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I've been playing around with normal trigonometry making circles and waves. I make circles like this:

$y = \cos(t)$

$t \in [0,2\pi)$

$x = \sin(t)$

Simple. I was wondering, could I graph the top half of a circle ONLY using $x$, and graphing by $y$? As in something like

$y=\cos(x+\sin(x))$

Which I understand doesn't work. Something that would make a wave, but have it warped like the top/bottom half of a circle If you could give me an equation with reasoning, or a answer of why it wouldn't work, that would be great! Thanks!

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    $y=\sqrt {1-x^2}$ for $x\in [-1,1]$. – 2017-01-28
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    Thanks for the fast answer! And cool, it does work. Any reasoning why? – 2017-01-28
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    A circle is the locus of points equidistant from a given point. If the given point is $(0,0)$ and the radius is $1$ we get $x^2+y^2=1$ or $y^2=1-x^2$. As you want $y≥0$ just take the positive square root. – 2017-01-28
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    @Blake Define distance in the $x,y$ plane. A circle is just the special case of when the distance from $(0,0)$ is fixed. – 2017-01-28
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    Awesome. Good to know. – 2017-01-28

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$\DeclareMathOperator{\Asin}{Arcsin}$You've got a good answer in the comments, but in the spirit of your question: The equation $$ x = \sin t,\quad -1 \leq x \leq 1, $$ can be written $$ t = \Asin x,\quad -\pi/2 \leq t \leq \pi/2. $$ This principal branch of arcsine satisfies $\sin(\Asin x) = x$ for $-1 \leq x \leq 1$.

Substituting into $y = \cos t$ and using $$ \cos t = \sqrt{1 - \sin^{2}t},\quad -\pi/2 \leq t \leq \pi/2, $$ gives $$ y = \cos(\Asin x) = \sqrt{1 - x^{2}},\quad -1 \leq x \leq 1. $$

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    That really helps. Thanks for explaining in detail :) – 2017-01-29