What are the components of the vector between the origin and the polar coordinate #(25, (-7pi)/12)#?

1 Answer
Feb 24, 2018

The coordinates are

#-=(x,y)=-=(-25xx(sqrt3-1)/(2sqrt2)),(-25xx(sqrt3-1)/(2sqrt2))#

Explanation:

The polar coordinates are #(r,theta)-=(25,(-7pi)/12)#

Comparing,

#r=25#

#theta=-(7pi)/12#

#pi^C=180^@#

#-(7pi)/12=-(7xx180)/12#

#=-7xx15=-105#

#x=rcostheta#

#rcostheta=25cos(-105^@)#
#=25cos105^@=25cos(180^@-75^@)#
#=25xx(-cos75^@)=25xx(-(sqrt3-1)/(2sqrt2))#
#x=-25xx(sqrt3-1)/(2sqrt2)#

#y=rsintheta#

#rsintheta=25sin(-105^@)#
#=-25sin105^@=-25sin(180^@-75^@)#
#y=-25sin75^@=-25xx(sqrt3+1)/(2sqrt2)#
#y=-25xx(sqrt3+1)/(2sqrt2)#

The coordinates are

#-=(x,y)=-=(-25xx(sqrt3-1)/(2sqrt2)),(-25xx(sqrt3-1)/(2sqrt2))#