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

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

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Popular Trigonometry >

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

Solution

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

Solution

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

Solution steps

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

Use the following trivial identity:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{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{π}{3}) = \frac{3}{2}

sin (\frac{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})^{2}

Simplify

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

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

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})^{2}

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})^{2}

Apply exponent rule:

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

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

(1 + 3 i)^{2} = - 2 + 23 i

(1 + 3 i)^{2}

Apply Perfect Square Formula:

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

a = 1, b = 3 i

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

Simplify

1^{2} + 2 \cdot 1 \cdot 3 i + (3 i)^{2} : - 2 + 23 i

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

Apply rule

1^{a} = 1

1^{2} = 1

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

2 \cdot 1 \cdot 3 i = 23 i

2 \cdot 1 \cdot 3 i

Multiply the numbers:

2 \cdot 1 = 2

= 23 i

(3 i)^{2} = - 3

(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}

i^{2} = - 1

i^{2}

Apply imaginary number rule:

i^{2} = - 1

= - 1

= 3 (- 1)

Remove parentheses:

(- a) = - a

= - 3 \cdot 1

Multiply the numbers:

3 \cdot 1 = 3

= - 3

= 1 + 23 i - 3

Subtract the numbers:

1 - 3 = - 2

= - 2 + 23 i

= - 2 + 23 i

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

Remove parentheses:

(- a) = - a

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

Factor

- 2 + 23 i : 2 (- 1 + 3 i)

- 2 + 23 i

Rewrite as

= - 2 \cdot 1 + 23 i

Factor out common term

2

= 2 (- 1 + 3 i)

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

Cancel the common factor:

2

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

Rewrite

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

in standard complex form:

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

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

Apply the fraction rule:

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

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

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

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

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

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Frequently Asked Questions (FAQ)

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