How do you find the coordinate points for $\pi/8$, then $3\pi/8$, $5\pi/8$ and $-7\pi/8$?

Question

I am really struggling with this unit circle problem.

How do you find the coordinate points for $\pi/8$ then $3\pi/8$, $5\pi/8$ and $-7\pi/8$?

I am familiar with other terminal points up to $\pi/6$, and I guess $\pi/8$ must be below $\pi/6$ in the same quadrant, but how do you find the coordinates?

Are there any steps or formula involved?

In the lecture all we asked to do is to try to memorize the coordinates with the common terminal points and it didn't include $\pi/8$. I have gone through the entire lecture notes but still don't know how to find it.

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I have attached a screenshot of the problem. Thank you.

Answer 1

$\frac{\pi}{8}$ is $\frac{1}{2}$ of $\frac{\pi}{4}$ and you should know $cos(\pi/4)= sin(\pi/4)= \frac{\sqrt{2}}{2}$. So use the "half angle" formulas!

$cos(\theta/2)= \pm\sqrt{\frac{1+ cos(\theta)}{2}}$ and $sin(\theta/2)= \pm\sqrt{\frac{1- cos(\theta)}{2}}$.

$\pi/8$ is in the first quadrant so both sine and cosine are positive- we use the positive sign.

$cos(\pi/8)=$$ \sqrt{\frac{1+ \frac{\sqrt{2}}{2}}{2}}= $$\sqrt{\frac{2+ \sqrt{2}}{4}}= \frac{\sqrt{2+ \sqrt{2}}}{2}$

$sin(\pi/8)=$$ \sqrt{\frac{1- \frac{\sqrt{2}}{2}}{2}}=$$ \sqrt{\frac{2- \sqrt{2}}{4}}= \frac{\sqrt{2- \sqrt{2}}}{2}$

Answer 2

Lets divide the arc formed in the first quadrant of a unit circle into two at a point (P) $(x,y)$ on a unit circle. We will have two arcs:
Let (AP) be arc formed between (y) axis and point (P),where point (A)'s coordinate is (0,1) and the arc length will be $\frac{\pi}{4}$.
Let (PB) be arc formed between point (P),where point (B)'s coordinate is (1,0), and (x) axis. The arc length will be $\frac{\pi}{4}$.
From this it follows that arc (AP) is equal to arc (PB) => line(AP) is equal to (PB).
Using the distance formula, we have:
=>$\sqrt{(y-1)^2 + (x-0)^2} = \sqrt{(y-0)^2 + (x-1)^2}$
=>$y^2 - 2y + 1 + x^2 = y^2 + x^2 - 2x + 1$
=>$y=x$
substitute the above equation into unit circle equation:$ y^2 + x^2 = 1$
=>$2y^2 = 1$ => $y=\pm\frac{1}{\sqrt{2}}$ = $\pm\frac{\sqrt{2}}{2}$
=>$x=\pm\frac{\sqrt{2}}{2}$
since we are in the first quadrant, $(x,y) = (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$.

Now we will split arc(PB) in half at point (Z). This will give us arcs, (PZ) and (ZB). They both have length$\frac{\pi}{8}$ => line (PZ) is equal to (ZB).
Again using the distance formula, we have:
=>$\sqrt{(y - \frac{\sqrt{2}}{2})^2 + (x - \frac{\sqrt{2}}{2})^2} = \sqrt{(y - 0)^2 + (x - 1)^2}$
=>$y^2 - \frac{2\sqrt{2}y}{2} + \frac{2}{4} + x^2 - \frac{2\sqrt{2}x}{2} + \frac{2}{4} = y^2 + x^2 - 2x + 1$
=>$y^2 + x^2 - \frac{2\sqrt{2}y}{2} - \frac{2\sqrt{2}x}{2} + \frac{2}{4} + \frac{2}{4} = y^2 + x^2 - 2x + 1$
=>$- \sqrt{2}y - \sqrt{2}x =- 2x $
=>$\sqrt{2}y = 2x - \sqrt{2}x $
=>$\sqrt{2}y = x(2 - \sqrt{2})$
=>$y = \pm \frac{(2 - \sqrt{2})x}{\sqrt{2}}$
Substituting $y$ into unit circle equation: $x^2 + (\frac{(2 - \sqrt{2})x}{\sqrt{2}})^2 = 1$
=>$x^2 + x^2(\frac{(2 - \sqrt{2})^2}{2}) = 1$
=>$2x^2 + x^2(2 - \sqrt{2})^2 = 2$
=>$x^2 = \frac{2}{ 2 + (2 - \sqrt{2})^2} = \frac{2}{ 2 + 4 - 4\sqrt{2}+2} = \frac{2}{ 8 - 4\sqrt{2}} = \frac{1}{ 4 - 2\sqrt{2}} = (\frac{1}{ 4 - 2\sqrt{2}})(\frac{4 + 2\sqrt{2}}{ 4 + 2\sqrt{2}}) = (\frac{4 + 2\sqrt{2}}{8}) = (\frac{2 + \sqrt{2}}{4})$
=>$x = \pm {\frac{\sqrt{2 + \sqrt{2}}}{2}}$
=>$x ={\frac{\sqrt{2 + \sqrt{2}}}{2}}$ since we are looking at the first quadrant.

Substitute value of (x) into the unit circle to find the value of (y). The remaining arcs can be found by finding the reference number with respect to $\frac{\pi}{8}$.