What are the components of the vector between the origin and the polar coordinate #(-2, (23pi)/12)#?

1 Answer
Dec 17, 2015

#(2cos((23pi)/12),2sin((23pi)/12))#

Explanation:

You can use complex numbers to simplify the problem. And by the way, a module is a non-negative number so #-2# is probably #2#.

Here, you're looking for the component of the complex number #2e^(i(23pi)/12)#. Its module is #2# and in general, #e^(itheta) = cos(theta) + isin(theta)#, so we can say that #2e^(i(23pi)/12) = 2(cos((23pi)/12) + isin((23pi)/12))#.

Unfortunately, #(23pi)/12# is not a usual value for #cos# and #sin# so I can only tell you that the components of this vector are #(2cos((23pi)/12),2sin((23pi)/12))#.