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.
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$\begingroup$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 $$\cos\left(\frac{\pi}5\right)=\frac{1+\sqrt5}4$$
$\endgroup$ 1 $\begingroup$On the unit circle, the coordinates $(x,y)=(\cos \theta, \sin \theta)$
From the half angle formula for sine and cosine, found here,
$$\cos \frac{a}{2}=\sqrt{\frac{1+\cos a}{2}}$$
$$\sin \frac{a}{2}=\sqrt{\frac{1-\cos a}{2}}$$
Now let $a=\frac{\pi}{4}$ to find $\sin \frac{\pi}{8}$ and $\cos \frac{\pi}{8}$. This same strategy applies to many other angles.
$\endgroup$ 1 $\begingroup$Since the examples you cited had angles that were odd integral multiples of $\pi/8$, I will confine my answer to those cases.
If we set $\alpha = \beta = \theta$ in the sum of angle identities \begin{align*} \cos(\alpha + \beta) & = \cos\alpha\cos\beta - \sin\alpha\sin\beta\\ \sin(\alpha + \beta) & = \sin\alpha\cos\beta + \cos\alpha\sin\beta \end{align*} we obtain the double angle identities \begin{align*} \cos(2\theta) & = \cos^2\theta - \sin^2\theta \tag{1}\\ \sin(2\theta) & = 2\sin\theta\cos\theta \end{align*} We can use the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ to substitute $1 - \cos^2\theta$ for $\sin^2\theta$ in equation 1, which yields \begin{align*} \cos(2\theta) & = \cos^2\theta - (1 - \cos^2\theta)\\ & = 2\cos^2\theta - 1 \tag{2} \end{align*} Similarly, we can substitute $1 - \sin^2\theta$ for $\cos^2\theta$ in equation 1, which yields \begin{align*} \cos(2\theta) & = 1 - \sin^2\theta - \sin^2\theta\\ & = 1 - 2\sin^2\theta \tag{3} \end{align*} If we set $\varphi = 2\theta$ in equation 2, we obtain $$\cos\varphi = 2\cos^2\left(\frac{\varphi}{2}\right) - 1 \tag{4}$$ Solving equation 4 for $\cos(\varphi/2)$ yields \begin{align*} 1 + \cos\varphi & = 2\cos^2\left(\frac{\varphi}{2}\right)\\ \frac{1 + \cos\varphi}{2} & = \cos^2\left(\frac{\varphi}{2}\right)\\ \pm \sqrt{\frac{1 + \cos\varphi}{2}} & = \cos\left(\frac{\varphi}{2}\right) \tag{5} \end{align*} where the sign of the radical in equation 5 depends on the quadrant of the angle $\varphi/2$.
If we set $\varphi = 2\theta$ in equation 3, we obtain $$\cos\varphi = 1 - 2\sin^2\left(\frac{\varphi}{2}\right) \tag{6}$$ If we solve equation 6 for $\sin(\varphi/2)$, we obtain \begin{align*} 2\sin^2\left(\frac{\varphi}{2}\right) & = 1 - \cos\varphi\\ \sin^2\left(\frac{\varphi}{2}\right) & = \frac{1 - \cos\varphi}{2}\\ \sin\left(\frac{\varphi}{2}\right) & = \pm\sqrt{\frac{1 - \cos\varphi}{2}} \tag{8} \end{align*} where the sign of the radical in equation 8 depends on the quadrant of the angle $\varphi/2$.
Substituting $\pi/4$ for $\varphi$ in equation 6 yields
\begin{align*}
\cos\left(\frac{\pi}{8}\right) & = \sqrt{\frac{1 + \cos\left(\frac{\pi}{4}\right)}{2}}\\
& = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}}\\
& = \sqrt{\frac{2 + \sqrt{2}}{4}}\\
& = \frac{\sqrt{2 + \sqrt{2}}}{2}
\end{align*}
where we take the positive root since $\pi/8$ is a first quadrant angle.
Substituting $\pi/4$ for $\varphi$ in equation 8 yields
\begin{align*}
\sin\left(\frac{\pi}{8}\right) & = \sqrt{\frac{1 - \cos\left(\frac{\pi}{4}\right)}{2}}\\
& = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}}\\
& = \sqrt{\frac{2 - \sqrt{2}}{4}}\\
& = \frac{\sqrt{2 - \sqrt{2}}}{2}
\end{align*}
where we take the positive root since $\pi/8$ is a first quadrant angle.
Since the terminal side of angle $\theta$ intersects the unit circle at the point $(\cos\theta, \sin\theta)$, we conclude that the terminal side of the angle $\pi/8$ intersects the unit circle at the point $$\left(\frac{\sqrt{2 + \sqrt{2}}}{2}, \frac{\sqrt{2 - \sqrt{2}}}{2}\right)$$ To determine the sine and cosine of $3\pi/8$, observe that $$\frac{\pi}{8} + \frac{3\pi}{8} = \frac{\pi}{2}$$ Since the angles $\pi/8$ and $3\pi/8$ are complementary, we can use the complementary angle identities \begin{align*} \cos\theta & = \sin\left(\frac{\pi}{2} - \theta\right)\\ \sin\theta & = \cos\left(\frac{\pi}{2} - \theta\right) \end{align*} from which we obtain \begin{align*} \cos\left(\frac{3\pi}{8}\right) & = \sin\left(\frac{\pi}{2} - \frac{3\pi}{8}\right)\\ & = \sin\left(\frac{\pi}{8}\right)\\ & = \frac{\sqrt{2 - \sqrt{2}}}{2} \end{align*} and \begin{align*} \sin\left(\frac{3\pi}{8}\right) & = \cos\left(\frac{\pi}{2} - \frac{3\pi}{8}\right)\\ & = \cos\left(\frac{\pi}{8}\right)\\ & = \frac{\sqrt{2 + \sqrt{2}}}{2} \end{align*} Finally, we can determine the cosine and sine of the angles $5\pi/8$, $7\pi/8$, $9\pi/8$, $11\pi/8$, $13\pi/8$, and $15\pi/8$ by using symmetry and taking into consideration that the cosine is positive in the first and fourth quadrants and negative in the second and third quadrants, while the sine is positive in the first and second quadrants and negative in the third and fourth quadrants.
Hint: $$\sin(A+B) = \sin A\cos B + \cos A\sin B$$ $$\cos(A+B) = \cos A\cos B - \sin A\sin B$$ $$\frac{3}{8}\pi = \frac{1}{8}\pi + \frac{1}{4}\pi$$ $$\frac{5}{8}\pi = \frac{1}{8}\pi + \frac{1}{2}\pi$$ $$\dots$$
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