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    hi everyone!
    I'm doing a question in C4 and looking at the markscheme has really confused me.
    i've uploaded it and if someone could explain the step(s?) between the first and second line, i'd really appreciate it!
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  1. File Type: doc jun 05 markscheme.doc (138.0 KB, 79 views)
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    Why has the index gone from 1/2 to 3/2? What was the question?

    If it should be 3/2:

     \\ \frac{\cos{\theta}}{(1-\sin^2{\theta})^{\frac{3}{2}}} \\

\\ = \frac{\cos{\theta}}{(\cos^2{\the  ta})^{\frac{3}{2}}} \\

\\ = \frac{\cos{\theta}}{((\cos^2{\th  eta})^{\frac{1}{2}})^3} \\

\\ = \frac{\cos{\theta}}{\cos^3{\thet  a}} \\

\\ = \frac{1}{\cos^2{\theta}}
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    Well, 1 - sin^2 = cos^2, and they're raising that to the power -3/2. Multiply that by cos, (which is (cos^2)^1/2), and you get (cos^2)^-1.

     \displaystyle \frac{\cos \theta}{(\cos^2 \theta)^{3/2}} = \frac{{(\cos^2 \theta)}^{1/2}}{(\cos^2 \theta)^{3/2}} = \frac{1}{\cos^2 \theta} .
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    (Original post by sheena18)
    hi everyone!
    I'm doing a question in C4 and looking at the markscheme has really confused me.
    i've uploaded it and if someone could explain the step(s?) between the first and second line, i'd really appreciate it!
    Remember sin^2x=1-cos^2x

    If you don't mind me asking, why are you in your utility room? :confused:
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    daniel: i tried to upload the question, but it didn't work. i found the link:

    http://eiewebvip.edexcel.org.uk/Repo...e_20050628.pdf

    its question 4

    adje: thanks for replying, i follow the trig identity part, but it's the powers that confuses me. i don't get this bit that you explained

    "Multiply that by cos, (which is (cos^2)^1/2), and you get (cos^2)^-1." ??


    muppety kid: lol, erm well at the moment my study is the utility room - tis a nightmare to concentrate when the washing machine's on lol!

    EDIT:

    to adje: i just your your edit, and the bit i don't get is how you get from the 2nd bit to the 3rd bit. how does 1/2 and 3/2 just cancel?
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    (Original post by sheena18)
    daniel: i tried to upload the question, but it didn't work. i found the link:

    http://eiewebvip.edexcel.org.uk/Repo...e_20050628.pdf

    its question 4
    Ahh so it is 3/2. Are you happy with everything in my above post? (I've edited it)
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    daniel, i get what you've done up to the 3rd line, i dont see how you got to the 4th liine :s
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    (Original post by sheena18)
    daniel, i get what you've done up to the 3rd line, i dont see how you got to the 4th liine :s
    Ok, remember that  \cos^2{\theta} is just the short hand way of writing  (\cos{\theta})^2 . Also remember that  (a^b)^c = a^{bc} .

    You have:

     (((\cos{\theta})^2)^{\frac{1}{2}  })^3 = ((\cos{\theta})^{\frac{2}{2}})^3 = (\cos{\theta})^3 = \cos^3{\theta}
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    i just realised that before u posted it, silly me!
    thanks so much adje/daniel for helping me out, u stars!
    how are u both feeling for thursday da da DA?
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    (Original post by sheena18)
    i just realised that before u posted it, silly me!
    thanks so much adje/daniel for helping me out, u stars!
    how are u both feeling for thursday da da DA?
    No problem.

    I'm looking forward to it, although I'm more looking forward to finishing it
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    im looking forward to it for the same reason, can't wait to put away the maths folders forever after thursday mwahaha!
 
 
 
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