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    Why do these two questions have a negative for the sine coefficient?

    ie why is it cos(11theta) - i sin(11theta) as opposed to cos(11theta) + i sin(11theta)

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    (Original post by 雷尼克)
    Why do these two questions have a negative for the sine coefficient?

    ie why is it cos(11theta) - i sin(11theta) as opposed to cos(11theta) + i sin(11theta)

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    There seems to be quite a few mistakes in this

    \sin 3\theta \neq \sin (-3\theta)
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    (Original post by Zacken)
    There seems to be quite a few mistakes in this

    \sin 3\theta \neq \sin (-3\theta)
    That's not a mistake, in C2 we are taught that sin(-theta) = -sin(theta), and cos(-theta) = cos(theta), this is brought up again in FP2 chapter 3.

    What you typed is true, but the sign coefficient to sine is also reversed in the answer from solutionbank which is in the OP.
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    (Original post by 雷尼克)
    Why do these two questions have a negative for the sine coefficient?

    ie why is it cos(11theta) - i sin(11theta) as opposed to cos(11theta) + i sin(11theta)

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    Should be \displaystyle \frac{e^{5i\theta}}{e^{6i\theta}  } = e^{-i\theta} = \cos \theta - i \sin \theta
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    (Original post by 雷尼克)
    That's not a mistake, in C2 we are taught that sin(-theta) = -sin(theta), and cos(-theta) = cos(theta), this is brought up again in FP2 chapter 3.

    What you typed is true, but the sign coefficient to sine is also reversed in the answer from solutionbank which is in the OP.
    No it's not - look at the second line, it goes from + i sin (3theta) to + i sin (-3 theta)

    Anyway, the whole thing is wrong, look at my post.
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    (Original post by 雷尼克)
    That's not a mistake, in C2 we are taught that sin(-theta) = -sin(theta), and cos(-theta) = cos(theta), this is brought up again in FP2 chapter 3.

    What you typed is true, but the sign coefficient to sine is also reversed in the answer from solutionbank which is in the OP.
    Second question/part is wrong as well:

    Should be \dfrac{e^{-i\theta}}{e^{-6i\theta}} = e^{5i\theta} = \cos 5\theta + i \sin 5\theta
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    (Original post by 雷尼克)
    That's not a mistake, in C2 we are taught that sin(-theta) = -sin(theta), and cos(-theta) = cos(theta), this is brought up again in FP2 chapter 3.

    What you typed is true, but the sign coefficient to sine is also reversed in the answer from solutionbank which is in the OP.
    ?

    It clearly goes from  \sin3\theta to  \sin(-3\theta) without a change of sign. There is a mistake in the thing you linked.
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    (Original post by Louisb19)
    ?

    It clearly goes from  \sin3\theta to  \sin(-3\theta) without a change of sign. There is a mistake in the thing you linked.
    cant trust solutionbank then T_T
 
 
 
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