anonymous
  • anonymous
medal award please help!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
hint for the first part \[\Large \frac{13\pi}{12} = \frac{\pi+12\pi}{12}\] \[\Large \frac{13\pi}{12} = \frac{\pi}{12} + \frac{12\pi}{12}\] \[\Large \frac{13\pi}{12} = \frac{\pi}{12} + \pi\]
anonymous
  • anonymous
I'm not sure what it is that they want.. they are asking for the coordinates i'm a little lost :/

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amistre64
  • amistre64
for a unit circle ... x = cos(t) y = sin(t)
amistre64
  • amistre64
but working them into their radical values might be a bit daunting :/
jim_thompson5910
  • jim_thompson5910
|dw:1444012468609:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1444012481543:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1444012492630:dw|
jim_thompson5910
  • jim_thompson5910
add pi to pi/12 to get 13pi/12 this is the same as doing a 180 degree rotation |dw:1444012532760:dw|
jim_thompson5910
  • jim_thompson5910
compare the two points marked on the unit circle the x coordinates are the same in magnitude, but they differ in sign (the first is positive, the second is negative) same for y coordinates
anonymous
  • anonymous
that makes sense,.. so when they mention the terminal point p(x,y) how would I determine that?
jim_thompson5910
  • jim_thompson5910
it's essentially point P but the signs are different for each coordinate
jim_thompson5910
  • jim_thompson5910
|dw:1444012707085:dw|
jim_thompson5910
  • jim_thompson5910
\[\Large P = \left(\frac{\sqrt{2+\sqrt{3}}}{2}, \frac{\sqrt{2-\sqrt{3}}}{2}\right)\]
jim_thompson5910
  • jim_thompson5910
\[\Large Q = \left(\color{red}{-}\frac{\sqrt{2+\sqrt{3}}}{2}, \color{red}{-}\frac{\sqrt{2-\sqrt{3}}}{2}\right)\]
anonymous
  • anonymous
so for 13pi/12 the terminal point would be -sqrt2+sqrt3/2?
jim_thompson5910
  • jim_thompson5910
it would be point Q I wrote above
anonymous
  • anonymous
i'm a little confused :/
jim_thompson5910
  • jim_thompson5910
all I did was take the coordinates of point P and make them negative
jim_thompson5910
  • jim_thompson5910
point P is the point corresponding to pi/12 point Q is the point corresponding to 13pi/12
anonymous
  • anonymous
so how would I follow this pattern for 5pi/12?
jim_thompson5910
  • jim_thompson5910
Hint for part 2 5pi/12 = (pi + 4pi)/12 5pi/12 = pi/12 + 4pi/12 5pi/12 = pi/12 + pi/3 Then use these identities \[\Large \sin(x+y) = \sin(x)\cos(y)+\cos(x)\sin(y)\] \[\Large\cos(x+y) = \cos(x)\cos(y)-\sin(x)\sin(y)\]
anonymous
  • anonymous
I'm really lost..
jim_thompson5910
  • jim_thompson5910
I'm guessing you've never seen those identities before?

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