Why is $r = 3 cos 2 θ$ $r=3cos2theta$ not symmetric over $θ = \frac{π}{2}$ $theta=pi/2$ ?

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Why is $r = 3 cos 2 θ$ $r=3cos2theta$ not symmetric over $θ = \frac{π}{2}$ $theta=pi/2$ ?

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Answer 1 · 2017-07-19T12:06:42

The graph IS symmetric about that line.

Explanation:

You already see the graph, so you were able to observe its symmetry.
One test to determine symmetry about $θ = \frac{π}{2}$ $theta = pi/2$ is to substitute
$θ - π$ $theta - pi$ for $θ$ $theta$ .

$3 cos (2 (θ - π)) = 3 cos (2 θ - 2 π)$ $3cos(2(theta -pi)) = 3cos(2theta -2pi)$
$= 3 cos 2 θ cos 2 π + sin 2 θ sin 2 π$ $=3cos2thetacos2pi+sin2thetasin2pi$
$= 3 cos 2 θ$ $=3cos2theta$ .

Therefore, the function is symmetric about $θ = \frac{π}{2}$ $theta = pi/2$ .

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Why is $r = 3 cos 2 θ$ $r=3cos2theta$ not symmetric over $θ = \frac{π}{2}$ $theta=pi/2$ ?

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Explanation:

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Why is $r = 3 cos 2 θ$ $r=3cos2theta$ not symmetric over $θ = \frac{π}{2}$ $theta=pi/2$ ?