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# how does r=2cos3(theta) = r=2cose(theta)(cos(theta)^2-(3 sin(theta)^2))

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how does r=2cos3(theta) = r=2cos(theta)(cos(theta)^2-(3 sin(theta)^2)) in polar equations terms

Guest Jun 21, 2015

#4
+26640
+10

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Alan  Jun 21, 2015
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#1
+92221
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I only just noticed "in polar form" so I am not sure what you need.

I can certainly show that they are the same.

You are actually asking me to show that

$$\\2cos3\theta=2cos\theta( Cos^2\theta-3sin^2\theta)\\\\ LHS=2cos3\theta\\\\ LHS=2[cos2\theta cos\theta-sin2\theta sin\theta]\\\\ LHS=2[(cos^2\theta-sin^2\theta) cos\theta-(2sin\theta cos\theta sin\theta)]\\\\ LHS=2cos\theta [cos^2\theta -sin^2\theta -2sin^2\theta ]\\\\ LHS=2cos\theta( Cos^2\theta-3sin^2\theta)\\\\ LHS=RHS$$

Melody  Jun 21, 2015
#2
+5

Thanks i think that should work i just have an investigation on Polar equations and the question was show how r=2cos3(theta) is euqal to ^2=2x^3-6xy^2

Knowing that (x^2+y^2)=r^2

x=rcos(theta)

y=rsin(theta)

Guest Jun 21, 2015
#3
+92221
0

Yes, that is right

But I still don't know how the answer should be set out :/

Melody  Jun 21, 2015
#4
+26640
+10

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Alan  Jun 21, 2015
#5
+92221
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Thanks Alan,

So then you just have to show that the other equation is the same ? (Which is easy to do)

So you need to work on both equations and get to the same polar equation.  :/

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I am still confused though.  I understand the algebra, no problems, but I don't understand how it all goes together.

I am not explaining well because I am not quite sure what it is that I don't understand.

I just know that there is something that I don't quite get here.

Maybe a graph would help?

Melody  Jun 21, 2015

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