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

Guest Jun 21, 2015

#1**+5 **

**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

#5**0 **

**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