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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.