Respuesta :

now, let's keep in mind that cos(π) = -1, and that sin(π) = 0, thus cos(π)cos(θ) is just -cos(θ), and sin(π)sin(θ) is just 0sin(θ) or just 0.


Also let's recall that symmetry identity of cos(-θ) = cos(θ).


[tex] \bf r=-3-2cos(\theta )\\\\
-------------------------------\\\\
\stackrel{\textit{testing for the y-axis, }\theta =\pi -\theta }{r=-3-2cos(\pi -\theta )}\implies r=-3-2[cos(\pi )cos(\theta )+sin(\pi )sin(\theta )]
\\\\\\
r=-3-2[-cos(\theta )+0]\implies r=-3+2cos(\theta )\qquad \otimes\\\\
------------------------------- [/tex]


[tex] \bf \stackrel{\textit{testing for the x-axis, }\theta =-\theta }{r=-3-2cos(-\theta )}\implies r=-3-2cos(\theta )\qquad \checkmark\\\\
-------------------------------\\\\
\stackrel{\textit{testing for the origin, }\theta =\pi +\theta }{r=-3-2cos(\pi +\theta)}\implies r=-3-2[cos(\pi )cos(\theta )-sin(\pi )sin(\theta )]
\\\\\\
r=-3-2[-cos(\theta )+0]\implies r=-3+2cos(\theta )\qquad \otimes [/tex]

maqibkhan234

Answer:

Yes

Step-by-step explanation:

Since, it is a graph in polar coordinates, So we can start with the base equation of this given equation which should be:

[tex]r = cos(theta)[/tex]

and its graph is a circle with centre at (0.5,0), if we multiply it by 2, then its radius will be doubled and it will become:

[tex]r =2 cos(theta)[/tex]

Thus, giving a circle of radius 1. If we multiply it by -1, then:

[tex]r =-2 cos(theta)[/tex]

It will give us a circle shifted 1 units to the left. If we add 3 in it, we will get a cardioid which is passing through y = 3 and y = -3 at x = 0,

[tex]r =3-2 cos(theta)[/tex]

And this graph will be a symmetric graph about the x-axis. Hence the answer is Yes.

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