Question

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

Answer 1

See graph and explanation.

Explanation:

Note that

`sin 4theta = 2 sin 2theta cos theta`

`= 4 sin theta cos theta (cos^2theta - sin^2theta)`

Using conversion formula `r(cos theta, sin theta)=(x, y)`,

the cartesian form is obtained as

`16xy(x^2-y^2)=sqrt(x^2+y^2)(x^2+y^2)^2`.

`r = 4 sin 4theta >=0 to 4theta in Q_1 or Q_2 and theta in Q_1`

The period for the graph is `(2pi)/4 = pi/2`. Overall, in `[0, 2pi]`

covering 4 periods, 4 petals are drawn, @ one/(half period `pi/4`).

In the other half, r < 0 and the the graphic designers have rightly

avoided negative r that is unreal, for real-time applications in

rotations and revolutions, about the pole.

graph{(x^2+y^2)^2.5-16xy(x^2-y^2)=0 }

As a compliment to the interested viewers of this answer, I create

here a 10-petal sine-cosine combined rose of conjoined twins.

The equations used are `r = 4 sin 5theta` and `r = 4 cos 5theta`.

graph{(0.25( x^2 + y^2 )^3 - x^5 + 10x^3y^2-5 xy^4 )(0.25( x^2 + y^2 )^3-5x^4y+10x^2y^3-y^5)=0}