Polar to Rectangular Conversion

  • Thread starter Dooh
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  • #1
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Can someone help me with these problems? It's been bugging me i can't seem to solve it.

Lets assume T = theta

I can't seem to find a way to convert these polar equations into rectangular form.

r = 2 sin 3T

r = 6 / 2 - 3 sinT

If possible, can someone help me with this and list it in a step-by-step fashion so i can see how you get the answer. Thanks.
 
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Answers and Replies

  • #2
Pyrrhus
Homework Helper
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[tex] r = 2 \sin (3 \theta) [/tex]

[tex] r^2 = 2r \sin (3 \theta) [/tex]

then try the triple angle formulas to reduce [itex] \sin (3 \theta) [/itex]
 
  • #3
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Sorry but the furthest that our teacher had taught us in the double angle formula.
 
  • #4
Pyrrhus
Homework Helper
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It's not hard. Why don't you try it?

[tex] \sin (3 \theta) = \sin (\theta + 2 \theta) [/tex]

Use the sum of angles for the sin

[tex] \sin (\alpha + \beta) = \sin \alpha cos \beta + \cos \alpha \sin \beta [/tex]
 
  • #5
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Ok, so i got:


[tex] r^2 = 2r (\sin \theta \cos 2 \theta + \cos \theta \sin 2 \theta ) [/tex]

can i take out the [tex] \sin \theta \cos \theta [/tex]

and get

[tex] r^2 = 2r \sin \theta \cos \theta (\cos \theta + \sin \theta ) [/tex]

or should i further expand the equation?
 
  • #6
Pyrrhus
Homework Helper
2,178
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No that's wrong it will be

[tex] \sin (3 \theta) = 3 \sin \theta - 4 \sin^{3} \theta [/tex]

thus

[tex] r^2 = 2r (3 \sin \theta - 4 \sin^{3} \theta) [/tex]
 

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