Finding solutions for $\sin(z)=\sin(2)$

Question

having some difficulty wrapping my head around how to methodically do this style of question:

Find all solutions $z\in ℂ$ for $\sin(z)=\sin(2)$

I attempted to solve this by using the identity,

$\sin(x)\cosh(y)+i\cos(x)\sinh(y)=\sin(2)$,

so that,

$\sin(x)\cosh(y)=\sin(2)$ and $\cos(x)\sinh(y)=0$,

but this seemed to lead to a dead end, so I'm unsure as to what the next step should be...

Any help is much appreciated

Answer 1

Since OP seems confused, I am providing the details.

Let $z\in\mathbb{C}$ such that $\sin(z)=\sin(2)$. Then $\frac{e^{iz}-e^{-iz}}{2i}=\sin(2)$, therefore, if $z=x+iy$, $e^{ix-y}-e^{-ix+y}=2i\sin(2)$, hence $e^{-y}(\cos(x)+i\sin(x))-e^y(\cos(x)-i\sin(x))=2i\sin(2)$. Real and imaginary parts have to be equal, so one gets $(e^{-y}-e^y)\cos(x)=0$ and $e^{-y}\sin(x)+e^y\sin(x)=2\sin(2)$. The first one yields either $y=0$ or $x=k\pi+\pi/2$, but $x=k\pi+\pi/2$ is ruled out by the second equation, since $e^t+e^{-t}\geq2$ . So $y=0$ and the second equation gives $\sin(x)=\sin(2)$, so $x=2k\pi+2$ or $x=(2k+1)\pi-2$

Answer 2

Hint:

$$\frac{e^{iz}-e^{-iz}}{2i}=k$$ Let $x=e^{iz}$, $$x^2-2ikx-1=0$$ $$x=ik\pm \sqrt{1-k^2}$$ $$z=-i(\ln (ik\pm \sqrt{1-k^2})+2n\pi i)= -i\ln (ik\pm \sqrt{1-k^2})+2n\pi $$

For $k=\sin(m)$ $$ik\pm \sqrt{1-k^2}=\pm \cos(m)+i\sin(m)=\pm e^{\pm im}$$

So $$z=-i(\ln(\pm 1)\pm im)+2n\pi$$ where we define $\ln(1)=0$ and $\ln(-1)=i\pi$.

A quicker approach:

Make use of the identities:

$$\sin(x)=\sin(\pi-x)$$ $$\sin(x)=\sin(x+2n\pi)$$

Answer 3

This is a different approach to the other answers, which use complex exponentials to solve. I am following up on the comment I put.

As you said in a comment, if $y=0$, then $\sin x=\sin 2$. Since $\sin$ is $2\pi$-periodic, $\sin(x+2\pi)=\sin(x)$. So we have $x=2+2\pi n$ for $n\in\Bbb Z$.

It is worth check if the alternative solution works - $x=\frac{2n+1}2\pi,y\in\Bbb R$. In this case, $\sin(x)=(-1)^n$, so we need $\cosh y=\pm \sin2$. But $\cosh y>1\,\,\,\,\forall y$ and $|\sin 2|<1$. So there is no solution here. Hence the solutions above are the only ones.

Answer 4

$$\begin{align} & \text{This problem still can be solved following your procedure}: \\ & \cos \left( x \right)\sinh \left( y \right)=0\Rightarrow \cos \left( x \right)=0\ OR\ \sinh \left( y \right)=0 \\ & \\ & If\sinh \left( y \right)=0,\ then\ y=0\ ,and\ hence\ \\ & \sin \left( x \right)\cosh \left( y \right)=\sin \left( x \right)\cosh \left( 0 \right) \\ & \quad \quad \quad \quad \quad \quad =\sin \left( x \right)\times 1 \\ & \quad \quad \quad \quad \quad \quad =\sin \left( x \right) \\ & \quad \quad \quad \quad \quad \quad \Rightarrow \sin \left( x \right)=\sin \left( 2 \right)\Rightarrow x=2+2n\pi . \\ & \\ & If\ \cos \left( x \right)=0,\ then\ \sin \left( x \right)=\pm \sqrt{1-\cos {{\left( x \right)}^{2}}}=\pm \sqrt{1-{{0}^{2}}}=\pm 1,and\ hence \\ & \ \sin \left( x \right)\cosh \left( y \right)=\pm \cosh \left( y \right)=\sin \left( 2 \right) \\ & \quad \quad \quad \quad \quad \quad \,\,\Rightarrow \ \cosh \left( y \right)=\pm \sin \left( 2 \right)\ \\ & \quad \quad \quad \quad \quad \quad \,\,\Rightarrow \ \cosh \left( y \right)=\sin \left( 2 \right)\ since\ \cosh \left( y \right),\sin \left( 2 \right)>0.\ \\ & \quad \quad \quad \quad \quad \quad \,\,\Rightarrow y={{\cosh }^{-1}}\left( \sin \left( 2 \right) \right). \\ & \\ & Finally\ z=\left( 2+2n\pi \right)+i{{\cosh }^{-1}}\left( \sin \left( 2 \right) \right) \\ \end{align} $$