How do you convert #(-sqrt2, 3pi/4) # to rectangular form?

Question
Answer 1

#(-sqrt2,(3pi)/4)# in rectangular coordinates is #(1,-1)#

#(r,theta)# in polar coordinates is #(rcostheta,rsintheta)# in rectangular coordinates.
Hence, #(-sqrt2,(3pi)/4)# in rectangular coordinates is
#(-sqrt2xxcos((3pi)/4),-sqrt2xxsin((3pi)/4))# or
#((-sqrt2)xx(-sqrt2/2),(-sqrt2)xxsqrt2/2)# or
#((2/2),-(2/2))# or
#(1,-1)#
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