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Math Help - sin (z) = a and a\in [-1,1]

  1. #1
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    sin (z) = a and a\in [-1,1]

    If \sin{z}=a, \ a\in [-1,1], then what can be said about z\mbox{?}

    Since a is between -1 and 1, a=\cos{\theta}.

    \sin{z}=\sin{x}\cosh{y}+\mathbf{i}\cos{x}\sinh{y}

    Setting the real and imaginary parts equal we obtain:

    \displaystyle\sin{x}\cosh{y}=\cos{\theta}\Rightarr  ow x=\frac{\pi}{4}+\pi k, \ k\in\mathbb{Z}, \ \mbox{and} \ y=0

    and

    \displaystyle\cos{x}\sinh{y}=0 \Rightarrow x=\frac{\pi}{2}+\pi k, \ k\in\mathbb{Z}, \ \mbox{or when} \ y=-y\Rightarrow \ y=0

    However, looking at when \displaystyle \cosh{y}=\cos{\theta}\Rightarrow \frac{e^y+e^{-y}}{2}=\frac{e^{\mathbf{i}\theta}+e^{-\mathbf{i}\theta}}{2}\Rightarrow e^y+e^{-y}=e^{\mathbf{i}\theta}+e^{-\mathbf{i}\theta}

    it appears that solution is only when y=0

    When \displaystyle  x=\frac{\pi}{2}+\pi k, \ k\in\mathbb{Z}

    \displaystyle \sin{\left(\frac{\pi}{2}+\pi k\right)}\cosh{y}=\cos{\theta}\Rightarrow (-1)^k\cosh{y}=\cos{\theta}\Rightarrow k=2p, \ p\in\mathbb{Z}

    \displaystyle\Rightarrow\cosh{y}=\cos{\theta}\Righ  tarrow y=\arccosh{(\cos{\theta})} \ \mbox{and} \ x=\frac{\pi}{2}+2\pi p

    One solution is: \displaystyle \ \ \left(\frac{\pi}{2}+2\pi p,\cosh^{-1}{(\cos{\theta})}\right)

    When y=0

    \displaystyle \sin{x}\cosh{0}=\cos{\theta}\Rightarrow\sin{x}=\co  s{\theta}\Rightarrow x=\frac{\pi}{4}+\pi m, \ m\in\mathbb{Z}

    Another solution is: \displaystyle \ \ \left(\frac{\pi}{4}+\pi m, 0\right)

    \displaystyle z=x+y\mathbf{i}\Rightarrow x=\frac{\pi}{4}+\pi m \ \mbox{when} \ y=0 \ \mbox{and} \ \ x=\frac{\pi}{2}+2\pi \ \mbox{when} \ y=\cosh^{-1}{(\cos{\theta})} \ m,p\in\mathbb{Z}

    Is this what can be said about z?
    Last edited by dwsmith; December 23rd 2010 at 10:48 AM. Reason: Added solutions.
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    If \sin{z}=a, \ a\in [-1,1], then what can be said about z\mbox{?}

    Since a is between -1 and 1, a=\cos{\theta}.

    \sin{z}=\sin{x}\cosh{y}+\mathbf{i}\cos{x}\sinh{y}
    If z = x+iy then \sin{z}=\sin{x}\cosh{y}+i\cos{x}\sinh{y}. If that number lies in the interval [1,1] then its imaginary part must be 0. Therefore either \sinh y = 0 or \cos x = 0.

    If \sinh y = 0 then y=0 and hence  z is real. But every real number x satisfies \sin x \in [-1,1]. So  z can be any real number.

    Now suppose that \sinh y \ne0 and look at the other possibility, namely \cos x = 0. If \cos x = 0 then \sin x = \pm1; and if \sinh y \ne0 then \cosh y > 1. So in this case |\sin{x}\cosh{y}| > 1 and hence \sin z\notin[-1,1].

    Conclusion: \sin z \in [-1,1]\;\Longleftrightarrow\;z\in\mathbb{R}.
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    Quote Originally Posted by Opalg View Post
    If z = x+iy then \sin{z}=\sin{x}\cosh{y}+i\cos{x}\sinh{y}. If that number lies in the interval [–1,1] then its imaginary part must be 0. Therefore either \sinh y = 0 or \cos x = 0.

    If \sinh y = 0 then y=0 and hence  z is real. But every real number x satisfies \sin x \in [-1,1]. So  z can be any real number.

    Now suppose that \sinh y \ne0 and look at the other possibility, namely \cos x = 0. If \cos x = 0 then \sin x = \pm1; and if \sinh y \ne0 then \cosh y > 1. So in this case |\sin{x}\cosh{y}| > 1 and hence \sin z\notin[-1,1].

    Conclusion: \sin z \in [-1,1]\;\Longleftrightarrow\;z\in\mathbb{R}.
    One question. By setting a = cos, I can eliminate the problem of |sinx coshy|>1 since I solved for the values that make it true. Does doing this make the approach wrong?
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