What is the value of sin(420) ?

The value of sin(420) is (sqrt(3))/2

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sin(420)

sin (42 0^{\circ})

\frac{3}{2}

Decimal Notation

0.86602 \dots

Solution steps

sin (42 0^{\circ})

sin (42 0^{\circ}) = sin (6 0^{\circ})

sin (42 0^{\circ})

Rewrite

42 0^{\circ}

36 0^{\circ} + 6 0^{\circ}

= sin (36 0^{\circ} + 6 0^{\circ})

Apply the periodicity of

sin

sin (x + 36 0^{\circ}) = sin (x)

sin (36 0^{\circ} + 6 0^{\circ}) = sin (6 0^{\circ})

= sin (6 0^{\circ})

= sin (6 0^{\circ})

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}

tan (- 9 0^{\circ})

cos (4 5^{\circ} - 6 0^{\circ})

tan (6)

arcsin (5)

3 cos (π)