Year 13 Maths Help Thread

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    (Original post by RDKGames)
    1. Get everything under same denominator
    2. Tidy up the numerator
    3. Cancel any common factors

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    ok now im more confused. Wouldnt it be easier to multiply the first fraction by x+2 / x+2
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    (Original post by kiiten)
    ok now im more confused. Wouldnt it be easier to multiply the first fraction by x+2 / x+2
    Sure, but then you would need to multiply it by (x+1)/(x+1) because (x+2)(x+1) is the quadratic on the second fraction so you would only be multiplying by one of the factors when you need both.

    All I'm really doing is applying \frac{a}{b}+\frac{c}{d}=\frac{ad  }{bd}+\frac{cb}{db}=\frac{ad+cb}  {bd} and simplifying everything. Doesn't get any more complex than that.
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    (Original post by RDKGames)
    Sure, but then you would need to multiply it by (x+1)/(x+1) because (x+2)(x+1) is the quadratic on the second fraction so you would only be multiplying by one of the factors when you need both.

    All I'm really doing is applying \frac{a}{b}+\frac{c}{d}=\frac{ad  }{bd}+\frac{cb}{db}=\frac{ad+cb}  {bd} and simplifying everything. Doesn't get any more complex than that.
    But the first fraction already has x+1 as a denominator??
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    (Original post by kiiten)
    But the first fraction already has x+1 as a denominator??
    Ah right I see what you mean. Yes you can just ignore multiplying by (x+1) then. Though you should still follow that method if the denominator don't share anything in common.
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    I think my mind is going blank; how do I get the expansion for cosec(2x)? Do I just use binomial with (1+[sin(2x)-1])^{-1}? Seems a bit long winded. I found the expansion for sin(2x) but unsure how I can use that here.
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    (Original post by RDKGames)
    Ah right I see what you mean. Yes you can just ignore multiplying by (x+1) then. Though you should still follow that method if the denominator don't share anything in common.
    Yep. Thanks for your help
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    (Original post by kiiten)
    But the first fraction already has x+1 as a denominator??
    Yep, you've got the right idea, your method is much more suitable and shows understanding. :-)
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    (Original post by RDKGames)
    I think my mind is going blank; how do I get the expansion for cosec(2x)? Do I just use binomial with (1+[sin(2x)-1])^{-1}? Seems a bit long winded. I found the expansion for sin(2x) but unsure how I can use that here.
    How many terms do you need? Just differentiate and plug in x=0 into the derivative, same for the second derivative, etc... up till how many terms you need. If you want the general formula, express it in terms of complex exponentials and use their respective taylor expands, you'll get something in the form of Bernoulli numbers.
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    (Original post by Zacken)
    How many terms do you need? Just differentiate and plug in x=0 into the derivative, same for the second derivative, etc... up till how many terms you need. If you want the general formula, express it in terms of complex exponentials and use their respective taylor expands, you'll get something in the form of Bernoulli numbers.
    I just need the first 3, is there no direct way to use the first 3 terms from expansion for sin(2x) for this?
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    (Original post by RDKGames)
    I just need the first 3, is there no direct way to use the first 3 terms from expansion for sin(2x) for this?
    Heuristically, nope. Not worth it.
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    (Original post by Zacken)
    Heuristically, nope. Not worth it.
    Ah right.

    (Original post by Zacken)
    Just differentiate and plug in x=0 into the derivative, same for the second derivative, etc... up till how many terms you need.
    Do you mean carry out this? cosec(2x) doesn't exist at x=0 :/
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    (Original post by RDKGames)
    Do you mean carry out this? cosec(2x) doesn't exist at x=0 :/
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    Can you post a picture of the question? Maybe using \text{\cosec}\,  2x = (1 + \cot^2 x)^{1/2} might work if you know the series for \cot.
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    (Original post by Zacken)
    Can you post a picture of the question? Maybe using \cosec 2x = (1 + \cot^2 x)^{1/2} might work if you know the series for \cot.
    I only know the series expansion for cosx and sinx as far as trig is concerned so far.
    Q17 part b

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    (Original post by RDKGames)
    I only know the series expansion for cosx and sinx as far as trig is concerned so far.

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    Ah, okay. Then it's as simple as \sin 2x \csc 2x = 1, expand \sin 2x and then compare coefficients.

    i.e:

    \displaystyle

\begin{equation*} \left(2x - \frac{4x^3}{3} + \frac{4x^5}{15}\right) (a + bx + cx^2 + dx^3) = 1\end{equation*}

    dropping terms of degree >3.

    Also, would make the numbers nicer if you just used \sin x \csc x = 1 and then x \mapsto 2x.
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    (Original post by RDKGames)
    I only know the series expansion for cosx and sinx as far as trig is concerned so far.
    Q17 part b

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    You need to use part a.

    You know the expansion of sin2x so 1/sin2x = (sin2x)^-1 = (.....)^-1. Now use binomial expansion.
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    (Original post by Zacken)
    Ah, okay. Then it's as simple as \sin 2x \csc 2x = 1, expand \sin 2x and then compare coefficients.

    i.e:

    \displaystyle

\begin{equation*} \left(2x - \frac{4x^3}{3} + \frac{4x^5}{15}\right) (a + bx + cx^2 + dx^3) = 1\end{equation*}

    dropping terms of degree >3.

    Also, would make the numbers nicer if you just used \sin x \csc x = 1 and then x \mapsto 2x.
    That makes sense, but what coefficients do I exactly compare? If 2ax=1 \rightarrow a=\frac{1}{2x} which fits the answer but I'm not sure about others.
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    (Original post by RDKGames)
    That makes sense, but what coefficients do I exactly compare? If 2ax=1 \rightarrow a=\frac{1}{2x} which fits the answer but I'm not sure about others.
    Look at my post. I remember doing this question 2 year ago.
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    (Original post by Math12345)
    Look at my post. I remember doing this question 2 year ago.
    I attempted binomial expansion but it got too hairy when it came to replacing sin(2x) by its expansion.
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    (Original post by RDKGames)
    I attempted binomial expansion but it got too hairy when it came to replacing sin(2x) by its expansion.
    sin(2x)=2x-(4/3)x^3+... (You can fill in the missing third term)

    (sin(2x))^-1 = (2x-(4/3)x^3))^-1 = (2x(1-(2/3)x^2))^-1 = 1/(2x) [1+(2/3)x^2 +...]

    You can continue (I used the expansion (1-y)^-1 above).
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    (Original post by RDKGames)
    That makes sense, but what coefficients do I exactly compare? If 2ax=1 \rightarrow a=\frac{1}{2x} which fits the answer but I'm not sure about others.
    Math12345's answer is rather nicer, but anyways, for completeness:

    Coefficient of x^2 gives 2b - \frac{1}{2}\cdot \frac{4}{3}= 0 \Rightarrow b=\frac{1}{3}

    x^3 gives 2c = 0 \Rightarrow c=0

    x^4 gives 2d - \frac{4b}{3} + \frac{2}{15} = 0\Rightarrow d = \frac{7}{45}
 
 
 
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