How do you convert #(-sqrt(3),-sqrt(3))# to polar form?

1 Answer
Shwetank Mauria
May 4, 2016

#(-sqrt3,-sqrt3)# in polar form is #(sqrt6,pi/4)#

Explanation:

If #(r,theta)# is in polar form and #(x,y)# in Cartesian form the relation between them is as follows:

#x=rcostheta#, #y=rsintheta#, #r^2=x^2+y^2# and #tantheta=y/x#

As in Cartesian form, the point is #(-sqrt3,-sqrt3)#,

in polar form, #r=sqrt((-sqrt3)^2+(-sqrt3)^2)=sqrt(3+3)=sqrt6#

and #tantheta-(-sqrt3)/(-sqrt3)=1# or #theta=pi/4#

Hence, #(-sqrt3,-sqrt3)# in polar form is #(sqrt6,pi/4)#

Related questions
See all questions in Converting Between Systems
Impact of this question
2664 views around the world
You can reuse this answer
Creative Commons License