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    (Original post by unamed)
    I put it as 1/7 too, why is it 7/1? isn't cos(alpha) 7 and sin(alpha) 1? :con:
    It's the other way around.

    (7cosx + sinx) - 5 = (Rsin@.cosx + Rcos@.sinx) - 5

    Therefore, Rsin@ = 7, Rcos@ = 1.
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    (Original post by Doughboy)
    It's the other way around.

    (7cosx + sinx) - 5 = (Rsin@.cosx + Rcos@.sinx) - 5

    Therefore, Rsin@ = 7, Rcos@ = 1.
    Why's it [email protected] and nor [email protected]? isn't it normally that?
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    (Original post by unamed)
    Why's it [email protected] and nor [email protected]? isn't it normally that?
    I don't understand what you are asking, but I try not to "remember" what [email protected] normally is. I just expand whatever the question asks for.

    For example, in this question they told us to express the original equation in the form:

    Rsin(x+@)

    Just expand that to get:

    Rsinxcos@ + Rsin@cosx
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    (Original post by unamed)
    Why's it [email protected] and nor [email protected]? isn't it normally that?
    I think your getting confused with your compound angle formulae, revise over them again, you cannot afford to be making mistakes like that in the exam.
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    (Original post by Doughboy)
    I don't understand what you are asking, but I try not to "remember" what [email protected] normally is. I just expand whatever the question asks for.

    For example, in this question they told us to express the original equation in the form:

    Rsin(x+@)

    Just expand that to get:

    Rsinxcos@ + Rsin@cosx
    sorry for the atrocious wording. I just realised where I was going wrong! The [email protected] is for the sinx part of the equation and then that would mean that [email protected] is 1. < not sure if you understood that either, but now I know where I went wrong!

    Thank you!.
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    (Original post by uer23)
    I think your getting confused with your compound angle formulae, revise over them again, you cannot afford to be making mistakes like that in the exam.
    No, I'm pretty good with my compound angle formulae. It was a stupid mistake, really, and I couldn't see where I was going wrong.
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    (Original post by Doughboy)
    I don't understand what you are asking, but I try not to "remember" what [email protected] normally is. I just expand whatever the question asks for.

    For example, in this question they told us to express the original equation in the form:



    Just expand that to get:

    Just thought I'd let you know: you can do an alpha (was this the symbol you intended?) by writing \alpha in LaTeX, so it comes out like \alpha.

    Feel free to ignore this if you had no intention of writing an alpha!
    (Our syllabus usually uses \sin(\theta+\alpha) ).
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    good luck people
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    (Original post by unamed)
    No, I'm pretty good with my compound angle formulae. It was a stupid mistake, really, and I couldn't see where I was going wrong.
    You get them in the formulae booklet anyway, no?

    We do on OCR:MEI anyway.
    So I just look them up each time to ensure I got it right!
    Spoiler:
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    Yes, I realise that that is incredibly lazy of me
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    (Original post by placenta medicae talpae)
    You get them in the formulae booklet anyway, no?

    We do on OCR:MEI anyway.
    So I just look them up each time to ensure I got it right!
    Spoiler:
    Show
    Yes, I realise that that is incredibly lazy of me
    we get some of them. We don't get the standard. Just checked, we get all of them! But I like remembering stuff, I feel like I'm cheating if I don't.
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    (Original post by placenta medicae talpae)
    Just thought I'd let you know: you can do an alpha (was this the symbol you intended?) by writing \alpha in LaTeX, so it comes out like \alpha.

    Feel free to ignore this if you had no intention of writing an alpha!
    (Our syllabus usually uses \sin(\theta+\alpha) ).
    Thanks. I did mean to put alpha, but a lack of the correct latex codes prevented me from doing so :P


    \alpha \theta
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    Given that x = 4 sin(2y + 6), (dy/dx) find in terms of x.
    I'm comfortable with finding dx/dy and then reciprocating it to find dy/dx in terms of y. But when do you change the y values to make it in terms of x?
    Thanks
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    (Original post by unamed)
    we get some of them. We don't get the standard. Just checked, we get all of them! But I like remembering stuff, I feel like I'm cheating if I don't.
    Absolutely, great for you!
    I also like remembering things, and I think that after using them so much when doing past papers, they are now firmly rooted in my memory.

    I think that the only ones we don't get are the double angle ones (they took the "factor" formulae off the syllabus).
    But you can derive them really easily anyway, in your head probably!
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    (Original post by jimber)
    Given that x = 4 sin(2y + 6), (dy/dx) find in terms of x.
    I'm comfortable with finding dx/dy and then reciprocating it to find dy/dx in terms of y. But when do you change the y values to make it in terms of x?
    Thanks
    after you've made it dy/dx. So you basically sub in (arcsinx/4-6)/2 for your y.
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    (Original post by Doughboy)
    Thanks. I did mean to put alpha, but a lack of the correct latex codes prevented me from doing so :P


    No probs!

    How cool is this as well - if you write it with a capital, it does the upper equivalent (I hope that made sense).
    So like \Theta comes out like \Theta
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    (Original post by placenta medicae talpae)
    Absolutely, great for you!
    I also like remembering things, and I think that after using them so much when doing past papers, they are now firmly rooted in my memory.

    I think that the only ones we don't get are the double angle ones (they took the "factor" formulae off the syllabus).
    But you can derive them really easily anyway, in your head probably!
    I checked again - we don't get the double angle formula. (sin2A) But we do get the factor formulae - they're still on our syllabus (but it's debatable, apparently). I can't do much in my head, I nearly always have to scribble somewhere! when's your C3 then?
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    could some one plz help me differentiate: x= 2cos(2y+pi)
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    (Original post by unamed)
    I checked again - we don't get the double angle formula. (sin2A) But we do get the factor formulae - they're still on our syllabus (but it's debatable, apparently). I can't do much in my head, I nearly always have to scribble somewhere! when's your C3 then?
    I find the double one for cos easy to remember because it's like the cos-sin formula for 1, but with the wrong sign.
    So instead of being \sin^2(\theta)+\cos^2(\theta), it's \sin^2(\theta)-\cos^2(\theta).

    And then I remember the one for \sin(2\theta) as the one with the 2 in front
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    (Original post by unamed)
    after you've made it dy/dx. So you basically sub in (arcsinx/4-6)/2 for your y.
    Thanks, I prey for none of these q's on the exam!
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    (Original post by Inzamam99)
    could some one plz help me differentiate: x= 2cos(2y+pi)
    to differentiate cosf(x) you do: -f'(x)sinf(x)
    which would make it: -4sin(2y+pi)

    edit: I just realised it's x=...
    which means you're going to have to do 1/-4sin(2y+pi)
    as dy/dx=1/dx/dy/
 
 
 
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