What is the value of sin((17pi)/6) ?

The value of sin((17pi)/6) is 1/2

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sin((17pi)/6)

sin (\frac{17 π}{6})

\frac{1}{2}

Decimal Notation

0.5

Solution steps

sin (\frac{17 π}{6})

sin (\frac{17 π}{6}) = sin (\frac{5 π}{6})

sin (\frac{17 π}{6})

Rewrite

\frac{17 π}{6}

2 π + \frac{5 π}{6}

= sin (2 π + \frac{5 π}{6})

Apply the periodicity of

sin

sin (x + 2 π) = sin (x)

sin (2 π + \frac{5 π}{6}) = sin (\frac{5 π}{6})

= sin (\frac{5 π}{6})

= sin (\frac{5 π}{6})

Use the following trivial identity:

sin (\frac{5 π}{6}) = \frac{1}{2}

sin (\frac{5 π}{6})

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{1}{2}

= \frac{1}{2}

cos (2 θ) = \frac{1}{2}

tan (arcsin (- \frac{4}{5}))

sin^{2} (\frac{π}{3})

tan (x) = 3

csc (\frac{5 π}{4})