What is the value of sin((13pi)/2) ?

The value of sin((13pi)/2) is 1

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sin((13pi)/2)

sin (\frac{13 π}{2})

1

Solution steps

sin (\frac{13 π}{2})

sin (\frac{13 π}{2}) = sin (\frac{π}{2})

sin (\frac{13 π}{2})

Rewrite

\frac{13 π}{2}

2 π \cdot 3 + \frac{π}{2}

= sin (2 π 3 + \frac{π}{2})

Apply the periodicity of

sin

sin (x + 2 π \cdot k) = sin (x)

sin (2 π \cdot 3 + \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

sec^{2} (\frac{2 π}{3})

sin (\frac{π}{6}) + cos (\frac{π}{6})

sin (\frac{π}{7})

arccot (- 4)

2 cos (\frac{7 π}{4})