Identify the symmetry and the type of polar equation represented by these equations

Question

x=2cost+cos2t

y=2sint-sin2t

Name the types of symmetry

Name the types of polar equations

Answer 1

x=2cos(t)+cos(2t), y=2sin(t)-sin(2t).

x²+y²=r²=(4cos²(t)+4cos(t)cos(2t)+cos²(2t)) + (4sin²(t)-4sin(t)sin(2t)+sin²(2t)).

So r²=4+4cos(t+2t)+1=5+4cos(3t), and r=√(5+4cos(3t)). This is the polar equation.

The value of r varies between 1 and 3 because -1≤cos(3t)≤1. When 3t=π, cos(3t)=-1 and t=π/3, and r=1. When t=0, cos(3t)=1 and r=3. When t=2π/3, cos(3t)=1; t=π, cos(3t)=-1; t=4π/3, cos(3t)=1.

As t goes from 0 to 2π we observe that at t=0 and π the y values are negated for t between π and 2π but the x values are the same. So the graph is symmetrical about the x axis, which acts like a mirror. The reason for the symmetry is that x=2cos(t)+cos(2t)=2cos(t)+2cos²(t)-1. When we replace t with 2π-t cos(2π-t)=cos(t), so x remains the same, while y=2sin(t)-sin(2t)=2sin(t)-2sin(t)cos(t)=2sin(t)(1-cos(t)). When we replace t with 2π-t it becomes y=-2sin(t)(1-cos(t)).