Question

Find the cube roots of 27(cos 330Á + i sin 330Á). Please help c:

Answer 1

Use formula
$e^{ix}=\cos{x}+\sin{x}$

$\sqrt[3]{27(\cos{330^o}+i\sin{330^o})} \\\sqrt[3]{27e^{i330^o} } \\ \sqrt[3]{27} \sqrt[3]{e^{i330^o}} \\ \sqrt[3]{3^3} \sqrt[3]{e^{i110^o\times 3}} \\ \sqrt[3]{3^3} \sqrt[3]{(e^{i110^o})^ 3}} \\ 3e^{i110^o} \\3(\cos{110^o}+i\sin{110^o})$

Since $330^o=360^o+330^o=690^o$ and $330^o=2\times 360^o+330^o=1050^o$ there are two more solutions:
1. $3(\cos{230^o}+i\sin{230^o})$
2. $3(\cos{350^o}+i\sin{350^o})$

Answer 2

Answer:

The cube roots are

$3(cos{110^{\circ}}+isin{110^{\circ}})$

$3(cos{230^{\circ}}+isin{230^{\circ}})$

$3(cos{350^{\circ}}+isin{350^{\circ}})$

Step-by-step explanation:

Given the expression 27(cos 330° + i sin 330°)

we have to find the cube roots of the above expression.

By Euler's formula,

$e^{i\theta}=cos{\theta}+isin{\theta$

We can write

$cos{330^{\circ}}+isin{330^{\circ}}=e^{330i}$

$\sqrt[3]{27(cos{330^{\circ}}+isin{330^{\circ}})}\\\\=\sqrt[3]{3^3(cos{330^{\circ}}+isin{330^{\circ}})}\\\\=\sqrt[3]{3^3(e^{330i})}\\\\=\sqrt[3]{3^3(e^{110i\times3})}\\\\=\sqrt[3]{(3e^{110i})^3}\\\\=3e^{110i}=3(cos{110^{\circ}}+isin{110^{\circ}})$

Since, 330°=360°+330°=690° and 330°=2(360°)+330°=1050°

Other two solutions are

 $3(cos{230^{\circ}}+isin{230^{\circ}})$ and $3(cos{350^{\circ}}+isin{350^{\circ}})$