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I want to show that if $\sin(a^3) = \sin(b^3)$, then $a = b$. However, I can't take the arcsin of both sides without assuming it's injective, and it becomes a circular argument.

Is there any other way I can prove that is injective from the definition? Or do I need to use other theorems/logic? Maybe I can try to argue it is continuous on it's interval and that it is strictly increasing, so therefore it should pass through each point only once, but that feels more like hand-wavy arguments rather than a rigorous proof.

Sebastiano
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Yoshi
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1 Answers1

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Since $x \mapsto x^3$ is a bijective map $(-1,1)\to (-1,1)$ you only have to show that $\sin(x)$ is injective on $(-1,1)$. Its derivative is $\cos(x)$ which is positive on that interval so $\sin(x)$ is stringly monotone, hence injective.

principal-ideal-domain
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