the Maclaurin series of ln(1+e^x) by manipulation/substitution. I did it by differentiation and got the correct solution but took far too long and my working looked horrific so is there a better way to do this?

So if you write ln(1+e^x) as ln(2 + (e^x - 1)) = (ln 2) + ln(1 + (e^x-1)/2) then you can replace (e^x-1)/2 by a power series starting (x/2 + x^2/4 + x^3/12 + ...) and sub in).

I would still be surprised if this was quicker than differentiating.

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