What is the Cartesian form of #(4,(23pi )/12)#?

1 Answer
Rhys
Mar 31, 2018

#(x,y) -> (sqrt(6) +sqrt(2) , sqrt(2) - sqrt(6) ) #

Explanation:

![https://www.google.co.uk/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&ved=2ahUKEwjZh4aOqJbaAhVE7BQKHSYGCTgQjRx6BAgAEAU&url=https%3A%2F%2Fmathinsight.org%2Fimage%2Fpolar_coordinates_cartesian&psig=AOvVaw1BLPuvER19bm1YM21HxgXQ&ust=1522577163343344](useruploads.socratic.org)

As seen in the image, we see that the #x# coordinate is #rcostheta #, and the #y# is #rsintheta # obtained by simple trig

#r = 4 , theta = (23pi)/12 #

#=> x = 4cos( (23pi)/12 ) = sqrt(6) + sqrt(2) #

#=> y = 4sin( (23pi) / 12 ) = sqrt(2) - sqrt(6) #

#(r, theta ) -> (4, (23pi)/12 ) #

#color(blue)((x,y) -> (sqrt(6) +sqrt(2) , sqrt(2) - sqrt(6) ) #

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