What is the Cartesian form of #(-10,(14pi)/8))#?

2 Answers
Jan 14, 2018

See a solution process below:

Explanation:

The formula for converting Polar Coordinates to Cartesian coordinates is:

For, #(r, theta)#; #x = r xx cos(theta)#; #y = r xx sin(theta)#

Substituting the coordinates from the problem gives:

For, #(-10, (14pi)/8)#:

#x = -10 xx cos((14pi)/8) = -10 xx 0.707 = -70.7#

#y = -10 xx sin((14pi)/8) = -10 xx -0.707 = 70.7#

#(-10, (14pi)/8) =(-70.7, 70.7)#

Jan 14, 2018

The cartesian form is #(-5sqrt(2), 5sqrt(2))#.

Explanation:

To convert polar to rectangular we use:

#x=rcos(theta)#
#y=rsin(theta)#

We're given the polar point #(-10,(14pi)/8)#.

First let's simplify #(14pi)/8# to #(7pi)/4#.

From what we have we know:

#x=(-10)cos((7pi)/4)#
#y=(-10)sin((7pi)/4)#

#(7pi)/4# is a Unit Circle angle so we know its sine and cosine.

#cos((7pi)/4)=sqrt(2)/2#
#sin((7pi)/4)=-sqrt(2)/2#

so:

#x=(-10)(sqrt(2)/2)=-5sqrt(2)#
#y=(-10)(-sqrt(2)/2)=5sqrt(2)#

so the cartesian form is #(-5sqrt(2), 5sqrt(2))#.