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    its on the january 2006 c3 paper question 4b

    what does 8cos(arcsin(x/4)) convert to?

    it shows that it can convert to 2(16-x^2)^0.5

    i dont understand how this changes into this...someone explan please!!!
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    sin^2 x+cos^2 x=1
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    Are you sure it's January 2006? Cannot find this question.

    EDIT:

    Found it
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    (Original post by goldsilvy)
    Are you sure it's January 2006? Cannot find this question.

    EDIT:

    Found it
    What question. Can't find it.
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    (Original post by JTeighty)
    What question. Can't find it.
    It's on the MS. Question 4b.
    Attached Images
  1. File Type: pdf C3Jan06A.pdf (636.5 KB, 70 views)
  2. File Type: pdf C3Jan06Q.pdf (154.2 KB, 208 views)
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    (Original post by goldsilvy)
    It's on the MS. Question 4b.
    What a stupid way to answer the question.

    Just work out dx/dy and then take the inverse to make it dy/dx
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    (Original post by JTeighty)
    What a stupid way to answer the question.

    Just work out dx/dy and then take the inverse to make it dy/dx
    That's what they did in the MS. Still there is this one final step,

    For those interested,



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    x = 4 sin(2y + 6) -> x^2 = 16 sin^2(2y+6) -> (1-x^2)^0.5 = 4cos(2y + 6)

    you have a 1/8cos(2y+6) so be careful with the constant multiplier on the bottom.
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    (Original post by goldsilvy)
    That's what they did in the MS. Still there is this one final step,

    For those interested,



    Sorry didn't realise there was another stage.
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    Am really close. If you square the whole fraction you'll get 1/(64cos^2(arcsinx/4))
    use sin^2+cos^2=1
    so in the denominator u get 64-sin^2(arcsinx/4) which cancels to 64-x^2/16
    Now root it.

    Now I'm stuck but I think it's practically done
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    (Original post by JTeighty)
    Am really close. If you square the whole fraction you'll get 1/(64cos^2(arcsinx/4))
    use sin^2+cos^2=1
    so in the denominator u get 64(1-sin^2(arcsinx/4) which cancels to 64(1-x^2/16)=64/16(16-x^2)=4(16-x^2)
    Now root it.

    Now I'm stuck but I think it's practically done
    fixed
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    technically, since the answer's in brackets, it's an extra. You won't get marked down for not doing that, the question's not explicit enough.
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    (Original post by mathz)
    fixed
    Of course. What a silly mistake.

    Thanks for correcting me :-)
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    how does the (2y+6) translate to arcsin(x/4)
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    someone pls help ... if so rep+
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    Because x=4sin(2y+6), you can undo this by dividing by 4, and undoing the sin, so that arcsin(x/4)=2y+6. After differentiation, you're asked to put dy/dx in terms of x, which means you can just substitute in the known value of 2y+6 into it's place in the equation.

    Voila.
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    of course ... duh ... heres you rep
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    x= 4sin(2y+6)

    y=\frac{1}{2}sin^{-1}(\frac{x}{4})-3 By making y the subject.
    \frac{dy}{dx}=±\frac{1}{8\sqrt{  1-\frac{x^{2}}{16}}}
    \frac{dy}{dx}=±\frac{1}{2\sqrt{  16-x^{2}}}
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    dude i'd give you a rep but just used ... 2moro thank btw
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    (Original post by morebritishthanindian)
    how does the (2y+6) translate to arcsin(x/4)
    if you use the orginal equation: x =4sin(2y+6)

    x/4 = sin(2y+6)

    arcsin(x/4) = 2y+6 then sub it back in to dy/dx
 
 
 
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