What is the value of sin(120)+cos(120) ?

The value of sin(120)+cos(120) is (sqrt(3)-1)/2

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sin(120)+cos(120)

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

\frac{3 - 1}{2}

Decimal Notation

0.36602 \dots

Solution steps

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

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

Simplify

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

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

Apply rule

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

= \frac{3 - 1}{2}

= \frac{3 - 1}{2}

arctan (\frac{- 5}{- 3})

sin (59.4 9^{\circ})

arccos (0.9235)

cos (31 5^{\circ}) - sin (24 0^{\circ}) + sin (22 5^{\circ})

3.8 cos (\frac{18 π}{13} - \frac{12 π}{13})

What is the value of sin(120)+cos(120) ?
The value of sin(120)+cos(120) is (sqrt(3)-1)/2