sketching solution to y'=sin(y) without attempting to solve?

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So I'm being asked to sketch some solution curves to$$y'=sin(y)$$

And describe it's behavior as x tend to $\infty$ without attempting to solve im also asked what would change if the initial conditions were different i.e $y(0)=0$ or $y(0)=0.1$.

I know it probably has something to do with isoclines but for the life of me where do i even start?

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To start with take any point in the first quadrant where the $y$ coordinate is between $0$ and $\pi$.

Rather than mark a point, mark a short line of gradient $\sin y$. So the steepest line has gradient $1$.

Repeat over and over again.

As $y\rightarrow\pi$ or $\rightarrow 0$ the gradient $\rightarrow 0$