anonymous
  • anonymous
Show that the point on a unit circle associated to the angle t=pie/3 is (1/2, the square root of 3/2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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jdoe0001
  • jdoe0001
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 $$ c^2 = a^2+\color{red}{b}^2 \implies \color{red}{b}^2 = \sqrt{c^2-a^2} $$
jdoe0001
  • jdoe0001
\(\large \cfrac{\pi}{2} = 60^o\)
jdoe0001
  • jdoe0001
darn I meant \(\large \cfrac{\pi}{3} = 60^o\)

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anonymous
  • anonymous
pie/2 is 90 degrees i thought our teacher asked up to prove this with the distance formula.
jdoe0001
  • jdoe0001
my bad typo :/
anonymous
  • anonymous
its all good!
anonymous
  • anonymous
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.. :/
jdoe0001
  • jdoe0001
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
anonymous
  • anonymous
\[\sqrt{(x-0)^{2}+(y-1)^{2}}=\sqrt{(x-\sqrt{3}/2)^{2}+(y-1/2)^{2}}\]
jdoe0001
  • jdoe0001
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
jdoe0001
  • jdoe0001
$$ d = \sqrt{(x_2-x_1)+(y_\color{blue}{2}-y_1)}\\ d^2 = (x_2-x_1)+(y_\color{blue}{2}-y_1)\\ \text{now using the origin and 1/2 for }x_2\\ \text{keeping in mind the distance between then is "1"}\\ 1^2 = \cfrac{1}{4}+y_\color{blue}{2}^2 \implies 1 -\cfrac{1}{4} = y_2^2\\ y_\color{blue}{2}^2=\sqrt{\cfrac{3}{4}} \implies y_\color{blue}{2}^2=\cfrac{\sqrt{3}}{\sqrt{4}} = \cfrac{\sqrt{3}}{2} $$ that is, using a 2nd point of \(\pmatrix{\frac{1}{2}, y_2}\)
jdoe0001
  • jdoe0001
hmmm, yet another typo :/
jdoe0001
  • jdoe0001
$$ d = \sqrt{(x_2-x_1)+(y_\color{blue}{2}-y_1)}\\ d^2 = (x_2-x_1)+(y_\color{blue}{2}-y_1)\\ \text{now using the origin and 1/2 for }x_2\\ \text{keeping in mind the distance between then is "1"}\\ 1^2 = \cfrac{1}{4}+y_\color{blue}{2}^2 \implies 1 -\cfrac{1}{4} = y_2^2\\ y_\color{blue}{2}=\sqrt{\cfrac{3}{4}} \implies y_\color{blue}{2}=\cfrac{\sqrt{3}}{\sqrt{4}} = \cfrac{\sqrt{3}}{2} $$
jdoe0001
  • jdoe0001
there

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