The solution
from the given: Polar coordinates (1, (23pi)/8)
Let r=1 and theta=(23pi)/8
x=r cos theta and y=r sin theta
Let us solve for x
x=r cos theta
x=(1)cos ((23pi)/8)
x=(1)cos ((16pi)/8+(7pi)/8)
x=(1)cos (2pi+(7pi)/8)
use the sum formula cos (A+B)=cos A*cos B- sin A* sin B
x=(1)[cos (2pi)*cos((7pi)/8)-sin (2pi)*sin((7pi)/8)]
x=(1)[1*cos((7pi)/8)-0*sin((7pi)/8)]
x=(1)cos((7pi)/8)
x=cos((7pi)/8)=-0.9238795325
Let us solve for y
y=r sin theta
y=(1)sin ((23pi)/8)
y=(1)sin ((16pi)/8+(7pi)/8)
y=(1)sin (2pi+(7pi)/8)
use the sum formula sin (A+B)=sin A*cos B+ cos A*sin B
y=(1)[sin (2pi)*cos((7pi)/8)+cos (2pi)*sin((7pi)/8)]
y=(1)[0*cos((7pi)/8)+1*sin((7pi)/8)]
x=(1)*sin((7pi)/8)
x=sin((7pi)/8)=0.3826834324
God bless....I hope the explanation is useful.
