# How to prove that sine is surjective

I know that the sine function in the domain $[0,1)$ is surjective, I know that since I was a child, in these days I'm thinking how to prove it formally.

Anyone can help me with this?

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In English, the function is called "sine." Sinus means something else entirely in English. – Thomas Andrews Sep 29 '12 at 16:11

With the geometric definition of sine: For $0<y<1$ the parallel line at distance $y$ to the $x$-axis intersects the unit circle around the origin $O$ at a point $P$ in the first quadrant (admittedly, this fact is also less trivial than at first sight). Let $Q$ be the projection of $P$ on the $x$-axis. Then from triangle $OQP$ we read $\sin\angle QOP = y$.
Otherwise (i.e without geometrical reference), the surjectivity is a consequence of the sine function being continuous together with $\sin 0 = 0$ and $\sin\frac\pi2=1$.