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can someone please explain why the graph of $\sin (x) ,0\leq x\leq 2\pi$ is the same graph as $\sin (2x), 0\leq 2x\leq 4\pi$?

I don't understand why they are the same. If I let $\alpha=2x$, then the second equation becomes $\sin (\alpha), 0\leq \alpha \leq 4\pi$ looks different from the first one?

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  • $\begingroup$ They are not the same. The condition $0 \le 2x \le 4\pi$ is exactly the same as $0 \le x \le 2\pi$, but $\sin(2x)$ is not the same as $\sin(x)$. Did you mean $0 \le 2x \le 2\pi$? $\endgroup$ – TonyK Apr 26 '14 at 18:49
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Hint: Multiply all three parts of the inequality by $\alpha=2$. How does it relate to the argument of $\sin$?

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