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I have two function $c_1, c_2 \colon [0,2\pi] \to \mathbb{R}^2$ define by $c_1(t) = (\cos(t), \sin(t))$ and $c_2(t) = (\cos(2t), \sin(2t))$

I want to show that they have the same image. It is pretty obvious, but I don't know how to prove it.

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The key to problems like this is to carefully write down what you want to prove, namely that two sets are equal. In this case, to prove $image(c_1) = image(c_2)$, you first show that $image(c_1) \subset image(c_2)$, and then show the reverse containment.

To do the first, you have to know what $image(c_1)$ actually is. It's $$ image(c_1) = \{ (\cos t, \sin t) \mid t \in [0, 2\pi]\}. $$ Thus every element in the image is a cosine-sine pair for some argument. You can do the same for the second image, and then you're ready to go:

Take a point in the image of the first function; it must be $(\cos a, \sin a)$ for some $a \in [0, 2\pi]$. You want to show that it's also $(\cos 2b, \sin 2b)$ for at least one point $b \in [0, 2\pi]$. Hint: pick $b = a/2$. Then write out what you get. And confirm that $b$ really is in the specified domain, while you're at it.

Then you have to do the same thing in the other direction.

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