We can convert between polar and Cartesian via
#x = r cos(theta) #
#y = r sin(theta) #
I don't know the values of cosine and sine of #(5pi)/12# off the top of my head, so we can look them up or calculate them:
#(5pi)/12 = (2pi)/12 + (3pi)/12 = pi/6 + pi / 4#
#cos(A+B) = cos(A)cos(B) - sin(A)sin(B)#
#cos((5pi)/12) = cos(pi/6) cos(pi/4) - sin(pi/6)sin(pi/4)#
#cos((5pi)/12) = (sqrt(3))/2 * 1/sqrt(2) - 1/2 * 1/sqrt(2) = 1/4(sqrt6 - sqrt2) #
#sin(A+B) = sin(A)cos(B) + sin(B) cos(A)#
#sin((5pi)/12) = sin(pi/6)cos(pi/4) + sin(pi/4)cos(pi/6)#
#sin((5pi)/12) = 1/2 * 1/sqrt(2) + sqrt3/2 * 1/sqrt(2) = 1/2(sqrt(6) + sqrt(2)) #
Therefore, the exact values for #(x, y)# are
#x = -1 * 1/4(sqrt6 - sqrt2) approx -0.259#
#y = -1 * 1/4(sqrt6 + sqrt2) approx -0.966 #
