What is the value of 4(cos(120)+isin(120)) ?

The value of 4(cos(120)+isin(120)) is -2+2sqrt(3)i

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4(cos(120)+isin(120))

4 (cos (12 0^{\circ}) + i sin (12 0^{\circ}))

- 2 + 23 i

Solution steps

4 (cos (12 0^{\circ}) + i sin (12 0^{\circ}))

Use the following trivial identity:

cos (12 0^{\circ}) = - \frac{1}{2}

cos (12 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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 (12 0^{\circ}) = \frac{3}{2}

sin (12 0^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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}

= 4 (- \frac{1}{2} + i \frac{3}{2})

Simplify

4 (- \frac{1}{2} + i \frac{3}{2}) : - 2 + 23 i

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

= 4 (- \frac{1}{2} + \frac{3 i}{2})

Simplify

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

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

Apply rule

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

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

= 4 \cdot \frac{- 1 + 3 i}{2}

Multiply fractions:

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

= \frac{( - 1 + 3 i ) \cdot 4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 (- 1 + 3 i)

Rewrite

2 (- 1 + 3 i)

in standard complex form:

- 2 + 23 i

2 (- 1 + 3 i)

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = - 1, c = 3 i

= 2 (- 1) + 23 i

Apply minus-plus rules

+ (- a) = - a

= - 2 \cdot 1 + 23 i

Multiply the numbers:

2 \cdot 1 = 2

= - 2 + 23 i

= - 2 + 23 i

= - 2 + 23 i

arccos (900405)

1.5 \cdot cos (3 0^{\circ})

(cos (2 0^{\circ}))^{2}

cos (0.64)

cos (0.94)

What is the value of 4(cos(120)+isin(120)) ?
The value of 4(cos(120)+isin(120)) is -2+2sqrt(3)i