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Edexcel FP2 Official 2016 Exam Thread - 8th June 2016 watch

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    (Original post by economicss)
    Zacken Please could you upload a sketch, thanks so much!
    Step 1: You know the curves intersect at \theta = \pi/12, 5\pi/12 - draw these half-lines.

    Step 2: Find the area of the sector bounded between these two half lines and the circle. \frac{1}{2} \times a^2 \times \frac{\pi}{6}

    Step 3: Find the blue shaded bits using integration of C1 from 0 to pi/12 + integration of C1 from 5pi/12 to pi/2.

    Add step 2 + step 3.

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    (Original post by Zacken)
    Step 1: You know the curves intersect at \theta = \pi/12, 5\pi/12 - draw these half-lines.

    Step 2: Find the area of the sector bounded between these two half lines and the circle. \frac{1}{2} \times a^2 \times \frac{\pi}{6}

    Step 3: Find the blue shaded bits using integration of C1 from 0 to pi/12 + integration of C1 from 5pi/12 to pi/2.

    Add step 2 + step 3.

    Thank you!! Please could you explain a bit of this question to me, I got it right apart from I thought the angle for the sector of the circle would be pi/ 6 as if it's 2pi/3 plus the other area I worked out then wouldn't this be the top bit of the circle rather than the shaded bit? Or is my
    Method incorrect? Thanks
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    (Original post by economicss)
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    Thank you!! Please could you explain a bit of this question to me, I got it right apart from I thought the angle for the sector of the circle would be pi/ 6 as if it's 2pi/3 plus the other area I worked out then wouldn't this be the top bit of the circle rather than the shaded bit? Or is my
    Method incorrect? Thanks
    Zacken
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    (Original post by economicss)
    Zacken
    How many of your qs has the poor guy helped you with ahaha, he's probs got work of his own to do.
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    (Original post by BBeyond)
    How many of your qs has the poor guy helped you with ahaha, he's probs got work of his own to do.
    Sorry haha, I got rejected from Cambridge btw if that was why you had animosity towards me 😂
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    (Original post by economicss)
    Zacken
    Zacken's not going to be there in the exam. Try and do as much as possible yourself and use other resources such as examsolutions, mark schemes etc to get the answer, because you have to remember that even though zacken is insanely patient at some point his patience will run out if you keep spamming him with questions, so I'd suggest you try and rely on him a bit less.
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    (Original post by economicss)
    Sorry haha, I got rejected from Cambridge btw if that was why you had animosity towards me 😂
    What ahaha
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    (Original post by economicss)

    Thank you!! Please could you explain a bit of this question to me, I got it right apart from I thought the angle for the sector of the circle would be pi/ 6 as if it's 2pi/3 plus the other area I worked out then wouldn't this be the top bit of the circle rather than the shaded bit? Or is my
    Method incorrect? Thanks
    Surely the two half lines are \frac{\pi}{6} and \frac{5\pi}{6} so that the "top" angle between those two lines is \frac{2\pi}{3} and hence the "bottom" angle or the sector angle is 2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}.

    Caveat: I've barely looked at the question.
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    I've got a nice and short question:

    If the complex number \displaystyle z=\left ( 1+i\tan\theta \right )^{3},

    Show that \displaystyle 1-3\tan^{2}\theta \equiv \dfrac{\cos3\theta}{\cos^{3} \theta}.
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    (Original post by aymanzayedmannan)
    I've got a nice and short question:

    If the complex number \displaystyle z=\left ( 1+i\tan\theta \right )^{3},

    Show that \displaystyle 1-3\tan^{2}\theta \equiv \dfrac{\cos3\theta}{\cos^{3} \theta}.
    Side note, you could also just do:

    \displaystyle

\begin{equation*} 1 - \frac{3\sin^2 \theta}{\cos^2 \theta} \equiv \frac{\cos^2 \theta - 3\sin^2 \theta}{\cos^2 \theta} \equiv \frac{\cos^3 \theta - 3\cos \theta \sin^2 \theta}{\cos^3 \theta} \equiv \frac{\Re \left(\cos \theta + i\sin \theta\right)^3}{\cos^3 \theta} \equiv \frac{\cos 3\theta}{\cos^3 \theta} \end{equation*}

    Edit to add: I can't immediately see how to do it the way you intended. I get that \Re(1 + i\tan \theta)^3 = 1 - 3\tan^2 \theta, but then what?
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    (Original post by Zacken)
    Side note, you could also just do:

    \displaystyle

\begin{equation*} 1 - \frac{3\sin^2 \theta}{\cos^2 \theta} \equiv \frac{\cos^2 \theta - 3\sin^2 \theta}{\cos^2 \theta} \equiv \frac{\cos^3 \theta - 3\cos \theta \sin^2 \theta}{\cos^3 \theta} \equiv \frac{\Re \left(\cos \theta + i\sin \theta\right)^3}{\cos^3 \theta} \equiv \frac{\cos 3\theta}{\cos^3 \theta} \end{equation*}

    Edit to add: I can't immediately see how to do it the way you intended. I get that \Re(1 + i\tan \theta)^3 = 1 - 3\tan^2 \theta, but then what?
     \displaystyle (1 + i\tan \theta)^3

     \displaystyle \left(\frac{\cos \theta}{\cos \theta} + i\frac{\sin \theta}{\cos \theta}\right)^3

     \displaystyle \left(\frac{1}{\cos \theta}\right)^3 \: (\cos \theta + i\sin \theta)^3

    then regular de moivre theorem
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    (Original post by DylanJ42)
     \displaystyle (1 + i\tan \theta)^3

     \displaystyle \left(\frac{\cos \theta}{\cos \theta} + i\frac{\sin \theta}{\cos \theta}\right)^3

     \displaystyle \left(\frac{1}{\cos \theta}\right)^3 \: (\cos \theta + i\sin \theta)^3

    then regular de moivre theorem
    Oh, niiice. Obvious in hindsight. Thanks!
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    Hey everyone
    Does anyone know which paper is this?or how to get its markscheme? I tried looking it up online couldn't find anything. Thanks


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    (Original post by jkhan9625)
    Name:  ImageUploadedByStudent Room1459972510.981910.jpg
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    Hey everyone
    Does anyone know which paper is this?or how to get its markscheme? I tried looking it up online couldn't find anything. Thanks


    Posted from TSR Mobile
    june 2013 withdrawn paper
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    (Original post by DylanJ42)
    june 2013 withdrawn paper
    Thanks a lot


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    Hi, could anyone give any advice please on sketching the graph in question 5cii of this paper https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf Thanks
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    (Original post by economicss)
    Hi, could anyone give any advice please on sketching the graph in question 5cii of this paper https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf Thanks
    Literally just done this paper. Uhm, you should know that there are minimum points at (\pm \sqrt{2} ,4) and that the y-axis is an asymptote (since you have x^2 in the denominator) and then the graph tends to positive infinity as x approaches infinity. Then the curve tails off the positive infinity as x tends to positive and negative infinity. Also there is symmetry in the y-axis because it's an even function i.e f(-x) = f(x) since all your x terms are squared.

    Edit: also your function will always lie above your x-axis since it is always positive due to the squared terms.
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    (Original post by Zacken)
    Literally just done this paper. Uhm, you should know that there are minimum points at (\pm \sqrt{2} ,4) and that the y-axis is an asymptote (since you have x^2 in the denominator) and then the graph tends to positive infinity as x approaches infinity. Then the curve tails off the positive infinity as x tends to positive and negative infinity. Also there is symmetry in the y-axis because it's an even function i.e f(-x) = f(x) since all your x terms are squared.

    Edit: also your function will always lie above your x-axis since it is always positive due to the squared terms.
    How did you do? Thanks for that, very helpful
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    (Original post by economicss)
    How did you do? Thanks for that, very helpful
    I did okay, it was a nice one, I think. You?
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    (Original post by Zacken)
    I did okay, it was a nice one, I think. You?
    Yeah okay thanks Hoping we get a similar difficulty one this year!
 
 
 
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