How do you convert polar equations to Cartesian equation given #theta = pi/3#?

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Answer 1 · 2016-11-15T03:36:41

Please see the explanation

#pi/3# is in the first quadrant, therefore, we substitute #tan^-1(y/x)# for #theta# and add the restriction #x > 0 and y > 0#:

#tan^-1(y/x) = pi/3; x > 0 and y > 0#

Use the tangent function on both sides:

#tan(tan^-1(y/x)) = tan(pi/3); x > 0 and y > 0#

The tangent function "undoes" its inverse and #tan(pi/3)# has a well known value:

#y/x = sqrt(3); x > 0 and y > 0#

Multiply both sides by x and drop the restriction on y, because it is now dependent on x:

#y = sqrt(3)x; x > 0#