Question

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

Answer 1

Please see the explanation for the conversion process.

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

Explanation:

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)`