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

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

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

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

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

Setting the real and imaginary parts equal we obtain:

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

and

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

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

it appears that solution is only when $\displaystyle y=0$

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

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

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

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

When $\displaystyle y=0$

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

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

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

Is this what can be said about z?

2. Originally Posted by dwsmith
If $\displaystyle \sin{z}=a, \ a\in [-1,1]$, then what can be said about $\displaystyle z\mbox{?}$

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

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

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

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

Conclusion: $\displaystyle \sin z \in [-1,1]\;\Longleftrightarrow\;z\in\mathbb{R}$.

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

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

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

Conclusion: $\displaystyle \sin z \in [-1,1]\;\Longleftrightarrow\;z\in\mathbb{R}$.
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?