How do you graph $r = 2 sin 2 x$ $r=2sin2x$ ?

Question

4082 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-19T03:25:32

See graph and explanation.

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

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

The graph of $r = a sin (n θ - α)$ $r = a sin ( ntheta - alpha )$ is obtained by rotating

anticlockwise graph of $r = a sin θ$ $r = a sin theta$ , about $θ = 0$ $theta = 0$ ,

through $α$ $alpha$ .

Use $(x, y) = r (cos θ, sin θ), r = \sqrt{x^{2} + y^{2}}$ $( x, y ) = r ( cos theta, sin theta ), r = sqrt( x^2 + y^2 )$ and

$sin 2 θ = 2 sin θ cos θ$ $sin 2theta = 2 sin theta cos theta$ and get the Cartesian form of

$r = 2 sin 2 θ$ $r = 2 sin 2theta$ as

${(x^{2} + y^{2})}^{1.5} - 4 x y = 0$ $( x^2 + y^2 )^1.5 - 4 xy = 0$ .

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

For anticlockwise rotation, through $α = \frac{p}{2}$ $alpha = p/2$ , use

$r = 2 sin (2 (θ - \frac{π}{2})) = - 2 sin 2 θ$ $r = 2 sin ( 2 ( theta - pi/2)) = - 2 sin 2theta$

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

For clockwise rotation, through $α = \frac{π}{4}$ $alpha = pi/4$ , use

$r = 2 sin (2 (θ + \frac{π}{4})) = 2 cos 2 θ$ $r = 2 sin ( 2 ( theta + pi/4)) = 2 cos 2theta$ .

See the graph, using

$cos 2 θ = ({cos}^{2} θ - {sin}^{2} θ) = \frac{x^{2} - y^{2}}{r^{2}}$ $cos 2theta = (cos^2theta - sin^2theta) = ( x^2 - y^2 )/r^2$

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

How do you graph r=2sin2xr=2sin2x?