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Urgent help C4 trig addition formulae & integration Watch

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    (Original post by SeanFM)
    Great :borat:

    So yes, you can take the half outside of the integral and that leaves you with (1/sin(...)). What's the integral of that?
    ln(sin(theta-(pi/3)))? Not really sure..
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    (Original post by jordanwu)
    ln(sin(theta-(pi/3)))? Not really sure..
    No, what does cosec x integrate to? You have a formula booklet. Use it.
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    (Original post by jordanwu)
    ln(sin(theta-(pi/3)))? Not really sure..
    Not quite. You may be able to use something from your formula book or things that you know.

    How can you rewrite 1/sin(x) where x is theta-(pi/3) as something that you know the integral of?
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    (Original post by SeanFM)
    Not quite. You may be able to use something from your formula book or things that you know.

    How can you rewrite 1/sin(x) where x is theta-(pi/3) as something that you know the integral of?
    Name:  image.jpg
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Size:  310.1 KB This?
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    (Original post by jordanwu)
    Name:  image.jpg
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    Almost there, but the integral of cosecx is ln|cosecx + cotx|.
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    (Original post by SeanFM)
    Almost there, but the integral of cosecx is ln|cosecx + cotx|.
    From this http://filestore.aqa.org.uk/subjects/FORMULAE.PDF it says that it's -ln|cosecx + cotx|?
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    (Original post by jordanwu)
    From this http://filestore.aqa.org.uk/subjects/FORMULAE.PDF it says that it's -ln|cosecx + cotx|?
    Oops, forgot the -, sorry! (it was next to an equals sign where I was looking it up )
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    (Original post by SeanFM)
    Oops, forgot the -, sorry! (it was next to an equals sign where I was looking it up )
    From the Aqa formula book it also says that the -ln|cosecx + cotx| = ln|tan(1/2x)|?
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    (Original post by jordanwu)
    From the Aqa formula book it also says that the -ln|cosecx + cotx| = ln|tan(1/2x)|?
    Oh, sorry, then you are right! I've never seen that equality before
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    (Original post by SeanFM)
    Oh, sorry, then you are right! I've never seen that equality before
    Lol really? XD But you're studying maths at uni right?
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    (Original post by SeanFM)
    Oh, sorry, then you are right! I've never seen that equality before
    \displaystyle \csc x + \cot x = \frac{1}{\sin x} + \frac{\cos x}{\sin x} = \frac{1+ \cos x}{\sin x} = \frac{1 + 2 \cos^2 x/2 - 1}{2 \cos x/2 \sin x/2} = \frac{\cos x/2}{\sin x/2}

    It's one of those weird things nobody uses or notices.
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    (Original post by jordanwu)
    Lol really? XD But you're studying maths at uni right?
    Yes but there isn't a huge focus on manipulating trig. The stuff you see at A-level form small parts of the course.
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    (Original post by Zacken)
    \displaystyle \csc x + \cot x = \frac{1}{\sin x} + \frac{\cos x}{\sin x} = \frac{1+ \cos x}{\sin x} = \frac{1 + 2 \cos^2 x/2 - 1}{2 \cos x/2 \sin x/2} = \frac{\cos x/2}{\sin x/2}

    It's one of those weird things nobody uses or notices.
    I see, thanks

    Latex can be funny in that things disappear.
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    (Original post by SeanFM)
    Latex can be funny in that things disappear.
    It's one of those weird quirks of latex, it doesn't have \cosec x, just \csc x. Thanks for reminding me!
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    (Original post by Zacken)
    It's one of those weird quirks of latex, it doesn't have \cosec x, just \csc x. Thanks for reminding me!
    Thanks for editing my post too
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    (Original post by SeanFM)
    Thanks for editing my post too
    See your VM's, I'm so sorry!
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    (Original post by SeanFM)
    Thanks for editing my post too
    So I've ended up with this: Name:  image.jpg
Views: 25
Size:  313.3 KB, please could you check through it and tell me whether I've made any mistakes? Thanks!
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    (Original post by jordanwu)
    So I've ended up with this: Name:  image.jpg
Views: 25
Size:  313.3 KB, please could you check through it and tell me whether I've made any mistakes? Thanks!
    Your values of tan(pi/3) and tan(pi/6) aren't quite right
 
 
 
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