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    Need help with q3. I cant seem to do the first part. Here is what I have tried to do.
    http://m.imgur.com/laIkAiy,0wlahrY,jriufqx
    Thanks.
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    (Original post by Super199)
    Need help with q3. I cant seem to do the first part. Here is what I have tried to do.
    http://m.imgur.com/laIkAiy,0wlahrY,jriufqx
    Thanks.
    What's the question?
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    (Original post by Super199)
    Need help with q3. I cant seem to do the first part. Here is what I have tried to do.
    http://m.imgur.com/laIkAiy,0wlahrY,jriufqx
    Thanks.
    You shouldn't have rationalised the w thingy, you should have done:

    \displaystyle \frac{(1+ \cos \theta) + i\sin \theta}{(1-\cos \theta) - i\sin \theta} \times \frac{(1-\cos \theta) + i\sin \theta}{(1-\cos \theta) + i \sin \theta}

    and then made use of the fact that \cos \theta = 2\cos^2 \frac{\theta}{2} - 1 = \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2} = 1 - 2\sin^2 \frac{\theta}{2} and \sin \theta = 2\sin \frac{\theta}{2}\cos \frac{\theta}{2}
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    (Original post by Zacken)
    You shouldn't have rationalised the w thingy, you should have done:

    \displaystyle \frac{(1+ \cos \theta) + i\sin \theta}{(1-\cos \theta) - i\sin \theta} \times \frac{(1-\cos \theta) + i\sin \theta}{(1-\cos \theta) + i \sin \theta}

    and then made use of the fact that \cos 2\theta = 2\cos^2 \frac{\theta}{2} - 1 = \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2} = 1 - 2\sin^2 \frac{\theta}{2} and \sin \theta = 2\sin \frac{\theta}{2}\cos \frac{\theta}{2}
    But havent you rationalized it but just not using w but the original thing.
    Do i plug in the things first or multiply everything?
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    (Original post by Super199)
    But havent you rationalized it but just not using w but the original thing
    Yes, but you rationalise things by multiplying by its conjugate. -w is not the conjugate of w since w = \cos \theta + i\sin \theta, the conjugate is only the sign before the i flipped.
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    (Original post by Zacken)
    Yes, but you rationalise things by multiplying by its conjugate. -w is not the conjugate of w since w = \cos \theta + i\sin \theta, the conjugate is only the sign before the i flipped.
    I see. Is it better to multiply things first or sub in the trig results.
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    (Original post by Super199)
    I see. Is it better to multiply things first or sub in the trig results.
    The trig identities he used are only applicable for the product.
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    I think this question would be easier if you used the exponential form of the complex number, w.
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    If  w=\cos \theta +i\sin \theta then  w=e^{i\theta}

    So you get  \displaystyle \frac{1+e^{i\theta}}{1-e^{i\theta}}

    If you divide through top and bottom by  e^{-i\theta/2} you get

     \displaystyle \frac{ e^{\frac{i\theta}{2}}+e^{-\frac{i\theta}{2}}}{e^{-\frac{i\theta}{2}}-e^{\frac{i\theta}{2}}} .
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    (Original post by B_9710)
    If  w=\cos \theta +i\sin \theta then  w=e^{i\theta}

    So you get  \displaystyle \frac{1+e^{i\theta}}{1-e^{i\theta}}

    If you divide through top and bottom by  e^{-i\theta/2} you get

     \displaystyle \frac{ e^{\frac{i\theta}{2}}+e^{-\frac{i\theta}{2}}}{e^{-\frac{i\theta}{2}}-e^{\frac{i\theta}{2}}} .
    Sorry I don't see where you are going with this?
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    (Original post by Super199)
    Sorry I don't see where you are going with this?
    Are you aware that  \displaystyle  e^{ix} - e^{-ix} = 2i\sin x

    And  \displaystyle e^{ix} + e^{-ix} = 2\cos x .
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    (Original post by B_9710)
    Are you aware that  \displaystyle  e^{ix} - e^{-ix} = 2i\sin x

    And  \displaystyle e^{ix} + e^{-ix} = 2\cos x .
    Nope. Where do these come from?
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    (Original post by Super199)
    Nope. Where do these come from?
    These are a direct result of Euler's formula: e^{ix}=\cos(x)+i\sin(x).
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    (Original post by Super199)
    Nope. Where do these come from?
    Well  \displaystyle \cos x + i\sin x = e^{ix} and you just use De Movire's theorem to prove the 2 results.

     \displaystyle e^{ix}+e^{-ix}= \cos x+i\sin x + cos x -i\sin x = 2\cos x .
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    (Original post by B_9710)
    Well  \displaystyle \cos x + i\sin x = e^{ix} and you just use De Movire's theorem to prove the 2 results.

     \displaystyle e^{ix}+e^{-ix}= \cos x+i\sin x + cos x -i\sin x = 2\cos x .
    Hmm i keep ****ing this up.
    The top gives 2cos1/2x
    The bottom gives -2isin1/2x?
    Is that right?
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    (Original post by Super199)
    Hmm i keep ****ing this up.
    The top gives 2cos1/2x
    The bottom gives -2isin1/2x?
    Is that right?
    Yes the 2 on top and bottom cancel. Then just move the imaginary part from denominator to numerator - how do yo think you do this?
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    (Original post by B_9710)
    Yes the 2 on top and bottom cancel. Then just move the imaginary part from denominator to numerator - how do yo think you do this?
    Rationalize?
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    (Original post by Super199)
    Rationalize?
    Exactly.
 
 
 
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