What is the value of (sin(2pi))/3 ?

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(sin(2pi))/3

\frac{sin ( 2 π )}{3}

0

Solution steps

\frac{sin ( 2 π )}{3}

Rewrite using trig identities:

sin (2 π) = 0

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

= \frac{0}{3}

Simplify

= 0

cos (arctan (- \frac{5}{12}))

arcsin (\frac{10}{15})

sin (\frac{π}{18}) cos (\frac{7 π}{36}) + cos (\frac{π}{18}) sin (\frac{7 π}{36})

4 \cdot cos (4 5^{\circ})

\frac{tan ( 7 9 ^{\circ} ) - tan ( 1 9 ^{\circ} )}{1 + tan ( 7 9 ^{\circ} ) tan ( 1 9 ^{\circ} )}