What is the value of sin(6 pi/9) ?

The value of sin(6 pi/9) is (sqrt(3))/2

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sin(6 pi/9)

sin (6 \frac{π}{9})

\frac{3}{2}

Decimal Notation

0.86602 \dots

Solution steps

sin (6 \cdot \frac{π}{9})

Simplify:

6 \cdot \frac{π}{9} = \frac{2 π}{3}

6 \cdot \frac{π}{9}

Multiply fractions:

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

= \frac{π 6}{9}

Cancel the common factor:

3

= \frac{2 π}{3}

= sin (\frac{2 π}{3})

Use the following trivial identity:

sin (\frac{2 π}{3}) = \frac{3}{2}

sin (\frac{2 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{π}{5}) cos (\frac{2 π}{15}) - sin (\frac{π}{5}) sin (\frac{2 π}{15})

3 cos (5.3 π)

(2 (cos (\frac{π}{4}) + i sin (\frac{π}{4})))^{6}

cos (\frac{π}{2} - \frac{π}{6})

\frac{1}{sin ( 3 5 ^{\circ} )}