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anonymous
 one year ago
medal award please help!
anonymous
 one year ago
medal award please help!

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1hint 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
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure what it is that they want.. they are asking for the coordinates i'm a little lost :/

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0for a unit circle ... x = cos(t) y = sin(t)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0but working them into their radical values might be a bit daunting :/

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444012468609:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444012481543:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444012492630:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1add pi to pi/12 to get 13pi/12 this is the same as doing a 180 degree rotation dw:1444012532760:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1compare 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
 one year ago
Best ResponseYou've already chosen the best response.0that makes sense,.. so when they mention the terminal point p(x,y) how would I determine that?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1it's essentially point P but the signs are different for each coordinate

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444012707085:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large P = \left(\frac{\sqrt{2+\sqrt{3}}}{2}, \frac{\sqrt{2\sqrt{3}}}{2}\right)\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large Q = \left(\color{red}{}\frac{\sqrt{2+\sqrt{3}}}{2}, \color{red}{}\frac{\sqrt{2\sqrt{3}}}{2}\right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for 13pi/12 the terminal point would be sqrt2+sqrt3/2?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1it would be point Q I wrote above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'm a little confused :/

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1all I did was take the coordinates of point P and make them negative

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1point P is the point corresponding to pi/12 point Q is the point corresponding to 13pi/12

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so how would I follow this pattern for 5pi/12?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Hint for part 2 5pi/12 = (pi + 4pi)/12 5pi/12 = pi/12 + 4pi/12 5pi/12 = pi/12 + pi/3 Then use these identities \[\Large \sin(x+y) = \sin(x)\cos(y)+\cos(x)\sin(y)\] \[\Large\cos(x+y) = \cos(x)\cos(y)\sin(x)\sin(y)\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I'm guessing you've never seen those identities before?
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