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Thread: Trigonometric equation

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    Trigonometric equation

    If $\displaystyle \cos x = \cos y$ and $\displaystyle \sin x = -\sin y$

    Then find value of $\displaystyle \sin (2006\; x)+\sin (2006\; y)$
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    Re: Trigonometric equation

    Quote Originally Posted by jacks View Post
    If $\displaystyle \cos x = \cos y$ and $\displaystyle \sin x = -\sin y$

    Then find value of $\displaystyle \sin (2006\; x)+\sin (2006\; y)$
    So what have you tried?

    Clearly y=-x satisfies the condition, but does anything else?

    CB
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    Re: Trigonometric equation

    Quote Originally Posted by jacks View Post
    If $\displaystyle \cos x = \cos y$ and $\displaystyle \sin x = -\sin y$

    Then find value of $\displaystyle \sin (2006\; x)+\sin (2006\; y)$
    Are x and y mutually exclusive?
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    Re: Trigonometric equation

    Thanks Got it

    Let $\displaystyle z=\cos x+i\sin x$ and $\displaystyle \omega = \cos y +i\sin y$

    Then from Given equation $\displaystyle z=\bar{\omega}$

    So $\displaystyle \cos (2006 \;x)+i\sin (2006\; x) = z^{2006}$

    OR $\displaystyle Im(z^{2006}) = \sin (2006\; x)$

    and $\displaystyle Im(\omega^{2006}) = \sin (2006\; y)$

    So $\displaystyle \sin (2006\; x)+\sin (2006\; y) = Im\left(z^{2006}+\omega^{2006}\right)=Im\left(\bar {\omega})^{2006}+\omega^{2006}\right) = 0$
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    Grand Panjandrum
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    Re: Trigonometric equation

    Quote Originally Posted by jacks View Post
    Thanks Got it

    Let $\displaystyle z=\cos x+i\sin x$ and $\displaystyle \omega = \cos y +i\sin y$

    Then from Given equation $\displaystyle z=\bar{\omega}$

    So $\displaystyle \cos (2006 \;x)+i\sin (2006\; x) = z^{2006}$

    OR $\displaystyle Im(z^{2006}) = \sin (2006\; x)$

    and $\displaystyle Im(\omega^{2006}) = \sin (2006\; y)$

    So $\displaystyle \sin (2006\; x)+\sin (2006\; y) = Im\left(z^{2006}+\omega^{2006}\right)=Im\left(\bar {\omega})^{2006}+\omega^{2006}\right) = 0$
    If you had indicated that you were familiar with complex analysis we could have pointed out that the conditions give:

    $\displaystyle e^{ix}=e^{-iy + 2n\pi i}, \ \ n=...-1, 0, 1, ...$

    and so:

    $\displaystyle e^{2006 ix}=e^{- 2006 iy} e^{4012 \pi i}=e^{- 2006 iy}$

    and the result follows immediately.

    CB
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