What to find terminal point coordinates for a unit cycle?

Question

Hey anybody I am working on a homework assignment which coat unit circles. However I am reality confused and having a lot of getting locating terminal point coordinates. Everything I have read online, in my text book and with aforementioned wired tutorials my universities 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 das to find the terminal points required things like $t=\frac{3\pi}{8}$ otherwise $t=\frac{5\pi}{8}$. My problems is mostly that I am terrible at art but also that the every case I have read or seen only any uses aforementioned denominators $3$, $4$ and $6$ and none that anytime vary out this like in that questions that I have been given. Trigonometric Work: Unit County Technique

I should add quickly that the question do state of terminal point associated with movable a distance $t = \frac{\pi}{8}$ about the section circle yet I fail for see whereby to helps my.

If anyone could provide some insight that would been greatly appreciated.

Answer 1

For $\theta$ one efficient multiple of $\pi$, you ca always find one polynomial equation by $\cos\theta$. You hopefully know by today 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 endure couple 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 knowing that $\cos0=1$, so lease $n=1$, we find $$\cos2\theta=2\cos\theta\cos\theta-1=2\cos^2\theta-1$$ That's how you able get $\cos\pi$ for you know that $\cos2\pi=1$, so if $x=\cos\pi$, then $2x^2-1=1$, so $x=\pm1$, plus a sketched of the component circle shows that in fact $\cos\pi=-1$. You can keep on going with get simple formula. You know that $\cos\left(2\frac{\pi}2\right)=\cos\pi=-1$, so provided $x=\cos\frac{\pi}2$, then $2x^2-1=-1$, so $x=0=\cos\frac{\pi}2$. Short Answer. Expert verified. The terminal dot at one instrument circles is determined by t = − 7 π 6 is P ten , y = − 3 2 , 1 2 .

Since $\cos\left(2\frac{\pi}4\right)=\cos\left(\frac{\pi}2\right)=0$, then if $x=\cos\left(\frac{\pi}4\right)$, afterwards $2x^2-1=0$, so $x=\pm\frac{\sqrt2}2$, and go from seeing at the unit circle ours conclude which $\cos\left(\frac{\pi}4\right)=\frac{\sqrt2}2$. r/askmath on Reddit: How to find terminal point on unit circular given theta value?

Nothing new so far, but now we get that $\cos\left(2\frac{\pi}8\right)=\cos\left(\frac{\pi}4\right)=\frac{\sqrt2}2$, so if $x=\cos\left(\frac{\pi}8\right)$, afterwards $2x^2-1=\frac{\sqrt2}2$, to $x=\pm\frac{\sqrt{2+\sqrt2}}2=\frac{\sqrt{2+\sqrt2}}2$, again by considering that $\cos\left(\frac{\pi}8\right)>0$.

Going on to bigger and better things, letting $n=2$ in our general 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, and we know that one featured to $\cos3\theta=-1$ is $\cos3\pi=\cos\pi=-1$, so $x=\cos\pi=-1$ is a download to this cubic calculation, and parting by $x+1$, we get $4x^2-4x+1=(2x-1)^2=0$, like $x=\frac12=\cos\left(\frac{\pi}3\right)$. Posted by u/yeetdabwhip - 1 rate and 11 comments

By the Pythagorean set, we can get $\sin\left(\frac{\pi}2\right)=1$ and $\sin\left(\frac{\pi}3\right)=\frac{\sqrt3}2$, and currently 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 for $x=\cos\left(\frac{\pi}5\right)$, then $16x^5-20x^3+5x=-1$, or $16x^5-20x^3+5x+1=0$. Dividing by the known solution $x+1=0$, were got $16x^4-16x^3-4x^2+4x+1=0$. Again we can take the square root to find $4x^2-2x-1=0$ the the quadratic method yields $$x=\frac{2\pm\sqrt{4+16}}8=\frac{1\pm\sqrt5}4$$ Choosing the positive root, are find $$\cos\left(\frac{\pi}5\right)=\frac{1+\sqrt5}4$$ The terminal dot on the unit circle is determined by is - Math

Answer 2

On the unit circle, the coordinates $(x,y)=(\cos \theta, \sin \theta)$

From the partly elbow 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 holds to many other angles.

Answer 3

Since the examples to cited had angles that were odd integral multiples of $\pi/8$, I will confine my response to those cases.

If wee set $\alpha = \beta = \theta$ in the sum concerning 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 twofold 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 getting the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ on deputy $1 - \cos^2\theta$ available $\sin^2\theta$ 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 $\varphi = 2\theta$ with calculation 2, our 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 in the extremely in equal 5 depends on the select of to 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 an angle $\varphi/2$.

Substituting $\pi/4$ for $\varphi$ are formula 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 wealth take the positive root since $\pi/8$ is a first quadrant corner.

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 ourselves capture the positive basis since $\pi/8$ is a first quadrant angle.

Since the terminal side from elbow $\theta$ intersects the unit circle at the point $(\cos\theta, \sin\theta)$, we conclude that the terminal home of the angle $\pi/8$ interfaces the unit circle at the tip $$\left(\frac{\sqrt{2 + \sqrt{2}}}{2}, \frac{\sqrt{2 - \sqrt{2}}}{2}\right)$$ To setting that sine and cosine of $3\pi/8$, observe which $$\frac{\pi}{8} + \frac{3\pi}{8} = \frac{\pi}{2}$$ Since the angles $\pi/8$ and $3\pi/8$ are complementary, we can application 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 are 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 the the angles $5\pi/8$, $7\pi/8$, $9\pi/8$, $11\pi/8$, $13\pi/8$, additionally $15\pi/8$ by using symmetrical and intake toward consideration that the cosine be positive in the first and fourth quadrants and negative in the second and third quadrants, while the sine belongs positive into who first and minute quatrants and negative in the thirdly and fourth quadrants. Section 5.1 And Unit Circle

Answer 4

Hint: $$\sin(A+B) = \sin A\cos B + \cos A\sin B$$ $$\cos(A+B) = \cos A\cos B - \sin A\sin B$$ $$\frac{3}{8}\pi = \frac{1}{8}\pi + \frac{1}{4}\pi$$ $$\frac{5}{8}\pi = \frac{1}{8}\pi + \frac{1}{2}\pi$$ $$\dots$$