# What is the Cartesian form of #( -2 , (13pi)/6 ) #?

##### 1 Answer

I got:

#(sqrt3,1)#

assuming

Recall:

#x = rcostheta#

#y = rsintheta#

#r = sqrt(x^2 + y^2)#

The coordinates you are given are in polar, and are the

However...

#2 = sqrt(x^2 + y^2) => 4 = x^2 + y^2#

#y = rsin(theta) = 2sin((13pi)/6)#

#=> y^2 = 4sin^2((13pi)/6) = 4sin^2(390^@) = 4sin^2(30^@) = 1#

#=> color(green)(y = 1)# (to keep consistent with the equations above,

#y > 0# , since#costheta > 0# and#r > 0# .)

Therefore, we can solve for

#=> x^2 = 4 - y^2 = 4 - 1 = 3#

#=> color(green)(x = sqrt3)# (to keep consistent with the equations above,

#xy > 0# , since#costheta > 0# and#r > 0# .)

So, the cartesian coordinates are:

#color(blue)((x,y) = (sqrt3,1))#

To make sure it worked...

#r = sqrt(3 + 1) = 2# #color(blue)(sqrt"")#

#theta = arccos(x/r) = arcsin(y/r)#

#= arccos(sqrt3/2)# As

#costheta = sqrt3/2# when#theta = 30^@# ,#arccos(sqrt3/2) = theta = 30^@# , which is coterminal with#390^@# .#color(blue)(sqrt"")#