Question

How do you graph `r=5sin(3/4theta)`?

Answer 1

See explanation. The function for graph is quite lengthy. I had already paved way, for making idiosyncratic graphs of `r = sin ( m/n )theta` and `r = cos ( m/n )theta`

Explanation:

See
https://socratic.org/questions/how-do-you-create-a-graph-of-r-3-sin-7theta-5

`0 <= r <= 5`.

The period is `(8/3)pi`.. No graph for negative r., when

`(3/4)theta in ( pi, 2pi ) rArr theta in ((4/3)pi, (8/3)pi)`.

For four complete rotations, through `8pi`, three loop sought to

be created.

Upon setting `(8/3)pi = k(2pi)`, for least integer k, ,k = 3 #..

Use `sin 3theta = sin 4(3/4theta)` and expand both.

`3 cos^2theta sin theta - sin^3theta`

`= 4 (cos^3(3/4theta) sin(3/4theta)`

`- cos(3/4theta) sin^3(3/4theta)`

Now switch over to Cartesian form, using astutely

`sin(3/4theta) = r / 5, cos(3/4theta) = sqrt ( 1 - r^2/25),`

`r (cos theta, sin theta ) = ( x, y ) and r= sqrt ( x^2 + y^3 )`, to get

`3 x^2 y - y^3 = 4 ( x^2 + y^2 )^1.5(( 1 - (x^2 + y^2 )/25)^1.5`

`( (x^2 + y ^2)/25 )^0.5 -( 1 - (x^2 + y^2 )/25)^0.5`

`(( x^2 + y ^2 )/25)^1.5)`

I have paved the way. the Socratic graph is immediate.
graph{3x^2 y-y^3 - 0.0064 ( x^2 + y^2 )^2( 25 - (x^2 + y^2 ))^0.5( 25 - 2(x^2 + y^2 ) ) = 0[-10 10 -5 5]}
See graph of `r = cos (3/4)theta`:
graph{x^3-3xy^2-(x^2+y^2)^1.5(8(x^2+y^2)^2-8(x^2+y^2)+1)=0[-2 2 -1 1]}