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I have function $f:R\to R^2 , \ \ f(x)=<\cos 3x, \sin 3x>$ and I have to find image on the interval $(0, \pi]$ and inverse image $[0, +\infty) \times[0, +\infty)$

I think the image will be simply $[-1,1]\times [-1,1] $ but I have problem with proving it. I would try something like this: for $x \in(0, \pi]$ we have $-1\le\sin 3x \le 1$ and $-1\le\cos 3x\le1$ so it's one inclusion but I don't know how to do the other one.

For inverse image there will be such $x$ that both $\cos 3x$ and $\sin 3x$ are bigger or equal to $0$ ??

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Image of $f$ is not $[-1,1]\times [-1,1]$, since there is no $x$ such that $f(x)=(1,1)$.

The image is diameters on the axis of unit circle; or more formally:

$$ f[(0,\pi]]=\{ (x,y)\in\mathbb{R} \mid (x\in[-1,1] \wedge y=0)\vee (x=0 \wedge y\in[-1,1])\} $$

To see this, just take one element from image then relate with its angle and divide this angle by $3$.

You can apply the same procedure for inverse image..

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  • $\begingroup$ what is the answer to the inverse image ? $\endgroup$
    – tom
    Nov 26, 2014 at 19:13
  • $\begingroup$ @tom. Try for yourself and see what you can find first. $\endgroup$
    – Alistair
    Nov 26, 2014 at 20:43

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