How to find $(x,y)$ coordinates for $\pi/8$ of a unit circle by using the distance formula?

Question

I am having difficulty finding the terminal point for $\pi/8$ of a unit circle using the distance formula where we are asked to find the $x$- and $y$-coordinates given moving a distance $t=\pi/4$ units from the point $(1,0)$ anti-clockwise around the unit circle, $x^2 + y^2 = 1$, has coordinates $(x,y) = (\sqrt{2}/2, \sqrt{2}/2)$. So, how do you use the distance formula to find the coordinates for $\pi/8$? How do you substitute the square root integers into the equation and solved? Thank you.

Answer 1

If you don’t know bisection formula for sines you can try in this way:

-You find the middle point between $(\sqrt 2 /2,\sqrt 2 /2)$ and $(1,0)$, call it $M$;

-You write the equation of the straight line from the origin to $M$, call it $r$, and find his intersection with unit circle, call it $P$;

-Angle between $r$ and x-axis is $\pi /8$, infact $OM$ is median but also bisector, so $P$ is what you are searching for.

Answer 2

Here is what I suspect is the intended way to solve this:

The point at $\pi/8$ is halfway between $0$ and $\pi/4$. So, if its coordinates are $(x, y)$, then we have $$ d((x, y), (1,0)) = d((x, y), (\sqrt2/2, \sqrt2/2)) $$ which is to say $$ \sqrt{(x-1)^2 + (y-0)^2} = \sqrt{(x-\sqrt2/2)^2 + (y-\sqrt2/2)^2} $$ Also, it's on the unit circle, so we get $x^2 + y^2 = 1$.

Together, you now have two equations in two unknowns to solve.

Answer 3

If you want to use the distance formula, write two different equations: the first is the point you want to find (x,y) measured from the known point for 45 degrees (where x and y are $\frac{\sqrt 2}{2}$) and the second is from the point (1,0) to (x,y).

Since those two distances are equal, you can set those two equations equal to each other and solve for either x or y.

Then plug the result into your other known equation: $x^2 + y^2 = 1$ to find the other variable.