In order to transform the Polar coordinates of the point #P##(r,theta)# into the Cartesian coordinates #P##(x,y)#, you must look at the right angled triangle made by the origin, #P# and #P'#, where #P'# has coordinates #(x,0)# or #(r,0)#.
In this triangle,
#sin theta = y/r#
and
#cos theta = x/r#,
therefore
#x = r*cos theta#
and
#y=r*sin theta#.
Your example has #theta = (-21pi)/8# and #r = -8#, so
#x = -8*sin((-21pi)/8)#
#y= -8*cos((-21pi)/8)#
Using the property that the #sin# and #cos# functions are periodic with period #2npi#, where n is an integer.
This means that for any #a#,
#sin a = sin (a + 2npi)#
#cos a = cos(a+2npi)#.
#sin((-21pi)/8) = sin((-16pi)/8 + (-5pi)/8) = sin(-2pi + (-5pi)/8)#
So #sin((-21pi)/8) = sin ((-5pi)/8)#.
#sin((-5pi)/8) = sqrt(2+sqrt(2))/2#
and
#cos((-5pi)/8)=sqrt(2-sqrt(2))/2#.
Back to the system of equations :
#x=-4sqrt(2+sqrt(2))#
#y=-4sqrt(2-sqrt(2))#.
