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

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Answer 1 · 2016-12-21T04:40:17

See graph and explanation.

Note that

$sin 4 θ = 2 sin 2 θ cos θ$ $sin 4theta = 2 sin 2theta cos theta$

$= 4 sin θ cos θ ({cos}^{2} θ - {sin}^{2} θ)$ $= 4 sin theta cos theta (cos^2theta - sin^2theta)$

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

the cartesian form is obtained as

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

$r = 4 sin 4 θ \geq 0 \to 4 θ \in Q_{1} or Q_{2} and θ \in Q_{1}$ $r = 4 sin 4theta >=0 to 4theta in Q_1 or Q_2 and theta in Q_1$

The period for the graph is $\frac{2 π}{4} = \frac{π}{2}$ $(2pi)/4 = pi/2$ . Overall, in $[0, 2 π]$ $[0, 2pi]$

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

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

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}

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