How do you convert #r = 3 sec ( pi/3 - theta) # into cartesian form?

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Answer 1 · 2016-10-12T18:07:04

Please see the explanation for the conversion process.

#y = -sqrt(3)/3x + 2sqrt(3)#

Multiply both sides of the equation by #cos(pi/3 - theta)# (because #cos(A)sec(A) = 1#):

#rcos(pi/3 - theta) = 3#

Use the identity #cos(A - B) = cos(A)cos(B) + sin(A)sin(B)#:

#r(cos(pi/3)cos(theta) + sin(pi/3)sin(theta)) = 3#

We know the values for the sine and cosine of #pi/3#:

#r(1/2cos(theta) + sqrt(3)/2sin(theta)) = 3#

Distribute r:

#1/2rcos(theta) + sqrt(3)/2rsin(theta) = 3#

Substitute y for #rsin(theta)# and x for #rcos(theta)#

#1/2x + sqrt(3)/2y = 3#

# sqrt(3)/2y = -1/2x + 3#

#y = -sqrt(3)/3x + 2sqrt(3)#