What are the components of the vector between the origin and the polar coordinate #(-2, (5pi)/8)#?

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Answer 1 · 2016-02-12T23:24:53

#x=sqrt(2-sqrt2)=0.765367# and
#y=-sqrt(2+sqrt2)=-1.84776#

the x component

#x=r cos theta#
#x=-2*cos ((5pi)/8)#

but #cos ((5pi)/8)=-sin(pi/8)=-sqrt((1-cos(pi/4))/2)=-sqrt((1-1/sqrt2)/2)#

#cos ((5pi)/8)=-1/2sqrt(2-sqrt2)#

so that

#x=-2(-1/2sqrt(2-sqrt2))#

#x=sqrt(2-sqrt2)=0.765367#

the y component

#y=r sin theta#

#y=-2 sin ((5pi)/8)#

but #sin ((5pi)/8)=cos (pi/8)#

#cos (pi/8)=sqrt((1+cos(pi/4))/2)=sqrt((1+1/sqrt2)/2)=1/2sqrt(2+sqrt2)#

and

#y=-2*(1/2sqrt(2+sqrt2))#

#y=-sqrt(2+sqrt2)=-1.84776#

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