Question

Use De Moivre’s Theorem to compute the following: $[3(cos27+isin27)}^5$ Please explain step by step how it works!

Answer 1

hello: 
Use De Moivre’s Theorem : 
(3(cos27 +isin27)^5 = 3^5( cos(27 × 5) +isin(27 × 5))
                                = 3^5 ( cos(135)+i sin(135))
                                = 3^5(-√2/2+i √2/2) 
because :  cos(135) = -√2/2    and   sin(135) = √2/2
(3(cos27 +isin27)^5 = (- 3^5√2/2)+ i ( 3^5√2/2)  ...(form : a+ib  when
a= (- 3^5√2/2)  and  b = ( 3^5√2/2)