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

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1. Lolipop_Drum312
2. Lolipop_Drum312
3. triciaal

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4. UsukiDoll

what about letting theta be $\theta = \frac{\pi}{2}, \pi , \frac{3\pi}{2}, 2 \pi$ and figure out what r is and then plot $(r, \theta)$

5. triciaal

sorry, 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)

6. UsukiDoll

we're supposed to draw this arent' we r = 2sin(theta)sin(theta)

7. UsukiDoll

so 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.

8. triciaal

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9. UsukiDoll

similarly 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|

10. UsukiDoll

for 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)

11. UsukiDoll

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12. UsukiDoll

for 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

13. UsukiDoll

for theta = 30 degrees) r =2sin(30)sin(30) = 2(1/2)(1/2) = 2/4 = 1/2 (1/2, 30 degrees)

14. UsukiDoll

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15. UsukiDoll

for 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.

16. UsukiDoll

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17. UsukiDoll

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