Question

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

Answer 1

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`