What is the value of sin(120)cos(60)-cos(120)sin(60) ?

The value of sin(120)cos(60)-cos(120)sin(60) is (sqrt(3))/2

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sin(120)cos(60)-cos(120)sin(60)

sin (12 0^{\circ}) cos (6 0^{\circ}) - cos (12 0^{\circ}) sin (6 0^{\circ})

\frac{3}{2}

Decimal Notation

0.86602 \dots

Solution steps

sin (12 0^{\circ}) cos (6 0^{\circ}) - cos (12 0^{\circ}) sin (6 0^{\circ})

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}

Use the following trivial identity:

cos (6 0^{\circ}) = \frac{1}{2}

cos (6 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:

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 (6 0^{\circ}) = \frac{3}{2}

sin (6 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}

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

Simplify

\frac{3}{2} \cdot \frac{1}{2} - (- \frac{1}{2}) \frac{3}{2} : \frac{3}{2}

\frac{3}{2} \cdot \frac{1}{2} - (- \frac{1}{2}) \frac{3}{2}

Apply rule

- (- a) = a

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

Add similar elements:

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

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

Multiply fractions:

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

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

Cancel the common factor:

2

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

Multiply:

3 \cdot 1 = 3

= \frac{3}{2}

= \frac{3}{2}

2 arctan (0)

sin (2 \cdot 30)

arctan (\frac{- 1}{- 1})

- sin (3)

- 4 cos (\frac{π}{3})

What is the value of sin(120)cos(60)-cos(120)sin(60) ?
The value of sin(120)cos(60)-cos(120)sin(60) is (sqrt(3))/2