What are the components of the vector between the origin and the polar coordinate #(7, (-7pi)/12)#?

1 Answer

Rectangular Coordinates:
#x=7/4*(sqrt(6-3sqrt3)-sqrt(2+sqrt3))=-1.81173#
#y=-7/4*(sqrt(6+3sqrt3)+sqrt(2-sqrt3))=-6.76148#

Polar coordinates: #(1.81173, pi)#and#(6.76148, (3pi)/2)#

Explanation:

#x=rcos theta#
#x=7*cos((-7pi)/12)#
if we add 1 revolution #=2pi# to the angle then
#2pi+(-7pi)/12=(17pi)/12#
#x=7*cos((17pi)/12)#
using double angle formulas

#x=7*cos((17pi)/12)=7*cos(pi+(5pi)/12) #

#x=7*cos(pi+(5pi)/12)#
#x=7[cos pi *cos ((5pi)/12)- sin pi *sin ((5pi)/12)] #

#x=7*(-cos((5pi)/12)-0)#

#x=7*(-1)*cos ((5pi)/12)#
Using double angle again

#x=-7*cos(pi/3+pi/12)#
#x=-7*[cos (pi/3) *cos (pi/12)-sin (pi/3) *sin (pi/12)]#

Recall the special angles #60^@=pi/3# and #30^@=pi/6# and #pi/12=1/2*pi/6# Use Half-Angle formulas for functions of #pi/12#

#cos (pi/12)=sqrt((1+cos pi/6)/2)=sqrt((1+sqrt3/2)/2)=1/2sqrt(2+sqrt3)#
#sin (pi/12)=sqrt((1-cos pi/6)/2)=sqrt((1-sqrt3/2)/2)=1/2sqrt(2-sqrt3)#

so that

#x=-7*[cos (pi/3) *cos (pi/12)-sin (pi/3) *sin (pi/12)]#

becomes

#x=-7*[1/2 *(sqrt(2+sqrt3))/2-sqrt3/2 *sqrt(2-sqrt3)/2]#

#x=7/4*(sqrt(6-3sqrt3)-sqrt(2+sqrt3))=-1.81173#

Do the same for #y# and come up with

#y=-7/4*(sqrt(6+3sqrt3)+sqrt(2-sqrt3))=-6.76148#