Why is $\sin(\arccos(x))$ a semicircle with radius 1? It was unexpected to see that from looking at the equation. Is there an intuitive explanation for why it's a perfect semicircle?
 A: For $x\in [-1,1]$,
$$\theta=\cos^{-1} x\in [0,\pi] \implies \sin \theta \geq 0$$
Now,
\begin{align*}
  y &= \sin (\cos^{-1} x) \\
    &= \sin \theta \\
    &=+\sqrt{1-\cos^{2} \theta} \\
    &=\sqrt{1-x^{2}}
\end{align*}
A: Using Inverse trigonometric functions,  $$0\le\arccos x\le\pi$$
$$\implies\sin(\arccos x)\ge0$$
If $\arccos x=u,\cos u=x$ and consequently,  $$y=\sin u=+\sqrt{1-x^2}$$
A: This function is defined on the interval $[-1,1]$ and if $y = \sin(\arccos x)$ we have
$$
y^2 + x^2 = \sin^2(\arccos x) + x^2 = \\ 1 - \cos^2(\arccos x) + x^2 = 1 - x^2 + x^2 = 1.
$$
As $\arccos x \in [0, \pi]$ we have $\sin(\arccos x) \in [0,1]$ and thus this is a semicircle with center in in $(0,0)$ and radius 1.
A: Assume that $x=\cos(\theta)$ for some $\theta$. Then $y=\sin(\arccos(x))=\sin(\arccos(\cos(\theta)))=\sin(\theta)$.
As $x^2+y^2=\cos^2(\theta)+\sin^2(\theta)=1$, we have the equation of a unit circle at the origin. I leave it to you to discuss the quadrants.
A: $x=\cos(\theta),y=\sin(\theta)$ is obviousely a circle, parametrized by theta. Now eliminate $\theta$: $\theta=\pm\arccos(x)$, gives $y=\sin(\pm\arccos(x))$ which is still a circle.
A: Take a right triangle whose hypotenuse is $1$, and mark one of the acute angles as $\theta$. Let $\cos(\theta)=x$. Then, $\theta=\cos^{-1}(x)$, and consequently, $\sin(\theta)=\sqrt{1-x^{2}}$. Let $y=\sin(\theta)=\sqrt{1-x^{2}}$. Clearly, since $x=\cos(\theta)$, $-1\leq x\leq 1$, and consequently, $0\leq y\leq 1$. This marks the locus of the top half of a circle of radius $1$.
A: Components of a unit segment from origin along the axes are:
$$ x = 1 \cos\theta,  \, y = 1 \sin\theta $$
Eliminate $ \theta $
$$ y= 1 \sin (\arccos x )$$
$$ y= \sqrt {1- x^2}$$
$$ x^2+y^2 =  1 $$
which is the equation of unit circle.
Another way is to construct a right angled triangle with adjacent side $x$ and hypotenuse 1, opposite side will be $ \sqrt {1- x^2}.$
