Question

Show that the point on a unit circle associated to the angle t=pie/3 is (1/2, the square root of 3/2)

Answer 1

http://upload.wikimedia.org/wikipedia/commons/4/45/30-60-90_triangle.jpg

by 30-30-60 rule, the "hypotenuse" is twice the length of the "adjacent" side

the rules also points out that the "opposite" side would be "adjacent times root 3"

but once you have the adjacent and the hypotenuse, you can just use the pythagorean theorem to get the opposite side

$$
c^2 = a^2+\color{red}{b}^2 \implies \color{red}{b}^2 = \sqrt{c^2-a^2}
$$

Answer 2

\(\large \cfrac{\pi}{2} = 60^o\)

Answer 3

darn I meant \(\large \cfrac{\pi}{3} = 60^o\)

Answer 4

pie/2 is 90 degrees i thought
our teacher asked up to prove this with the distance formula.

Answer 5

my bad typo :/

Answer 6

its all good!

Answer 7

would you know how to make it work with the distance formula...? I have figure it out some the part that i an struggling with would be the (x-(the square root of 3/2) square part.. :/

Answer 8

the distance formula is a "renamed" version of the pythagorean theorem, they both do exactly the same, different name :), so use the distance formula, just don't call it pythagorean

Answer 9

\[\sqrt{(x-0)^{2}+(y-1)^{2}}=\sqrt{(x-\sqrt{3}/2)^{2}+(y-1/2)^{2}}\]

Answer 10

well, for the distance formula you'd just need to have to points to work on, for \(\frac{\pi}{3} \) you'd use the origin as one (0,0) and some other point, the endpoint for the \(\frac{\pi}{3}\) angle

Answer 11

$$
d = \sqrt{(x_2-x_1)+(y_\color{blue}{2}-y_1)}\\
d^2 = (x_2-x_1)+(y_\color{blue}{2}-y_1)\\
\text{now using the origin and 1/2 for }x_2\\
\text{keeping in mind the distance between then is "1"}\\
1^2 = \cfrac{1}{4}+y_\color{blue}{2}^2 \implies
1 -\cfrac{1}{4} = y_2^2\\
y_\color{blue}{2}^2=\sqrt{\cfrac{3}{4}} \implies
y_\color{blue}{2}^2=\cfrac{\sqrt{3}}{\sqrt{4}} = \cfrac{\sqrt{3}}{2}
$$

that is, using a 2nd point of \(\pmatrix{\frac{1}{2}, y_2}\)

Answer 12

hmmm, yet another typo :/

Answer 13

$$
d = \sqrt{(x_2-x_1)+(y_\color{blue}{2}-y_1)}\\
d^2 = (x_2-x_1)+(y_\color{blue}{2}-y_1)\\
\text{now using the origin and 1/2 for }x_2\\
\text{keeping in mind the distance between then is "1"}\\
1^2 = \cfrac{1}{4}+y_\color{blue}{2}^2 \implies
1 -\cfrac{1}{4} = y_2^2\\
y_\color{blue}{2}=\sqrt{\cfrac{3}{4}} \implies
y_\color{blue}{2}=\cfrac{\sqrt{3}}{\sqrt{4}} = \cfrac{\sqrt{3}}{2}
$$

Answer 14

there