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ra1l [238]
4 years ago
9

HELP!!! Determine if the graph is symmetric about the x-axis, the y-axis, or the origin. r = 4 cos 3θ

Mathematics
2 answers:
RoseWind [281]4 years ago
6 0
<h2>Answer:</h2>

                The graph is symmetric about the x-axis.

<h2>Step-by-step explanation:</h2>
  • The graph is symmetric about the polar axis(x-axis) if we replace (r,θ) with (r,-θ) then it is equivalent to the original equation.
  • The graph is symmetric about the y-axis if we replace (r,θ) with (-r,-θ) then it is equivalent to the original equation.
  • The graph is symmetric about the pole(origin) if we replace (r,θ) with (-r,θ) then it is equivalent to the original equation.

Here we have equation as:

           r=4\cos 3\theta

  • when we replace (r,θ) with (r,-θ) we have:

r=4\cos 3(-\theta)\\\\i.e.\\\\r=4\cos (-3\theta)\\\\i.e.\\\\r=4\cos 3\theta

(  Since, we know that:

\cos (-\tehta)=\cos \theta  )

Hence, the graph is symmetric about the x-axis.

  • when we replace  (r,θ) with (-r,-θ)

-r=4\cos 3(-\theta)\\\\i.e.\\\\-r=4\cos (-3\theta)\\\\i.e.\\\\-r=4\cos 3\theta\\\\i.e.\\\\r=-4\cos 3\theta

Hence, we observe that on replacing the function the two graphs are not equivalent.

Hence, the graph is not symmetric about the y-axis.

  • when we replace (r,θ) with (-r,θ) we have:

-r=4\cos 3\theta\\\\i.e.\\\\r=-4\cos 3\theta

Hence, we observe that on replacing the function the two graphs are not equivalent.

Hence, the graph is not symmetric about the pole(origin)

natulia [17]4 years ago
3 0
Testing for y-symmetry let's make r = 5 cos 3θ and set <span>θ=π−θ</span> so<span><span><span>r=5cos3θ</span><span>θ=(π−θ)</span><span>r=5cos(3(π−θ))⟹r=5cos(3π−3θ)</span></span></span>
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