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    I'm stuck at question 6 part c. At this point, I've established that tan3(theta)=1 and that the roots (theta) are tan(pi/12), tan(5pi/12), tan(9pi/12).

    I don't understand the "hence" because the marks cheme shows tan(pi/12)+tan(5pi/12)+tan(9pi/12)=3 but shouldn't it be tan3(pi/12)+tan3(5pi/12)+tan3(9pi/12)=3

    Yours Sincerely,
    A confused student
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    (Original post by Serendreamers)
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    MS

    I'm stuck at question 6 part c. At this point, I've established that tan3(theta)=1 and that the roots (theta) are tan(pi/12), tan(5pi/12), tan(9pi/12).

    I don't understand the "hence" because the marks cheme shows tan(pi/12)+tan(5pi/12)+tan(9pi/12)=3 but shouldn't it be tan3(pi/12)+tan3(5pi/12)+tan3(9pi/12)=3

    Yours Sincerely,
    A confused student
    No, because the cubic corresponds to the \tan(3\theta)
    identity in a (iii).

    x=\tan\theta
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    (Original post by Indeterminate)
    No, because the cubic corresponds to the \tan(3\theta)
    identity in a (iii).

    x=\tan\theta
    Argh why brain why?!

    Thanks
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    (Original post by Indeterminate)
    No, because the cubic corresponds to the \tan(3\theta)
    identity in a (iii).

    x=\tan\theta
    Actually wait,

    Now how does tan(pi/12)+tan(5pi/12)+tan(9pi/12)=1 ?

    I'm really confused as to how the "hence" comes into play here
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    (Original post by Serendreamers)
    Actually wait,

    Now how does tan(pi/12)+tan(5pi/12)+tan(9pi/12)=1 ?
    No, each of those are the theta for which

    \tan 3\theta = 1

    By the tan graph, your equation equals 3 and the sum with thrice those values also equals 3.
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    (Original post by Indeterminate)
    No, each of those are the theta for which

    \tan 3\theta = 1

    By the tan graph, your equation equals 3 and the sum with thrice those values also equals 3.
    I understand that tan(theta)=1 i.e. tan(3pi/12)+tan(15pi/12)+tan(27pi/12)=3

    But how could I "hence" show that the sum of the thetas with a third of the value also equal 3?

    Because I'm pretty sure (tan3A+tan3B)/(tanA+tanB)=1 is not an identity :P
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    (Original post by Serendreamers)
    I understand that tan(theta)=1 i.e. tan(3pi/12)+tan(15pi/12)+tan(27pi/12)=3

    But how could I "hence" show that the sum of the thetas with a third of the value also equal 3?

    Because I'm pretty sure (tan3A+tan3B)/(tanA+tanB)=1 is an identity :P
    I'm sure it would suffice to observe the identity you quoted.

    Alternatively, you may subtract multiples of pi to see why it works.
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    (Original post by Indeterminate)
    I'm sure it would suffice to observe the identity you quoted.

    Alternatively, you may subtract multiples of pi to see why it works.
    I'm still unsure as to why,

    tan(pi/12)+tan(5pi/12)+tan(9pi/12)=tan3(pi/12)+tan3(5pi/12)+tan3(9pi/12)
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    (Original post by Serendreamers)
    QP
    MS

    I'm stuck at question 6 part c. At this point, I've established that tan3(theta)=1 and that the roots (theta) are tan(pi/12), tan(5pi/12), tan(9pi/12).

    I don't understand the "hence" because the marks cheme shows tan(pi/12)+tan(5pi/12)+tan(9pi/12)=3 but shouldn't it be tan3(pi/12)+tan3(5pi/12)+tan3(9pi/12)=3

    Yours Sincerely,
    A confused student
    I think you've confused yourself here!

    The first part of the question lets you write tan(3X) in terms of t where t=tan X.

    If you take X = \pi/12 then tan(3X) = 1, so if t = tan(X) = tan(\pi/4) then t satisfies the cubic equation t^3 - 3t^2 - 3t + 1 = 0

    We can used this logic to find 2 other distinct roots of the cubic. We know that if X = 5\pi/12, 3X = 5\pi/4, tan(3X) = 1
    and
    X = 9\pi/12, 3X = 9\pi/4, tan(3X) = 1

    Therefore, the roots of the cubic are a= tan(\pi/12), b = tan(5\pi/12), c = tan(9\pi/12)

    Now factorize the cubic as (t-a)(t-b)(t-c) and compare coefficients of t^2:
    -(a+b+c) = -3
    so
    a+b+c = 3
    or
    tan(\pi/12) + tan(5\pi/12) + tan(9\pi/12) = 3

    But tan(9\pi/12) = tan(3\pi/4) = -1

    and the required result follows.
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    (Original post by davros)
    I think you've confused yourself here!

    The first part of the question lets you write tan(3X) in terms of t where t=tan X.

    If you take X = \pi/12 then tan(3X) = 1, so if t = tan(X) = tan(\pi/4) then t satisfies the cubic equation t^3 - 3t^2 - 3t + 1 = 0

    We can used this logic to find 2 other distinct roots of the cubic. We know that if X = 5\pi/12, 3X = 5\pi/4, tan(3X) = 1
    and
    X = 9\pi/12, 3X = 9\pi/4, tan(3X) = 1

    Therefore, the roots of the cubic are a= tan(\pi/12), b = tan(5\pi/12), c = tan(9\pi/12)

    Now factorize the cubic as (t-a)(t-b)(t-c) and compare coefficients of t^2:
    -(a+b+c) = -3
    so
    a+b+c = 3
    or
    tan(\pi/12) + tan(5\pi/12) + tan(9\pi/12) = 3

    But tan(9\pi/12) = tan(3\pi/4) = -1

    and the required result follows.
    Oh thank you! It never occurred to me that the topic of roots of polynomials would be thrown in there!
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    (Original post by Serendreamers)
    Oh thank you! It never occurred to me that the topic of roots of polynomials would be thrown in there!
    I must admit, the first time I opened your attachment I skimmed the question paper very quickly and thought this question followed on directly from the one about roots of polynomials, so it was fresh in my mind
 
 
 
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