Is sin((2pi)/3)=sin((-11pi)/3) ?

The answer to whether sin((2pi)/3)=sin((-11pi)/3) is True

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prove sin((2pi)/3)=sin((-11pi)/3)

Solution

prove

sin (\frac{2 π}{3}) = sin (\frac{- 11 π}{3})

Solution

True

Solution steps

sin (\frac{2 π}{3}) = sin (\frac{- 11 π}{3})

Manipulating left side

sin (\frac{2 π}{3})

Simplify

= \frac{3}{2}

Manipulating right side

sin (\frac{- 11 π}{3})

Simplify

sin (\frac{- 11 π}{3}) : \frac{3}{2}

sin (\frac{- 11 π}{3})

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= sin (- \frac{11 π}{3})

Use the following property:

sin (- x) = - sin (x)

sin (- \frac{11 π}{3}) = - sin (\frac{11 π}{3})

= - sin (\frac{11 π}{3})

sin (\frac{11 π}{3}) = sin (\frac{5 π}{3})

sin (\frac{11 π}{3})

Rewrite

\frac{11 π}{3}

2 π + \frac{5 π}{3}

= sin (2 π + \frac{5 π}{3})

Apply the periodicity of

sin

sin (x + 2 π) = sin (x)

sin (2 π + \frac{5 π}{3}) = sin (\frac{5 π}{3})

= sin (\frac{5 π}{3})

= - sin (\frac{5 π}{3})

Rewrite using trig identities:

sin (\frac{5 π}{3}) = - \frac{3}{2}

sin (\frac{5 π}{3})

Rewrite using trig identities:

sin (π) cos (\frac{2 π}{3}) + cos (π) sin (\frac{2 π}{3})

sin (\frac{5 π}{3})

Write

sin (\frac{5 π}{3})

sin (π + \frac{2 π}{3})

= sin (π + \frac{2 π}{3})

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (π) cos (\frac{2 π}{3}) + cos (π) sin (\frac{2 π}{3})

= sin (π) cos (\frac{2 π}{3}) + cos (π) sin (\frac{2 π}{3})

Use the following trivial identity:

sin (π) = 0

sin (π)

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

Use the following trivial identity:

cos (\frac{2 π}{3}) = - \frac{1}{2}

cos (\frac{2 π}{3})

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - \frac{1}{2}

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= (- 1)

Use the following trivial identity:

sin (\frac{2 π}{3}) = \frac{3}{2}

sin (\frac{2 π}{3})

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{3}{2}

= 0 \cdot (- \frac{1}{2}) + (- 1) \frac{3}{2}

Simplify

= - \frac{3}{2}

= - (- \frac{3}{2})

Simplify

= \frac{3}{2}

= \frac{3}{2}

We showed that the two sides could take the same form

\Rightarrow True

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Is sin((2pi)/3)=sin((-11pi)/3) ?
The answer to whether sin((2pi)/3)=sin((-11pi)/3) is True