Why is $\sin(\arccos(y)) = \sqrt{1-y^2}$? Why is: $\sin(\arccos(y)) = \sqrt{1-y^2}$ ?
I thought maybe transforming sin to cos:
$\cos(\frac{\pi}{2} - \arccos(y))$ but it doesn't get me anywhere. Hint?
 A: Let $\arccos(y)=t$. Then,
$$y=\cos(t)\implies y^2=\cos^2t=1-\sin^2t\implies \sin^2t=1-y^2\implies \sin t=\pm\sqrt{1-y^2}$$
But note that $t=\arccos(y)\in [0,\pi]~\forall~y\in\Bbb R$ by definition and sine is non-negative in this interval. Hence, the negative solution is an extraneous solution and we can therefore discard it to get,
$$\sin t=\sqrt{1-y^2}\implies \sin(\arccos y)=\sqrt{1-y^2}$$
A: $$
\sin^2 x + \cos^2 x = 1 \Rightarrow  \sin x = \sqrt{1 - \cos^2 x}
$$
Now let $ y = \cos x$. Then $x = \arccos(y)$, and the result follows.
A: 
$$\arccos y =\theta$$
$$\sin \theta =\sqrt{1-y^2}$$
A: While the other two answers are perfectly valid, I believe that viewing the problem from a general right triangle provides better intuition for this identity. 
It would help to draw a triangle and work from the inside out. The $\arccos(y)$ asks what angle, let's call it $\theta$, gives $\frac{y}{1}$ when we take the cosine of that angle. Thus, this can also be expressed by a right triangle with a hypotenuse of $1$ and a side length adjacent to the angle $\theta$ of y. 
Now, it is only a matter of finding the sine of this angle. Use the Pythagorean Theorem to find the opposite side and keep in mind that $\sin(\theta)= \frac{opposite}{hypotenuse}.$ Can you proceed?
A: Let $\arccos y =x$ so : $$\cos x=y$$ 
What you need to prove is $$\sin x=\sqrt{1-y^2}$$
Which should be obvious because :
$$\sin^2x+\cos^2 x=1$$
