Why can't I obtain values of Sine/Cosine with $f(x)=\sqrt{1-x^2}?$ Taking into account the circle equation $ x^2 + y^2 = 1$, I've made the following function on Mathematica : $ f (x) = \sqrt {1 - x^2}$ which yields this plot with the domain $ 0\leq x\leq 1$ : 

I' ve made an animated gif of this : 

Which seems pretty reasonable to yield the values for sine and cosine.But then $ f (1/2) = \frac {\sqrt {3}} {2}$ which should yield the Sin/Cos of 45 degrees, but it gets different of the value given by Mathematica when using the Cos and Sin functions ($\sin 45 = \cos 45 = \frac {1} {\sqrt {2}}$), So what's hapenning?

Notes:


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*I've read somewhere that the problem is that Sine is supposed to take an angle as an input and that my function takes a length as input. When I wrote Sine/Cosine I actually tried to express the points $(x,y)$ on the circle, I thought they were the same because of this - so if length is different of angle, perhaps one could find those points with with the function I provided, but it would have a different meaning. 

*I'm aware of the existence of trigonometric functions in Calculus, but at the current moment, I know nothing about them.

*The function seems to be not that similar to sine and cosine functions (as one can see here for example) but I feel it could be adapted to such task. 
 A: $f(1/2)=\frac{\sqrt{3}}{2}$ means in terms of sine and cosine:
$$f(\cos(\phi))=\sqrt{1-\cos(\phi)^2}=|\sin(\phi)|$$
where $\phi=\arccos(1/2)=\pi/3=60^°$, not $45^°$. Indeed, $\sin(\phi)=\sin(\pi/3)=\frac{\sqrt{3}}{2}$.
In general, the equation $f(x)=y$ expresses that the point $(x,y)$ lies on the upper unit semicircle at the angle $\phi=\arccos(x)=\arcsin(y)$.
A: Your function plots a semi-circle above the $x$ axis. The $x$ value you specify is $\cos(\theta)$ for some value of $\theta$, and the resulting value $f(x)$ is $\sin(\theta)$ for that same value of $\theta$. You should find that for a given value of $x\in[-1,1]$ the corresponding value of $\theta$ is $$\theta = \tan^{-1}\left(\frac{f(x)}{x}\right).$$
For example, $$\frac{f(x)}{x} = \frac{\sqrt{1-x^2}}{x}=\frac{\sqrt{1-\cos^2\theta}}{\cos\theta} = \frac{\sin\theta}{\cos\theta}=\tan\theta,$$ as required.
A: You've got $x=\cos\theta$ and $\sqrt{1-x^2}=\sin\theta$.  When $x=1/2$, then $\cos\theta=1/2$, meaning you've got a $60^\circ$ angle. The sine of $60^\circ$ is $\sqrt{3}/2$.
