How would I find the standard form of the complex number #9(cos((7pi)/6)=isin((7pi)/6))#?

1 Answer
Oct 9, 2015

Just evaluate the trigonometric functions and expand to get #-(9sqrt(3))/2-9/2 i# (standard form for a complex number is #a+bi#, where #a,b\in RR#)

Explanation:

Note that #cos((7pi)/6)=-sqrt(3)/2# and #sin((7pi)/6)=-1/2#. Therefore

#9(cos((7pi)/6)+i sin((7pi)/6))=9(-sqrt(3)/2-i 1/2)#

#=-(9sqrt(3))/2-9/2 i#