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    By writing the following equations as quadratics in tan θ/2,solve,in the
    interval 0 ≤ θ ≤ 360°:

    (i) sinθ + 2cosθ = 1

    So i was thinking of using this formula:
    tan2θ= 2tanθ/1-2tan^2θ
    which can be rewritten into:
    tanθ=2tanθ/2/1-tan^2θ/2

    Tanθ/2= t

    so 2t/1-t^2

    But when i checked the working out, it wanted:

    2t/1+t^2

    HELPP PLEASE!!

    Thanks
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    (Original post by Lilly1234567890)
    By writing the following equations as quadratics in tan θ/2,solve,in the
    interval 0 ≤ θ ≤ 360°:

    (i) sinθ + 2cosθ = 1

    So i was thinking of using this formula:
    tan2θ= 2tanθ/1-2tan^2θ
    which can be rewritten into:
    tanθ=2tanθ/2/1-tan^2θ/2

    Tanθ/2= t

    so 2t/1-t^2

    But when i checked the working out, it wanted:

    2t/1+t^2

    HELPP PLEASE!!

    Thanks
    what board are you doing please?
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    (Original post by TeeEm)
    what board are you doing please?
    Edexcel
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    (Original post by Lilly1234567890)
    Edexcel
    this will never come up in an edexcel maths or further maths paper.
    it comes from older/other boards and you require to use the "LITTLE t" identities

    I will post a few examples in a few minutes for you to see how these work
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    (Original post by TeeEm)
    this will never come up in an edexcel maths or further maths paper.
    it comes from older/other boards and you require to use the "LITTLE t" identities

    I will post a few examples in a few minutes for you to see how these work
    Thanks!
    the question is on pg.119 Q7
    so i thought it was part of the syllabus
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    (Original post by Lilly1234567890)
    Thanks!
    the question is on pg.119 Q7
    so i thought it was part of the syllabus
    these results are for enrichment material these days

    here are 3 questions

    the identities you have to prove in part (a) are common knowledge to mathematicians
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    (Original post by Lilly1234567890)
    By writing the following equations as quadratics in tan θ/2,solve,in the
    interval 0 ≤ θ ≤ 360°:

    (i) sinθ + 2cosθ = 1

    So i was thinking of using this formula:
    tan2θ= 2tanθ/1-2tan^2θ
    which can be rewritten into:
    tanθ=2tanθ/2/1-tan^2θ/2

    Tanθ/2= t

    so 2t/1-t^2

    But when i checked the working out, it wanted:

    2t/1+t^2

    HELPP PLEASE!!

    Thanks
    You have shown that \displaystyle \tan \theta = \frac{2t}{1-t^2} which is correct.

    The next step is to figure out what \sin \theta is.
 
 
 
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