How do you convert #(4, 2pi/3)# to rectangular form?

1 Answer
Aug 5, 2016

#((-2),(2sqrt3))#

Explanation:

To convert from #color(blue)"polar to cartesian form"#

#color(orange)"Reminder"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(x=rcostheta , y=rsintheta)color(white)(a/a)|)))#

here r = 4 and #theta=(2pi)/3#
#color(blue)"--------------------------------"#
#x=4cos((2pi)/3)#

#color(orange)"Reminder"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(cos((2pi)/3)=-cos(pi-(2pi)/3)=-cos(pi/3))color(white)(a/a)|)))#

#rArrx=-4cos(pi/3)=-4xx1/2=-2#
#color(blue)"------------------------------------------------------------"#
and #y=4sin((2pi)/3)#

#color(orange)"Reminder"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(sin((2pi)/3)=sin(pi-(2pi)/3)=sin(pi/3))color(white)(a/a)|)))#

#rArry=4sin(pi/3)=4xxsqrt3/2=2sqrt3#
#color(blue)"---------------------------------------------"#

Thus the components of the vector #=((-2),(2sqrt3))#