Suppose that r (cos(θ) + i sin(θ)) is a third root of 27 (cos(π/5) + i sin(π/5)), i.e.
[r (cos(θ) + i sin(θ))]³ = 27 (cos(π/5) + i sin(π/5))
Expand the left side using de Moivre's theorem:
r³ (cos(3θ) + i sin(3θ)) = 27 (cos(π/5) + i sin(π/5))
Matching up real and imaginary parts, we have
r³ cos(3θ) = 27 cos(π/5)
r³ sin(3θ) = 27 sin(π/5)
Right away we see that r³ = 27 ⇒ r = 3, which eliminates A.
We also have
(r³ sin(3θ)) / (r³ cos(3θ)) = (27 sin(π/5)) / (27 cos(π/5))
tan(3θ) = tan(π/5)
3θ = tan⁻¹(tan(π/5))
3θ = π/5 + nπ
(where n is an integer)
θ = π/15 + nπ/3
Now,
• n = 0 ⇒ θ = π/15 ⇒ D is a third root
• n = 1 ⇒ θ = π/15 + π/3 = 2π/5
• n = 2 ⇒ θ = π/15 + 2π/3 = 11π/15 ⇒ C is a third root
I think this video will help you so much, it's about how to solve it from Khan Academy
https://www.khanacademy.org/math/pre-algebra/pre-algebra-exponents-radicals/pre-algebra-scientific-notation/v/scientific-notation-old
Least. 0.41, 0.5, 3/5 Greatest
