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