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

Question

3648 views around the world

You can reuse this answer
Creative Commons License

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 $rcos(theta)$

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

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

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

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