What is the Cartesian form of #(1, (23pi)/8 ) #?

Question

What is the Cartesian form of #(1, (23pi)/8  ) #?

Emilia Aguilar · Accepted Answer

The Cartesian Form of #(1, (23pi)/8)# is
#(cos ((7pi)/8), sin ((7pi)/8))=(-0.9238795325, 0.3826834324)# 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.

Luke Alvarez · Answer

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