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    Has anyone completed the C3 specimen paper?
    I'm stuck on questions 6(f) and 7 because the mark scheme doesn't show the full working out. :mad:
    Please help!
    Thanks.
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    Post the question and your attempt at answering it, and then more people (e.g. me) will be able to help than just the few who have the C3 Edexcel specimen paper.
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    Sorry!
    I just attached the paper.
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  1. File Type: pdfC3%20Specimen.pdf (64.1 KB, 680 views)
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    (Original post by rafaelvdv93)
    Sorry!
    I just attached the paper.
    For 7), you can rewrite  tan \theta as \frac{sin \theta} {cos \theta} . And ofcourse you've always got the quotient rule under your sleeve when things start to misbehave lol.

    Does that help?
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    For 6f, I've not read through the question but I believe it's telling you - from my glance - that \frac{dT}{dt}=-1.8
    you've already found \frac{dT}{dt} for part e, so just set it equal to -1.8 and solve for little t. Then, sub back into your equation and solve for big T.
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    7)

    i) I assume you can do this
    ii) Find dx/dy. Take the inverse, use a trig identity and replace tan^2(y/2) with x
    iii) I get dy/dx = e^{-x}(2 \cos 2x - \sin 2x)

    So R cos \alpha = 2 and R sin \alpha = 1

    Therefore \alpha = arctan (1/2)
    So \sin \alpha = \frac{1}{\sqrt{5}}
    Which means  R = \sqrt{5}
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    Thanks everyone!
    I definitely understand 6(f) and 7(i)(ii) now!
    But I'm still unsure about 7(iii)
    I don't understand how you got the values of R cos(alpha) and R sin(alpha).
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    (Original post by electriic_ink)
    7)

    i) I assume you can do this
    ii) Find dx/dy. Take the inverse, use a trig identity and replace tan^2(y/2) with x
    iii) I get dy/dx = e^{-x}(2 \cos 2x - \sin 2x)

    So R cos \alpha = 2 and R sin \alpha = 1

    Therefore \alpha = arctan (1/2)
    So \sin \alpha = \frac{1}{\sqrt{5}}
    Which means  R = \sqrt{5}
    where does sin alpha =1/5 come from?
 
 
 
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