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anonymous
 one year ago
r = 2sin^2theta How can I draw it and what shape is that graph?
anonymous
 one year ago
r = 2sin^2theta How can I draw it and what shape is that graph?

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triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433054853663:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1what about letting theta be \[\theta = \frac{\pi}{2}, \pi , \frac{3\pi}{2}, 2 \pi\] and figure out what r is and then plot \[(r, \theta)\]

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0sorry, don't have the solution but how some of this helps r^2 = 4sin^2. Sin^2 r^2 = 4 sin^2 (1  cos^2) = 4sin^2  4sin^2 Cos^2 square root of 2sin^2 = 2sin and of 4Sin^2cos^2 = 2Sin.Cos difference of 2 squares r^2 = (2 sin 2 Sin.Cos)(2 Sin + 2 Sin. Cos) 2Sin Cos = Sin (2theta)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1we're supposed to draw this arent' we r = 2sin(theta)sin(theta)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1so if we let theta = 0 r = 2sin0sin0 = 2(0)(0) = 0 so r = 0 since r = 0 and theta = 0 we have origin. we are graphing polar coordinates.

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433056445519:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1similarly if we have theta = 90 degrees because degree mode is easier r = 2sin90sin90 sin90 = 1 r=2(1)(1) = 2 r = 2 (2, 90 degrees) dw:1433052916154:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1for theta = pi (180) r=2sin(180)sin(180) = 2(0)(0) = 0 (r, theta) = (0,180)dw:1433053067791:dw for theta = 3pi/2 (270) r=2sin(270)sin(270) = 2(1)(1) = 2 (2, 270 degrees)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1dw:1433053155766:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1for theta = 2pi r = 2sin(360)sin(360) r=2(0)(0) = 0 (0, 360 degrees) so when we're done with the main degrees from the four quadrants, we can focus on the more specific ones, pi/3, pi/4, pi/6

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1for theta = 30 degrees) r =2sin(30)sin(30) = 2(1/2)(1/2) = 2/4 = 1/2 (1/2, 30 degrees)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1dw:1433053334467:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1for theta = 60 degrees r = 2sin(60)sin(60) = 2(\sqrt{3}/2)(\sqrt{3}/2) = 2(3/4)=(6/4) = (3/2) (3/2, 60) the graph is getting whacky on os... but since sin is an odd function there should be symmetry around the origin.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1dw:1433053550001:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1dw:1433053562404:dw
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