Trig Equation of a Circle

Question

I was trying out derivatives of inverse functions(I'm new to that) which led me to this. I graphed the equation $y=\sin\left(\cos^{-1}(x)\right)$. This is a half circle and squaring both sides gives a full unit circle. Except for $y=\sin\left(\cos^{-1}(x)\right)$ and $x^2+y^2=1$, are there any other interesting forms of the equation for a circle with radius 1 and the center at the origin. Also, do these two forms relate to each other in any way?

Answer 1

Draw a right triangle. One side (adjacent) is $x$, the hypotenuse is $1$, and, by Pythagorean Theorem, the remaining side (opposite) is $\sqrt{1-x^2}$.

Therefore, $y = \sin(\cos^{-1}x) = \frac{\sqrt{1-x^2}}1=\sqrt{1-x^2}$.

Squaring both sides, the result is $y^2 = 1-x^2$ or $y^2+x^2=1$.

Edit: Here is a picture:

Answer 2

It's easy to see that if $a = \cos^{-1}(x)$, then $\sin a = \sqrt{1-x^2}$, so you are looking at $y = \sqrt{1-x^2}$, which is indeed the upper semicircle.

For the other forms, you can picture any of the basic semicircles (left, upper, right, lower) in a similar way, e.g. the left semicircle is the relation $x = -\sqrt{1-y^2}$...

Answer 3

The simplest and simultaniously the deepest equation for the unit circle is $$r=1$$ in polar coordinates.

Answer 4

In context: Thales circle.

$x,y$ axes, $A(-1,0)$, $B(0,1)$.

The locus of points $P(x,y)$ s.t. $\angle APB=90°$.

1)Line $AP$: $y =m(x-(-1))$

2) Line PB : $y =-(1/m)(x-1)$.

Multiply: 1)$\cdot$ 2) to eliminate $m$.

$y^2=-(x+1)(x-1)=-x^2+1$;

$y^2+x^2=1$.