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 $=<-((sqrt2+sqrt6))/2, ((sqrt2-sqrt6))/2 >$

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