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    Im stuck on this c4 question normlyy i just find t and then sub in but for this one im stuck can anyone help plssName:  1368720360422.jpg
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Size:  439.5 KB i cant do it
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    (Original post by annaridgway95)
    1 - 2sin^2x but I don't understand where that fits in?


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    \dfrac{\cos(x) - 2\sin^2(x)\cos(x)}{\sin(x)}
    Now you need to work in a  1 - 2\sin^2(x)
    Can you see how?
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    (Original post by joostan)
    \dfrac{\cos(x) - 2\sin^2(x)\cos(x)}{\sin(x)}
    Now you need to work in a  1 - 2\sin^2(x)
    Can you see how?
    Yeah I get it now, I'm being stupid! Hahaha.
    Thank you


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    (Original post by annaridgway95)
    Yeah I get it now, I'm being stupid! Hahaha.
    Thank you


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    No problem
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    (Original post by Revisionbug)
    Im stuck on this c4 question normlyy i just find t and then sub in but for this one im stuck can anyone help plssName:  1368720360422.jpg
Views: 162
Size:  439.5 KB i cant do it
    Add x and y then substitute in
    Negging me is somewhat unnecessary, no?
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    Hi! I found something called the "common mistake "section in the c2 maths CD , there's one part that i dont understand

    Whats meant by "not actually a mistake"?

    I didnt know how to tackle this question ( to use the integration and equation L )

    Can someone explain? Thanks loads!!!





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    (Original post by Revisionbug)
    Im stuck on this c4 question normlyy i just find t and then sub in but for this one im stuck can anyone help plssName:  1368720360422.jpg
Views: 162
Size:  439.5 KB i cant do it
     x=(t+\frac{2}{t})
    y=(t-\frac{2}{t})
     x^2=(t+\frac{2}{t})^2=t^2+4+ \frac{4}{t^2}
    y^2=(t-\frac{2}{t})^2=t^2-4+\frac{4}{t^2}
     x^2 - y^2 = 8
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    GAAH nevermind! Blonde moment there! sorry for wasting your time, people!
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    (Original post by ScarlettFierce)
    GAAH nevermind! Blonde moment there! sorry for wasting your time, people!
    i) \mathbb{P} (X = 5) = \displaystyle\binom{11}{5} \left(\dfrac{3}{4} \right)^{5} \left(1-\dfrac{3}{4} \right)^{6}

    ii) \mathbb{P} (X = 0) =  \displaystyle\binom{11}{0} \left(p \right)^{0} \left(1-p \right)^{11} = 0.05

    Var(X) for a binomial distribution = np(1-p)
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    2 weird but urgent questions but how do you know when to solve a quadratic equation using two brackets - also in a log table - is x at the top row or is it the other way round ?


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    (Original post by krishkmistry)
    2 weird but urgent questions but how do you know when to solve a quadratic equation using two brackets - also in a log table - is x at the top row or is it the other way round ?


    Posted from TSR Mobile
    For the first one, if you are able to factorise it, then factorise it. But not all quadratics will be able to be solved that way. I'm not sure what you mean by the next part.
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    (Original post by brittanna)
    For the first one, if you are able to factorise it, then factorise it. But not all quadratics will be able to be solved that way. I'm not sure what you mean by the next part.
    Basically I'm asking how many terms do you need to solve quadratically - via brackets of qp - I'm also asking that when you get a table on a logs graph q - is the top row the x value or the y value when plotting it on a graph ?


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    (Original post by krishkmistry)
    Basically I'm asking how many terms do you need to solve quadratically - via brackets of qp - I'm also asking that when you get a table on a logs graph q - is the top row the x value or the y value when plotting it on a graph ?


    Posted from TSR Mobile
    If you have something like x^2+5x+4=0, you can write this as (x+4)(x+1)=0 \Rightarrow x=-4, x=-1. However, if you have something like x^2+2x+2=0, you would need to solve this by completing the square or by substituting into the quadratic formula. Only quadratics with at least two terms can be factorised.

    Does it not tell you in the table which row is which?
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    (Original post by MathsNerd1)
    ..
    Following on from my suggestion,, have you tried any STEP DE's/Induction? If so, how have you found it?
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    (Original post by DJMayes)
    Following on from my suggestion,, have you tried any STEP DE's/Induction? If so, how have you found it?
    I'm going to be honest with you and say I haven't yet as I've been practising on my Maths and Chemistry for the mocks I have but I'm devoting 3 hours tomorrow to STEP either for a mock or some questions and I'll make sure that I'll do one of those topics and let you know how I found them.
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    How do you work out normal distribution with quartiles?
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    Quick question:

    Given

    x=\frac{1}{5}(2sint-cost+e^{-2t})

    show that x varies between

    -\frac{1}{5}\sqrt 5 and \frac{1}{5}\sqrt 5 for large positive values of t

    I understand that the term involving e will tend to 0, and I understand what they have done to obtain the answer

    Spoiler:
    Show
    \sqrt {2^2 + (-1)^2}


    Why do you do this? Apologise for the vague question, I just can't wrap my head around it.
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    TSR Support Team
    I've suddenly started getting email updates whenever someone posts on this thread, any idea how to stop it?
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    (Original post by NJam)
    Quick question:

    Given

    x=\frac{1}{5}(2sint-cost+e^{-2t})

    show that x varies between

    -\frac{1}{5}\sqrt 5 and \frac{1}{5}\sqrt 5 for large positive values of t

    I understand that the term involving e will tend to 0, and I understand what they have done to obtain the answer

    Spoiler:
    Show
    \sqrt {2^2 + (-1)^2}


    Why do you do this? Apologise for the vague question, I just can't wrap my head around it.
    This in an idea called the harmonic functions. Essentially, we can express the sum of any sine and cosine as a single trig function:

     asinx+bcosx = Rsin(x+c)

    In order to do this, expand the RHS:

     asinx+bcosx = Rsinxcosc+Rcosxsinc

    Knowing your values of a and b, you can then solve for values of R and c in order to express it as a single function:

     Rcosc = a  Rsinc = b

    You can then use a bit of re-arrangement to find a and b:

     R^2(cos^2c+sin^2c) = a^2 + b^2

    So  R = \sqrt{a^2+b^2}

    And you can work out C by dividing the two equations to get something involving tan only. This is how they've got the root 5; by expressing the function as a single trigonometric function.
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    (Original post by DJMayes)
    This in an idea called the harmonic functions. Essentially, we can express the sum of any sine and cosine as a single trig function:

     asinx+bcosx = Rsin(x+c)

    In order to do this, expand the RHS:

     asinx+bcosx = Rsinxcosc+Rcosxsinc

    Knowing your values of a and b, you can then solve for values of R and c in order to express it as a single function:

     Rcosc = a  Rsinc = b

    You can then use a bit of re-arrangement to find a and b:

     R^2(cos^2c+sin^2c) = a^2 + b^2

    So  R = \sqrt{a^2+b^2}

    And you can work out C by dividing the two equations to get something involving tan only. This is how they've got the root 5; by expressing the function as a single trigonometric function.
    Fantastic, exactly what I needed! +rep
 
 
 
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