Hey everyone I am working on a homework assignment which covers unit circles. However I am really confused and having a lot of trouble locating terminal point coordinates. Everything I have read online, in my text book and in the online tutorials my university provides seems to only cover coordinates when $t=\frac{\pi}{3}$,$\frac{3\pi}{3}$, $\frac{\pi}{4}$, $\frac{5\pi}{4}$ or $\frac{\pi}{6}$ etc.
However all my questions are asking me to find the terminal points for things like $t=\frac{3\pi}{8}$ or $t=\frac{5\pi}{8}$. My problems is mostly that I am terrible at math but also that the every example I have read or seen only ever uses the denominators $3$, $4$ and $6$ and none that ever vary from this like in the questions that I have been given.
I should add quickly that the question does state the terminal point associated with moving a distance $t = \frac{\pi}{8}$ around the unit circle but I fail to see how this helps me.
If anyone could provide some insight that would be greatly appreciated.
Best Answer
For $\theta$ a rational multiple of $\pi$, you can always find a polynomial equation for $\cos\theta$. You hopefully know by now that $$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$$ Then $$\cos(n+1)\theta=\cos(n\theta+\theta)=\cos n\theta\cos\theta-\sin n\theta\sin\theta$$ and $$\cos(n-1)\theta=\cos(n\theta-\theta)=\cos n\theta\cos\theta+\sin n\theta\sin\theta$$ Adding these last two identities, we have $$\cos(n+1)\theta+\cos(n-1)\theta=2\cos n\theta\cos\theta$$ Thus finally $$\cos(n+1)\theta=2\cos n\theta\cos\theta-\cos(n-1)\theta$$ We know that $\cos0=1$, so letting $n=1$, we find $$\cos2\theta=2\cos\theta\cos\theta-1=2\cos^2\theta-1$$ That's how you can get $\cos\pi$ because you know that $\cos2\pi=1$, so if $x=\cos\pi$, then $2x^2-1=1$, so $x=\pm1$, and a sketch of the unit circle shows that in fact $\cos\pi=-1$. You can keep on going with this simple formula. You know that $\cos\left(2\frac{\pi}2\right)=\cos\pi=-1$, so if $x=\cos\frac{\pi}2$, then $2x^2-1=-1$, so $x=0=\cos\frac{\pi}2$.
Since $\cos\left(2\frac{\pi}4\right)=\cos\left(\frac{\pi}2\right)=0$, then if $x=\cos\left(\frac{\pi}4\right)$, then $2x^2-1=0$, so $x=\pm\frac{\sqrt2}2$, and again from looking at the unit circle we conclude that $\cos\left(\frac{\pi}4\right)=\frac{\sqrt2}2$.
Nothing new so far, but now we know that $\cos\left(2\frac{\pi}8\right)=\cos\left(\frac{\pi}4\right)=\frac{\sqrt2}2$, so if $x=\cos\left(\frac{\pi}8\right)$, then $2x^2-1=\frac{\sqrt2}2$, so $x=\pm\frac{\sqrt{2+\sqrt2}}2=\frac{\sqrt{2+\sqrt2}}2$, again by considering that $\cos\left(\frac{\pi}8\right)>0$.
Going on to bigger and better things, letting $n=2$ in our equation for $\cos(n+1)\theta$, we get $$\cos3\theta=2(2\cos^2\theta-1)-\cos\theta=4\cos^3\theta-3\cos\theta$$ Since $\cos\left(3\frac{\pi}3\right)=\cos\pi=-1$, if $x=\cos\left(\frac{\pi}3\right)$, then $4x^3-3x=-1$ or $4x^3-3x+1=0$. Now, this looks intimidating, but we know that one solution to $\cos3\theta=-1$ is $\cos3\pi=\cos\pi=-1$, so $x=\cos\pi=-1$ is a solution to this cubic equation, and dividing by $x+1$, we get $4x^2-4x+1=(2x-1)^2=0$, so $x=\frac12=\cos\left(\frac{\pi}3\right)$.
By the Pythagorean theorem, we can get $\sin\left(\frac{\pi}2\right)=1$ and $\sin\left(\frac{\pi}3\right)=\frac{\sqrt3}2$, and now we can do things like $\frac{\pi}2-\frac{\pi}3=\frac{\pi}6$, so $$\cos\left(\frac{\pi}6\right)=\cos\left(\frac{\pi}2-\frac{\pi}3\right)=\cos\left(\frac{\pi}2\right)\cos\left(\frac{\pi}3\right)+\sin\left(\frac{\pi}2\right)\sin\left(\frac{\pi}3\right)=(0)(\frac12)+(1)(\frac{\sqrt3}2)=\frac{\sqrt3}2$$ One last bit of fun. $$\cos4\theta=2(4\cos^3\theta-3\cos\theta)\cos\theta-(2\cos^2\theta-1)=8\cos^4\theta-8\cos^2\theta+1$$ $$\cos5\theta=2(8\cos^4\theta-8\cos^2\theta+1)\cos\theta-(4\cos^3\theta-3\cos\theta)=16\cos^5\theta-20\cos^3\theta+5\cos\theta$$ So $\cos\left(5\frac{\pi}5\right)=\cos\pi=-1$, so if $x=\cos\left(\frac{\pi}5\right)$, then $16x^5-20x^3+5x=-1$, or $16x^5-20x^3+5x+1=0$. Dividing by the known solution $x+1=0$, we get $16x^4-16x^3-4x^2+4x+1=0$. Again we can take the square root to find $4x^2-2x-1=0$ and the quadratic formula yields $$x=\frac{2\pm\sqrt{4+16}}8=\frac{1\pm\sqrt5}4$$ Choosing the positive root, we find $$\cos\left(\frac{\pi}5\right)=\frac{1+\sqrt5}4$$