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If $f:\mathbb{R}\to\mathbb{R}$ and $g:\mathbb{R}\to\mathbb{R}$ are functions such that $f = \sin \circ \, h$ and $g = \cos \circ \, h$ for some function $h:\mathbb{R}\to\mathbb{R}$ it follows that $f^2(x)+ g^2(x) = 1$ for all $x\in\mathbb{R}$. I'm interested in whether the converse is true (and if so, how would one prove it?). A related question arising in complex analysis can be found here but I'm still learning calculus so I lack the background to understand it. I know this may be a very stupid question but I'm genuinely curious about it so any hint on how to approach this problem would suffice. Thanks for reading! :)

E. Assael
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  • There is no need for continuity of $f,g,h$? So just choose for every $x$ e.g., the unique value $h(x)\in[0,2\pi)$ that works. – user10354138 Sep 04 '21 at 16:44
  • Sure: the linked question has differentiability in both hypothesis and conclusion and this question does not, so the questions are logically unrelated (an answer to any of them does not imply an answer to the other). – Adayah Sep 04 '21 at 16:53
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    @Adayah: https://math.stackexchange.com/a/4055387/42969 gives a result under the assumption of continuity only. But you are right, without that assumption the result may be different, therefore I'll reopen the question. – Martin R Sep 04 '21 at 17:03
  • Without continuity it is actually quite simple: For any $x$ define $h(x) = \phi$ where $\phi$ is chosen such that $\sin(\phi) = f(x)$ and $\cos(\phi) = g(x)$. – Martin R Sep 04 '21 at 17:07

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Sure, all points $(f(x),g(x))$ lie on a curve (the unit circle), which is parameterized by $\theta\mapsto (\sin(\theta),\cos(x))$.

In other words, for all $x\in\mathbb{R}$, there exists some $\theta_x\in\mathbb{R}$ with $(f(x),g(x))=(\sin(\theta_x),\cos(\theta_x))$. Now simply set $h(x):=\theta_x$.

Mastrem
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