# 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:

• 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.

• You've got $x=\cos\theta$ and $\sqrt{1-x^2}=\sin\theta$. Your function takes $\cos\theta$ rather than $\theta$ as its input. See my answer below. – Michael Hardy Feb 8 '14 at 18:23

$f(1/2)=\frac{\sqrt{3}}{2}$ 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 $\phi=\arccos(x)=\arcsin(y)$.

• @PristineKavalostka Yes this function generates points on the circle: $f(x)$ gives you the correct value for $y$ such that $(x,y)$ lies on the (upper half) of the unit circle. – J.R. Feb 8 '14 at 16:01
• @PristineKavalostka Your function doesn't take "a length" as input, it takes the $x$-coordinate of the corresponding point on the unit circle as an input. – J.R. Feb 8 '14 at 16:04
• @PristineKavalostka It does give the value of sine and cosine. But the function does not give the angle which you have to plug into sine and cosine to get this point. To recover the angle you can use any reverse trig function. – J.R. Feb 8 '14 at 16:12
• @PristineKavalostka You know if you have some $x$ coordinate. Then you can calculate $y=f(x)$ and you then know that $(x,y)$ is on the unit circle. On the other hand, also every point on the unit circle is of the form $(\cos(\phi),\sin(\phi))$ for some angle $\phi$... that means $x=\cos(\phi)$, $f(x)=y=\sin(\phi)$, so $f(x)$ is the value of sine at the angle of the point $(x,f(x))$. – J.R. Feb 8 '14 at 16:13
• Now it makes sense: $\sin \theta= \frac{Opp(\theta)}{Hyp(\theta)}$, the hypotenuse in this case was always 1. Then I was doing $\frac{f(x)}{1}=f(x)$. The idea of having to use the reverse trig function to get the angle is a revolution to me. Thanks. You taught me more trigonometry with the answer and comments than my school in 3 years. – Billy Rubina Feb 10 '14 at 4:12

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.

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 $\sqrt{3}/2$.