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

Dec 21, 2016

See graph and explanation.

#### Explanation:

Note that

$\sin 4 \theta = 2 \sin 2 \theta \cos \theta$

$= 4 \sin \theta \cos \theta \left({\cos}^{2} \theta - {\sin}^{2} \theta\right)$

Using conversion formula $r \left(\cos \theta , \sin \theta\right) = \left(x , y\right)$,

the cartesian form is obtained as

$16 x y \left({x}^{2} - {y}^{2}\right) = \sqrt{{x}^{2} + {y}^{2}} {\left({x}^{2} + {y}^{2}\right)}^{2}$.

$r = 4 \sin 4 \theta \ge 0 \to 4 \theta \in {Q}_{1} \mathmr{and} {Q}_{2} \mathmr{and} \theta \in {Q}_{1}$

The period for the graph is $\frac{2 \pi}{4} = \frac{\pi}{2}$. Overall, in $\left[0 , 2 \pi\right]$

covering 4 periods, 4 petals are drawn, @ one/(half period $\frac{\pi}{4}$).

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 $r = 4 \sin 5 \theta$ and $r = 4 \cos 5 \theta$.

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}