What are the components of the vector between the origin and the polar coordinate #(2, (13pi)/12)#?

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Answer 1 · 2017-12-15T08:17:08

The vector is # =<-((sqrt2+sqrt6))/2, ((sqrt2-sqrt6))/2 >#

To convert from polar coordinates #(r, theta)# to rectangular coordinates , we apply the following equations

#x=rcostheta#

#y=rsintheta#

Here,

The polar coordinates are

#r=2#

and

#theta=13/12pi#

Therefore,

The rectangular coordinates are

#x=2cos(13/12pi)=2cos(1/3pi+3/4pi)=2(cos(1/3pi)cos(3/4pi)-sin(1/3pi)sin(3/4pi))#

#= 2((1/2) * (-sqrt2/2)-(sqrt3/2) * (sqrt2/2)) #

#=2(-sqrt2-sqrt6)/4#

#=-(sqrt2+sqrt6)/2#

#y=2sin(13/12pi)=2sin(1/3pi+3/4pi)=2(sin(1/3pi)cos(3/4pi)+cos(1/3pi)sin(3/4pi))#

#= 2((sqrt3/2) * (-sqrt2/2)+(1/2) * (sqrt2/2)) #

#=2(sqrt2-sqrt6)/4#

#=(sqrt2-sqrt6)/2#

Finally,

The vector is # =<-((sqrt2+sqrt6))/2, ((sqrt2-sqrt6))/2 >#