Respuesta :
[tex]\bf \stackrel{\textit{testing for the x-axis symmetry, }\theta =-\theta }{r=8cos(3\theta )\implies r=8cos[3(-\theta)]}\implies r=8cos(-3\theta)
\\\\\\
r=8cos(3\theta)~~\boxed{\checkmark}~~\impliedby \stackrel{\textit{trigonometry symmetry identities}}{cos(\theta)=cos(-\theta)}\\\\
-------------------------------[/tex]
[tex]\bf \stackrel{\textit{testing for the y-axis symmetry, }\theta =\pi -\theta }{r=8cos(3\theta )\implies r=8cos[3(\pi -\theta)]}\implies r=8cos(3\pi -3\theta)\boxed{\otimes}\\\\ -------------------------------\\\\ \stackrel{\textit{testing for the origin symmetry, }\theta =\pi +\theta }{r=8cos(3\theta )\implies r=8cos[3(\pi +\theta)]}\implies r=8cos(3\pi +3\theta)\boxed{\otimes}[/tex]
so as you can see, since the x-axis test yielded the same original expression, it has symmetry with the x-axis, or namely the "polar axis".
[tex]\bf \stackrel{\textit{testing for the y-axis symmetry, }\theta =\pi -\theta }{r=8cos(3\theta )\implies r=8cos[3(\pi -\theta)]}\implies r=8cos(3\pi -3\theta)\boxed{\otimes}\\\\ -------------------------------\\\\ \stackrel{\textit{testing for the origin symmetry, }\theta =\pi +\theta }{r=8cos(3\theta )\implies r=8cos[3(\pi +\theta)]}\implies r=8cos(3\pi +3\theta)\boxed{\otimes}[/tex]
so as you can see, since the x-axis test yielded the same original expression, it has symmetry with the x-axis, or namely the "polar axis".
Answer:
The graph is symmetric about the x-axis.
Step-by-step explanation:
The given polar equation is
[tex]r=8\cos 3\theta[/tex]
If the polar equation satisfy by the point (r,-θ), then it is symmetric about the x-axis.
If the polar equation satisfy by the point (-r,-θ), then it is symmetric about the y-axis.
If the polar equation satisfy by the point (-r,θ), then it is symmetric about the origin.
Substitute (r,-θ) for (r,θ) in the given equation.
[tex]r=8\cos 3(-\theta)[/tex]
[tex]r=8\cos 3\theta[/tex]
[tex]r=r[/tex]
Therefore the graph is symmetric about the x-axis.
Substitute (-r,-θ) for (r,θ) in the given equation.
[tex]-r=8\cos 3(-\theta)=8\cos 3\theta=r\neq -r[/tex]
The graph is not symmetric about y-axis.
Substitute (-r,θ) for (r,θ) in the given equation.
[tex]-r=8\cos 3\theta=r\neq -r[/tex]
The graph is not symmetric about the origin.
