What is the value of sin((-9pi)/2) ?

The value of sin((-9pi)/2) is -1

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sin((-9pi)/2)

sin (\frac{- 9 π}{2})

- 1

Solution steps

sin (\frac{- 9 π}{2})

Simplify

= sin (- \frac{9 π}{2})

Use the following property:

sin (- x) = - sin (x)

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

= - 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

\frac{sin ( 3 0 ^{\circ} )}{sin ( 19. 3 ^{\circ} )}

3 sin (6)

arcsin (0.225)

arcsin (8 \cdot \frac{sin ( 6 0 ^{\circ} )}{7})

(\frac{1 \cdot ( 9.81 )}{cos ( 2 7 ^{\circ} )}) \cdot sin (2 7^{\circ})