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

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sin(-2pi)

sin (- 2 π)

0

Solution steps

sin (- 2 π)

Use the following property:

sin (- x) = - sin (x)

sin (- 2 π) = - sin (2 π)

= - sin (2 π)

sin (2 π) = sin (0)

sin (2 π)

Rewrite

2 π

2 π + 0

= sin (2 π + 0)

Apply the periodicity of

sin

sin (x + 2 π) = sin (x)

sin (2 π + 0) = sin (0)

= sin (0)

= - sin (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

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}

= 0

= - 0

Simplify

= 0

cos (2 x) = 1, 0 \leq x \leq 2 π

- sin (x) = - cos^{2} (x) - 1

2 sin (θ) - 1 = 0

sin (\frac{19 π}{12})

sin (36 0^{\circ})