What is the value of sin((9pi)/4) ?

The value of sin((9pi)/4) is (sqrt(2))/2

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sin((9pi)/4)

sin (\frac{9 π}{4})

\frac{2}{2}

Decimal Notation

0.70710 \dots

Solution steps

sin (\frac{9 π}{4})

sin (\frac{9 π}{4}) = sin (\frac{π}{4})

sin (\frac{9 π}{4})

Rewrite

\frac{9 π}{4}

2 π + \frac{π}{4}

= sin (2 π + \frac{π}{4})

Apply the periodicity of

sin

sin (x + 2 π) = sin (x)

sin (2 π + \frac{π}{4}) = sin (\frac{π}{4})

= sin (\frac{π}{4})

= sin (\frac{π}{4})

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{2}{2}

sin (\frac{π}{4})

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{2}{2}

= \frac{2}{2}

sin (13 5^{\circ})

arcsec (2)

2 cos (\frac{x}{3}) - 3 = 0

2 csc^{2} (x) + 3 csc (x) - 2 = 0

2 sin^{2} (x) + 3 sin (x) + 1 = 0