anonymous
  • anonymous
r = 2sin^2theta How can I draw it and what shape is that graph?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Lolipop_Drum312
  • Lolipop_Drum312
http://www.wolframalpha.com/pro?src=pro-banner-5 <--
Lolipop_Drum312
  • Lolipop_Drum312
http://www.wolframalpha.com/pro?src=pro-banner-5 <--
triciaal
  • triciaal
|dw:1433054853663:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

UsukiDoll
  • 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 \[(r, \theta)\]
triciaal
  • 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)
UsukiDoll
  • UsukiDoll
we're supposed to draw this arent' we r = 2sin(theta)sin(theta)
UsukiDoll
  • 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.
triciaal
  • triciaal
|dw:1433056445519:dw|
UsukiDoll
  • 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|
UsukiDoll
  • 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)
UsukiDoll
  • UsukiDoll
|dw:1433053155766:dw|
UsukiDoll
  • 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
UsukiDoll
  • UsukiDoll
for theta = 30 degrees) r =2sin(30)sin(30) = 2(1/2)(1/2) = 2/4 = 1/2 (1/2, 30 degrees)
UsukiDoll
  • UsukiDoll
|dw:1433053334467:dw|
UsukiDoll
  • 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.
UsukiDoll
  • UsukiDoll
|dw:1433053550001:dw|
UsukiDoll
  • UsukiDoll
|dw:1433053562404:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.