What is the value of sin(1500) ?

The value of sin(1500) is (sqrt(3))/2

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sin(1500)

sin (150 0^{\circ})

\frac{3}{2}

Decimal Notation

0.86602 \dots

Solution steps

sin (150 0^{\circ})

sin (150 0^{\circ}) = sin (6 0^{\circ})

sin (150 0^{\circ})

Rewrite

150 0^{\circ}

36 0^{\circ} \cdot 4 + 6 0^{\circ}

= sin (36 0^{\circ} 4 + 6 0^{\circ})

Apply the periodicity of

sin

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

sin (36 0^{\circ} \cdot 4 + 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}

\frac{20}{sin ( 3 8 ^{\circ} )}

sin^{2} (2 1^{\circ}) + sin^{2} (6 1^{\circ}) + sin^{2} (6 9^{\circ}) + sin^{2} (2 9^{\circ})

1 - sec^{2} (4)

tan (3.14 \cdot 1.5)

\frac{574 - 67 \cdot 9.8 ( sin ( 3 9 ^{\circ} ) + 0.39 \cdot cos ( 3 9 ^{\circ} ) )}{67}