How do you graph #r=2sin2x#?

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Answer 1 · 2018-07-19T03:25:32

See graph and explanation.

The graph of # r = a sin ( ntheta - alpha )#, n = 1, 2, 3, 4, ...# shows

n equal and symmetrical loops, around the pole r = 0.

The graph of #r = a sin ( ntheta - alpha )# is obtained by rotating

anticlockwise graph of #r = a sin theta#, about #theta = 0#,

through #alpha#.

Use #( x, y ) = r ( cos theta, sin theta ), r = sqrt( x^2 + y^2 )# and

#sin 2theta = 2 sin theta cos theta# and get the Cartesian form of

#r = 2 sin 2theta# as

#( x^2 + y^2 )^1.5 - 4 xy = 0#.

Now, the Socratic graph is immediate.
graph{ (x^2 + y^2 )^1.5 - 4 xy = 0[-4 4 -2 2 ]}

For anticlockwise rotation, through # alpha = p/2#, use

# r = 2 sin ( 2 ( theta - pi/2)) = - 2 sin 2theta#

The graph is immediate.
graph{ (x^2 + y^2 )^1.5 + 4 xy = 0[-4 4 -2 2 ]}

For clockwise rotation, through # alpha = pi/4#, use

# r = 2 sin ( 2 ( theta + pi/4)) = 2 cos 2theta#.

See the graph, using

#cos 2theta = (cos^2theta - sin^2theta) = ( x^2 - y^2 )/r^2#

graph{(x^2+y^2)^1.5-2(x^2-y^2)=0[-4 4 -2 2]}