What is the value of sin(140)cos(20)-cos(140)sin(20) ?

The value of sin(140)cos(20)-cos(140)sin(20) is (sqrt(3))/2

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sin(140)cos(20)-cos(140)sin(20)

sin (14 0^{\circ}) cos (2 0^{\circ}) - cos (14 0^{\circ}) sin (2 0^{\circ})

\frac{3}{2}

Decimal Notation

0.86602 \dots

Solution steps

sin (14 0^{\circ}) cos (2 0^{\circ}) - cos (14 0^{\circ}) sin (2 0^{\circ})

Rewrite using trig identities:

sin (12 0^{\circ})

sin (14 0^{\circ}) cos (2 0^{\circ}) - cos (14 0^{\circ}) sin (2 0^{\circ})

Use the Angle Difference identity:

sin (s) cos (t) - cos (s) sin (t) = sin (s - t)

= sin (14 0^{\circ} - 2 0^{\circ})

Simplify

= sin (12 0^{\circ})

= sin (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}

= \frac{3}{2}

tan (0 π)

arccos (0.69)

arccos (0.82)

arccos (0.81)

arccos (\frac{2}{14})

What is the value of sin(140)cos(20)-cos(140)sin(20) ?
The value of sin(140)cos(20)-cos(140)sin(20) is (sqrt(3))/2