What is the value of sin(1/((\frac{2){9pi})}) ?

The value of sin(1/((\frac{2){9pi})}) is 1

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sin(1/((\frac{2){9pi})})

sin (\frac{1}{( \frac{2}{9 π} )})

1

Solution steps

sin (\frac{1}{( \frac{2}{9 π} )})

= sin (\frac{1}{\frac{2}{9 π}})

Simplify

= sin (\frac{9 π}{2})

sin (\frac{9 π}{2}) = sin (\frac{π}{2})

sin (\frac{9 π}{2})

Rewrite

\frac{9 π}{2}

2 π \cdot 2 + \frac{π}{2}

= sin (2 π 2 + \frac{π}{2})

Apply the periodicity of

sin

sin (x + 2 π \cdot k) = sin (x)

sin (2 π \cdot 2 + \frac{π}{2}) = sin (\frac{π}{2})

= sin (\frac{π}{2})

= sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

sin (\frac{π}{2})

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}

= 1

= 1

3 sin (24 0^{\circ})

\frac{4 sin ( 6 0 ^{\circ} )}{sin ( 3 0 ^{\circ} )}

arccos (\frac{3}{12})

sec (90 0^{\circ})

0 cos (6 0^{\circ})

What is the value of sin(1/((\frac{2){9pi})}) ?
The value of sin(1/((\frac{2){9pi})}) is 1