How do you convert #(2sqrt(2), pi/6)# into cartesian form?

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How do you convert #(2sqrt(2), pi/6)# into cartesian form?

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Answer 1 · 2016-10-19T14:16:27

The answer is #(sqrt6,sqrt2)#

Explanation:

#r=2sqrt2# and #theta=pi/6#
So #x=rcostheta#=>#x=2sqrt2*cos(pi/6)#
and#y=rsintheta#=>#y=2sqrt2sin(pi/6)#
#cos (pi/6)=sqrt3/2#
and#sin(pi/6)=1/2#
So the answer is #(2sqrt2*sqrt3/2,2sqrt2*1/2)#
=#(sqrt6,sqrt2)#

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How do you convert #(2sqrt(2), pi/6)# into cartesian form?

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