What is the value of (cos((2pi)/3)+isin((2pi)/3))^3 ?

The value of (cos((2pi)/3)+isin((2pi)/3))^3 is 1

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(cos((2pi)/3)+isin((2pi)/3))^3

Solution

(cos (\frac{2 π}{3}) + i sin (\frac{2 π}{3}))^{3}

Solution

1

Solution steps

(cos (\frac{2 π}{3}) + i sin (\frac{2 π}{3}))^{3}

Use the following trivial identity:

cos (\frac{2 π}{3}) = - \frac{1}{2}

cos (\frac{2 π}{3})

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - \frac{1}{2}

Use the following trivial identity:

sin (\frac{2 π}{3}) = \frac{3}{2}

sin (\frac{2 π}{3})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

= (- \frac{1}{2} + i \frac{3}{2})^{3}

Simplify

(- \frac{1}{2} + i \frac{3}{2})^{3} : 1

(- \frac{1}{2} + i \frac{3}{2})^{3}

Multiply

i \frac{3}{2} : \frac{3 i}{2}

i \frac{3}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 i}{2}

= (- \frac{1}{2} + \frac{3 i}{2})^{3}

Combine the fractions

- \frac{1}{2} + \frac{3 i}{2} : \frac{- 1 + 3 i}{2}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 + 3 i}{2}

= (\frac{- 1 + 3 i}{2})^{3}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( - 1 + 3 i ) ^{3}}{2 ^{3}}

(- 1 + 3 i)^{3} = 8

(- 1 + 3 i)^{3}

Apply Perfect Cube Formula:

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

a = - 1, b = 3 i

= (- 1)^{3} + 3 (- 1)^{2} 3 i + 3 (- 1) (3 i)^{2} + (3 i)^{3}

Simplify

(- 1)^{3} + 3 (- 1)^{2} 3 i + 3 (- 1) (3 i)^{2} + (3 i)^{3} : 8

(- 1)^{3} + 3 (- 1)^{2} 3 i + 3 (- 1) (3 i)^{2} + (3 i)^{3}

Remove parentheses:

(- a) = - a

= (- 1)^{3} + 3 (- 1)^{2} 3 i - 3 \cdot 1 \cdot (3 i)^{2} + (3 i)^{3}

(- 1)^{3} = - 1

(- 1)^{3}

Apply exponent rule:

(- a)^{n} = - a^{n},

n

is odd

(- 1)^{3} = - 1^{3}

= - 1^{3}

Apply rule

1^{a} = 1

= - 1

= - 1 + (- 1)^{2} \cdot 33 i - 3 \cdot 1 \cdot (3 i)^{2} + (3 i)^{3}

3 (- 1)^{2} 3 i = 33 i

3 (- 1)^{2} 3 i

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

= 3 \cdot 1 \cdot 3 i

Multiply the numbers:

3 \cdot 1 = 3

= 33 i

3 \cdot 1 \cdot (3 i)^{2} = - 9

3 \cdot 1 \cdot (3 i)^{2}

(3 i)^{2} = 3 i^{2}

(3 i)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= i^{2} (3)^{2}

(3)^{2} : 3

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

= 3 i^{2}

= 3 \cdot 1 \cdot 3 i^{2}

Multiply the numbers:

3 \cdot 1 \cdot 3 = 9

= 9 i^{2}

i^{2} = - 1

i^{2}

Apply imaginary number rule:

i^{2} = - 1

= - 1

= 9 (- 1)

Remove parentheses:

(- a) = - a

= - 9 \cdot 1

Multiply the numbers:

9 \cdot 1 = 9

= - 9

(3 i)^{3} = - 33 i

(3 i)^{3}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= i^{3} (3)^{3}

(3)^{3} : 3^{\frac{3}{2}}

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{3}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 3}

\frac{1}{2} \cdot 3 = \frac{3}{2}

\frac{1}{2} \cdot 3

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 3}{2}

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3}{2}

= 3^{\frac{3}{2}}

= 3^{\frac{3}{2}} i^{3}

3^{\frac{3}{2}} = 33

3^{\frac{3}{2}}

3^{\frac{3}{2}} = 3^{1 + \frac{1}{2}}

= 3^{1 + \frac{1}{2}}

Apply exponent rule:

x^{a + b} = x^{a} x^{b}

= 3^{1} \cdot 3^{\frac{1}{2}}

Refine

= 33

= 33 i^{3}

i^{3} = - i

i^{3}

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

i^{3} = i^{2} i

= i^{2} i

Apply imaginary number rule:

i^{2} = - 1

= - 1 i

Multiply:

1 i = i

= - i

= 33 (- i)

Remove parentheses:

(- a) = - a

= - 33 i

= - 1 + 33 i - (- 9) - 33 i

Apply rule

- (- a) = a

= - 1 + 33 i + 9 - 33 i

Group like terms

= 33 i - 33 i - 1 + 9

Add similar elements:

33 i - 33 i = 0

= - 1 + 9

Add/Subtract the numbers:

- 1 + 9 = 8

= 8

= 8

= \frac{8}{2 ^{3}}

Simplify

\frac{8}{2 ^{3}}

Factor

8 : 2^{3}

Factor

8 = 2^{3}

= \frac{2 ^{3}}{2 ^{3}}

Cancel the common factor:

2^{3}

= 1

= 1

= 1

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Frequently Asked Questions (FAQ)

What is the value of (cos((2pi)/3)+isin((2pi)/3))^3 ?
The value of (cos((2pi)/3)+isin((2pi)/3))^3 is 1