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    (Original post by ama1234a)
    Thank you so much for clearing that up! Can I ask why it doesn't matter if the angle is 90?
    When the angle is 90 degrees, a.b = 0. So the angle will be 90 degrees on every side where the vectors cross (think of it like a square).
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    How to do 3b
    http://www.mrbartonmaths.com/resourc...20core%204.pdf

    ??
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    (Original post by Dapperblook22)
    When the angle is 90 degrees, a.b = 0. So the angle will be 90 degrees on every side where the vectors cross (think of it like a square).
    Wicked, thank you!! I have one more question if thats okay.. Sometimes in the mark scheme, you have to square the vector to answer the question but I've never understood why we need to do that? When do you usually have to square a vector and what kind of question is it/what am I looking for usually?
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    Multiply both sides by (2x-1), then compare coefficients to find a and b.
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    long division i use the bus method
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    (Original post by Dapperblook22)
    Multiply both sides by (2x-1), then compare coefficients to find a and b.
    mark scheme does it some weird way.. would ur way still be viable
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    (Original post by fpmaniac)
    mark scheme does it some weird way.. would ur way still be viable
    The two methods usually accepted are long division and comparing coefficients. Another method is subsituting in values and solving simultaneously. Any of these will work, in recent years and examiners reports they have accepted all three.
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    (Original post by Dinasaurus)
    Can someone explain how to do 5c from 2015's paper for me as well,
    You have your equation: y-0.5=2(x-0.5)

    x=cos2q, use double angle identity to get x=1-2sin^2(q)

    Plug x=1-2sin^2(q) and y=sinq and you get

    sinq-0.5=2(-2sin^2q + 0.5)
    rearrange
    4sin^2q + sinq - 1.5 = 0

    Solve
    sinq = 0.5, sinq = -3/4

    sinq = 0.5 is P so the answer is sinq = -3/4, x = 1-2(3/4)^2 = -1/8
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    (Original post by bartbarrow)
    You have your equation: y-0.5=2(x-0.5)

    x=cos2q, use double angle identity to get x=1-2sin^2(q)

    Plug x=1-2sin^2(q) and y=sinq and you get

    sinq-0.5=2(-2sin^2q + 0.5)
    rearrange
    4sin^2q + sinq - 1.5 = 0

    Solve
    sinq = 0.5, sinq = -3/4

    sinq = 0.5 is P so the answer is sinq = -3/4, x = 1-2(3/4)^2 = -1/8
    Thanks a lot. The trigonometric identities that we get in the formula booklet, is it safe to say we can use them to derive the double angle identities?
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    (Original post by Dinasaurus)
    Thanks a lot. The trigonometric identities that we get in the formula booklet, is it safe to say we can use them to derive the double angle identities?
    Yup. You can also use them for triple angle identities too.
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    (Original post by Dapperblook22)
    Yup. You can also use them for triple angle identities too.
    How would you do that, I understand for double angle you just let there be 2 As instead of A and B.

    But for triple angle, how?
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    (Original post by Dinasaurus)
    How would you do that, I understand for double angle you just let there be 2 As instead of A and B.

    But for triple angle, how?
    For example, let sin(3x) = sin(2x + x), you then use the compound and double angle identity to derive the triples.
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    (Original post by Dinasaurus)
    How would you do that, I understand for double angle you just let there be 2 As instead of A and B.

    But for triple angle, how?
    Sounds more difficult than it is, if you get sin3x you could set A=2x and B=x to split that up, then you can use double angle as normal to split the 2x up again
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    Can someone help with this proof.
    Prove (-2sin2x)/(2cos2x) is equal to (2tanx)/((sec^2x)-2)
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    (Original post by Dapperblook22)
    For example, let sin(3x) = sin(2x + x), you then use the compound and double angle identity to derive the triples.
    (Original post by bartbarrow)
    Sounds more difficult than it is, if you get sin3x you could set A=2x and B=x to split that up, then you can use double angle as normal to split the 2x up again
    Thanks to you both, seeing triple angle made me think there'd be a third part instead of just letting one be 2x and the other x.
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    (Original post by 15150776)
    Can someone help with this proof.
    Prove (-2sin2x)/(2cos2x) is equal to (2tanx)/((sec^2x)-2)
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    (Original post by bartbarrow)
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    Thank you very much! Didnt spot the divide by sec^2x
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    Just to clarify, for max rate of growth you differentiate the differential equation and equate it to 0?
    would we be expected to find the least rate of growth?
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    (Original post by Arima)
    Just to clarify, for max rate of growth you differentiate the differential equation and equate it to 0?
    would we be expected to find the least rate of growth?
    Say you have du/dx and you need to find the maximum or minimum rate of growth:

    Imagine setting y=du/dx and plotting y

    The maximum/minimum rate of growths are going to be stationary points so you can differentiate y (or in other words, differentiate du/dx) and equate to 0 to find stationary points. Where d^2u/dx^2 = 0 will give you those points, then you can differentiate it again (d^3u/dx^3) and plug those values in, if it is < 0 you have found the maximum and if it is > 0 you have found the minimum.
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    (Original post by Arima)
    Just to clarify, for max rate of growth you differentiate the differential equation and equate it to 0?
    would we be expected to find the least rate of growth?
    Possibly.
 
 
 
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