What is the value of sin(2(pi/6)) ?

The value of sin(2(pi/6)) is (sqrt(3))/2

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sin(2(pi/6))

sin (2 (\frac{π}{6}))

\frac{3}{2}

Decimal Notation

0.86602 \dots

Solution steps

sin (2 (\frac{π}{6}))

= sin (2 \cdot \frac{π}{6})

Simplify:

2 \cdot \frac{π}{6} = \frac{π}{3}

2 \cdot \frac{π}{6}

Multiply fractions:

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

= \frac{π 2}{6}

Cancel the common factor:

2

= \frac{π}{3}

= sin (\frac{π}{3})

Use the following trivial identity:

sin (\frac{π}{3}) = \frac{3}{2}

sin (\frac{π}{3})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} 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}

cos (\frac{23 π}{4})

sin (14 5^{\circ})

tan (\frac{- 11 π}{6})

\frac{1}{cos ( 3 3 ^{\circ} ) cos ( 1 2 ^{\circ} ) - sin ( 3 3 ^{\circ} ) sin ( 1 2 ^{\circ} )}

arcsin (\frac{19}{32})