Help with surjectivity of a function

Question

My function $f$ is as follows: $$f: \mathbb{R} \rightarrow \mathbb{R^2}: t \rightarrow (2cos(t), - sin(t))$$

Now, I'm fairly certain that the function isn't injective, as both sine and cosine are periodic, but I'm not quier sure about the surjectivity. A surjective function would imply that I cover all elements of the codomain, which in this case would be \mathbb{R^2}, so this shouldn't be the case as sine and cosine are 'restricted' on the y-axis.

Now that I've made my assumptions I also need to back them up, and this is where my question arises. Prooving that the function isn't injective is pretty straight forward (plug in $2\pi*t$ for $t$ and you get the same result), but proofing that it's not surjective is a bit more complicated.

My approach would be to try and disprove that $f(t) = (a, b): a,b \in \mathbb{R}$, but I'm not quiet sure about how to do this mathematically. Is it enough to show that the $2cos(t)=a$ and $-sin(t)=b$ don't result in the same $t$? Or am I approaching the problem incorrectly?

Thank you very much!

Answer 1

Your approach is a bit non-concrete (though I can appreciate that to recognise this might take some more experience).

For a more concrete approach: there are at least two possible ways to approach this.

• Restrict one of the coordinates to something well-behaved. $2\cos(t)$ is equal to $0$ precisely when… what? And how does $-\sin(t)$ behave under those conditions?
• Are the projections $2 \cos(t)$ and $-\sin(t)$ surjective? (This is a necessary condition for $(2 \cos(t), -\sin(t))$ to be surjective.)

(On second thoughts, the second method is a degenerate version of the first.)