Question

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

Answer 1

Please see the explanation

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`