What is the Cartesian form of #(-4,(-21pi)/4))#?

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Answer 1 · 2018-07-26T22:54:41

#(2sqrt2, -2sqrt2)#

We have the coordinate #(-4, (-21pi)/4)# in polar form.

Coordinates in polar form have the standard form #(color(green)(r), color(purple)(Θ))#.

To convert from polar form to Cartesian form, we use the following formulas:

Now, let's plug stuff in. We know #color(green)(r) = -4# and #color(purple)(Θ) = (-21pi)/4#

#color(red)(x) = (-4)*cos((-21pi)/4)#

#color(red)(x) = (-4)*(-sqrt2/2)#

#color(red)(x) = 2sqrt2#

#color(blue)(y) = (-4)*sin((-21pi)/4)#

#color(blue)(y) = (-4)*(sqrt2/2)#

#color(blue)(y) = -2sqrt2#

We get #color(red)(x) = 2sqrt2# and #color(blue)(y) = -2sqrt2#, making our Cartesian coordinate #(2sqrt2, -2sqrt2).#

Answer 2 · 2018-07-26T22:56:41

#(2\sqrt2, -2\sqrt2)#

Cartesian coordinates #(x, y)# of given point #(r, \theta)\equiv(-4, {-21\pi}/4)# are given as

#x=r\cos\theta#

#=-4\cos(-{21\pi}/4)#

#=-4\cos({5\pi}/4)#

#=-4(-1/\sqrt2)#

#=2\sqrt2#

#y=r\sin\theta#

#=-4\sin(-{21\pi}/4)#

#=4\sin({5\pi}/4)#

#=4(-1/\sqrt2)#

#=-2\sqrt2#

hence the cartesian coordinates of given point are

#(x, y)\equiv(2\sqrt2, -2\sqrt2)#