Is tan((4pi)/3-pi/4)=2-sqrt(3) ?

The answer to whether tan((4pi)/3-pi/4)=2-sqrt(3) is True

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prove tan((4pi)/3-pi/4)=2-sqrt(3)

Solution

prove

tan (\frac{4 π}{3} - \frac{π}{4}) = 2 - 3

Solution

True

Solution steps

tan (\frac{4 π}{3} - \frac{π}{4}) = 2 - 3

Manipulating left side

tan (\frac{4 π}{3} - \frac{π}{4})

Rewrite using trig identities

tan (\frac{4 π}{3} - \frac{π}{4})

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{sin ( \frac{4 π}{3} - \frac{π}{4} )}{cos ( \frac{4 π}{3} - \frac{π}{4} )}

Use the Angle Difference identity:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

= \frac{sin ( \frac{4 π}{3} ) cos ( \frac{π}{4} ) - cos ( \frac{4 π}{3} ) sin ( \frac{π}{4} )}{cos ( \frac{4 π}{3} - \frac{π}{4} )}

Use the Angle Difference identity:

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

= \frac{sin ( \frac{4 π}{3} ) cos ( \frac{π}{4} ) - cos ( \frac{4 π}{3} ) sin ( \frac{π}{4} )}{cos ( \frac{4 π}{3} ) cos ( \frac{π}{4} ) + sin ( \frac{4 π}{3} ) sin ( \frac{π}{4} )}

\frac{sin ( \frac{4 π}{3} ) cos ( \frac{π}{4} ) - cos ( \frac{4 π}{3} ) sin ( \frac{π}{4} )}{cos ( \frac{4 π}{3} ) cos ( \frac{π}{4} ) + sin ( \frac{4 π}{3} ) sin ( \frac{π}{4} )} = 2 - 3

\frac{sin ( \frac{4 π}{3} ) cos ( \frac{π}{4} ) - cos ( \frac{4 π}{3} ) sin ( \frac{π}{4} )}{cos ( \frac{4 π}{3} ) cos ( \frac{π}{4} ) + sin ( \frac{4 π}{3} ) sin ( \frac{π}{4} )}

sin (\frac{4 π}{3}) cos (\frac{π}{4}) - cos (\frac{4 π}{3}) sin (\frac{π}{4}) = \frac{2}{2} \cdot \frac{- 3 + 1}{2}

sin (\frac{4 π}{3}) cos (\frac{π}{4}) - cos (\frac{4 π}{3}) sin (\frac{π}{4})

sin (\frac{4 π}{3}) cos (\frac{π}{4}) = - \frac{3}{2} \cdot \frac{2}{2}

sin (\frac{4 π}{3}) cos (\frac{π}{4})

sin (\frac{4 π}{3}) = - \frac{3}{2}

sin (\frac{4 π}{3})

Rewrite using trig identities:

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

sin (\frac{4 π}{3})

Write

sin (\frac{4 π}{3})

sin (π + \frac{π}{3})

= sin (π + \frac{π}{3})

Use the Angle Sum identity:

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

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

= sin (π) cos (\frac{π}{3}) + cos (π) sin (\frac{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{π}{3}) = \frac{1}{2}

cos (\frac{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{π}{3}) = \frac{3}{2}

sin (\frac{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} cos (\frac{π}{4})

Simplify

cos (\frac{π}{4}) : \frac{2}{2}

cos (\frac{π}{4})

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

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{2}{2}

= - \frac{3}{2} \cdot \frac{2}{2}

= - \frac{3}{2} \cdot \frac{2}{2} - sin (\frac{π}{4}) cos (\frac{4 π}{3})

cos (\frac{4 π}{3}) sin (\frac{π}{4}) = - \frac{1}{2} \cdot \frac{2}{2}

cos (\frac{4 π}{3}) sin (\frac{π}{4})

cos (\frac{4 π}{3}) = - \frac{1}{2}

cos (\frac{4 π}{3})

Rewrite using trig identities:

cos (π) cos (\frac{π}{3}) - sin (π) sin (\frac{π}{3})

cos (\frac{4 π}{3})

Write

cos (\frac{4 π}{3})

cos (π + \frac{π}{3})

= cos (π + \frac{π}{3})

Use the Angle Sum identity:

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

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

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

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:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{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:

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:

sin (\frac{π}{3}) = \frac{3}{2}

sin (\frac{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}

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

Simplify

= - \frac{1}{2}

= - \frac{1}{2} sin (\frac{π}{4})

Simplify

sin (\frac{π}{4}) : \frac{2}{2}

sin (\frac{π}{4})

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{2}{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}

= \frac{2}{2}

= - \frac{1}{2} \cdot \frac{2}{2}

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

Apply rule

- (- a) = a

= - \frac{3}{2} \cdot \frac{2}{2} + \frac{1}{2} \cdot \frac{2}{2}

Factor out common term

\frac{2}{2}

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

- \frac{3}{2} + \frac{1}{2} = \frac{- 3 + 1}{2}

- \frac{3}{2} + \frac{1}{2}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 3 + 1}{2}

= \frac{2}{2} \cdot \frac{1 - 3}{2}

= \frac{\frac{2}{2} \cdot \frac{1 - 3}{2}}{cos ( \frac{π}{4} ) cos ( \frac{4 π}{3} ) + sin ( \frac{π}{4} ) sin ( \frac{4 π}{3} )}

cos (\frac{4 π}{3}) cos (\frac{π}{4}) + sin (\frac{4 π}{3}) sin (\frac{π}{4}) = \frac{2}{2} \cdot \frac{- 1 - 3}{2}

cos (\frac{4 π}{3}) cos (\frac{π}{4}) + sin (\frac{4 π}{3}) sin (\frac{π}{4})

cos (\frac{4 π}{3}) cos (\frac{π}{4}) = - \frac{1}{2} \cdot \frac{2}{2}

cos (\frac{4 π}{3}) cos (\frac{π}{4})

cos (\frac{4 π}{3}) = - \frac{1}{2}

cos (\frac{4 π}{3})

Rewrite using trig identities:

cos (π) cos (\frac{π}{3}) - sin (π) sin (\frac{π}{3})

cos (\frac{4 π}{3})

Write

cos (\frac{4 π}{3})

cos (π + \frac{π}{3})

= cos (π + \frac{π}{3})

Use the Angle Sum identity:

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

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

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

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:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{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:

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:

sin (\frac{π}{3}) = \frac{3}{2}

sin (\frac{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}

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

Simplify

= - \frac{1}{2}

= - \frac{1}{2} cos (\frac{π}{4})

Simplify

cos (\frac{π}{4}) : \frac{2}{2}

cos (\frac{π}{4})

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

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{2}{2}

= - \frac{1}{2} \cdot \frac{2}{2}

= - \frac{1}{2} \cdot \frac{2}{2} + sin (\frac{π}{4}) sin (\frac{4 π}{3})

sin (\frac{4 π}{3}) sin (\frac{π}{4}) = - \frac{3}{2} \cdot \frac{2}{2}

sin (\frac{4 π}{3}) sin (\frac{π}{4})

sin (\frac{4 π}{3}) = - \frac{3}{2}

sin (\frac{4 π}{3})

Rewrite using trig identities:

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

sin (\frac{4 π}{3})

Write

sin (\frac{4 π}{3})

sin (π + \frac{π}{3})

= sin (π + \frac{π}{3})

Use the Angle Sum identity:

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

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

= sin (π) cos (\frac{π}{3}) + cos (π) sin (\frac{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{π}{3}) = \frac{1}{2}

cos (\frac{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{π}{3}) = \frac{3}{2}

sin (\frac{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} sin (\frac{π}{4})

Simplify

sin (\frac{π}{4}) : \frac{2}{2}

sin (\frac{π}{4})

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{2}{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}

= \frac{2}{2}

= - \frac{3}{2} \cdot \frac{2}{2}

= - \frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2}

Factor out common term

\frac{2}{2}

= \frac{2}{2} (- \frac{1}{2} - \frac{3}{2})

- \frac{1}{2} - \frac{3}{2} = \frac{- 1 - 3}{2}

- \frac{1}{2} - \frac{3}{2}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 - 3}{2}

= \frac{2}{2} \cdot \frac{- 1 - 3}{2}

= \frac{\frac{2}{2} \cdot \frac{1 - 3}{2}}{\frac{2}{2} \cdot \frac{- 1 - 3}{2}}

Simplify

\frac{\frac{- 3 + 1}{2} \cdot \frac{2}{2}}{\frac{- 1 - 3}{2} \cdot \frac{2}{2}}

Multiply

\frac{- 1 - 3}{2} \cdot \frac{2}{2} : \frac{- 1 - 3}{2 2}

\frac{- 1 - 3}{2} \cdot \frac{2}{2}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 ( - 1 - 3 )}{4}

Factor

4 : 2^{2}

Factor

4 = 2^{2}

= \frac{2 ( - 1 - 3 )}{2 ^{2}}

Cancel

\frac{( - 1 - 3 ) 2}{2 ^{2}} : \frac{- 1 - 3}{2 ^{\frac{3}{2}}}

\frac{( - 1 - 3 ) 2}{2 ^{2}}

Apply radical rule:

n a = a^{\frac{1}{n}}

2 = 2^{\frac{1}{2}}

= \frac{2 ^{\frac{1}{2}} ( - 1 - 3 )}{2 ^{2}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{1}{2}}}{2 ^{2}} = \frac{1}{2 ^{2 - \frac{1}{2}}}

= \frac{- 1 - 3}{2 ^{2 - \frac{1}{2}}}

Subtract the numbers:

2 - \frac{1}{2} = \frac{3}{2}

= \frac{- 1 - 3}{2 ^{\frac{3}{2}}}

= \frac{- 1 - 3}{2 ^{\frac{3}{2}}}

2^{\frac{3}{2}} = 22

2^{\frac{3}{2}}

2^{\frac{3}{2}} = 2^{1 + \frac{1}{2}}

= 2^{1 + \frac{1}{2}}

Apply exponent rule:

x^{a + b} = x^{a} x^{b}

= 2^{1} \cdot 2^{\frac{1}{2}}

Refine

= 22

= \frac{- 1 - 3}{2 2}

= \frac{\frac{2}{2} \cdot \frac{1 - 3}{2}}{\frac{- 1 - 3}{2 2}}

Multiply

\frac{- 3 + 1}{2} \cdot \frac{2}{2} : \frac{- 3 + 1}{2 2}

\frac{- 3 + 1}{2} \cdot \frac{2}{2}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 ( 1 - 3 )}{4}

Factor

4 : 2^{2}

Factor

4 = 2^{2}

= \frac{2 ( 1 - 3 )}{2 ^{2}}

Cancel

\frac{( - 3 + 1 ) 2}{2 ^{2}} : \frac{- 3 + 1}{2 ^{\frac{3}{2}}}

\frac{( - 3 + 1 ) 2}{2 ^{2}}

Apply radical rule:

n a = a^{\frac{1}{n}}

2 = 2^{\frac{1}{2}}

= \frac{2 ^{\frac{1}{2}} ( 1 - 3 )}{2 ^{2}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{1}{2}}}{2 ^{2}} = \frac{1}{2 ^{2 - \frac{1}{2}}}

= \frac{- 3 + 1}{2 ^{2 - \frac{1}{2}}}

Subtract the numbers:

2 - \frac{1}{2} = \frac{3}{2}

= \frac{- 3 + 1}{2 ^{\frac{3}{2}}}

= \frac{- 3 + 1}{2 ^{\frac{3}{2}}}

2^{\frac{3}{2}} = 22

2^{\frac{3}{2}}

2^{\frac{3}{2}} = 2^{1 + \frac{1}{2}}

= 2^{1 + \frac{1}{2}}

Apply exponent rule:

x^{a + b} = x^{a} x^{b}

= 2^{1} \cdot 2^{\frac{1}{2}}

Refine

= 22

= \frac{- 3 + 1}{2 2}

= \frac{\frac{- 3 + 1}{2 2}}{\frac{- 1 - 3}{2 2}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

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

Cancel the common factor:

2

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

Cancel the common factor:

2

= \frac{- 3 + 1}{- 1 - 3}

Apply the fraction rule:

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

\frac{- 3 + 1}{- 1 - 3} = \frac{- ( 3 - 1 )}{- ( 1 + 3 )}

= \frac{3 - 1}{1 + 3}

Rationalize

\frac{3 - 1}{1 + 3} : 2 - 3

\frac{3 - 1}{1 + 3}

Multiply by the conjugate

\frac{1 - 3}{1 - 3}

= \frac{( 3 - 1 ) ( 1 - 3 )}{( 1 + 3 ) ( 1 - 3 )}

(3 - 1) (1 - 3) = 23 - 4

(3 - 1) (1 - 3)

Apply FOIL method:

(a + b) (c + d) = a c + a d + b c + b d

a = 3, b = - 1, c = 1, d = - 3

= 3 \cdot 1 + 3 (- 3) + (- 1) \cdot 1 + (- 1) (- 3)

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

= 1 \cdot 3 - 33 - 1 \cdot 1 + 1 \cdot 3

Simplify

1 \cdot 3 - 33 - 1 \cdot 1 + 1 \cdot 3 : 23 - 4

1 \cdot 3 - 33 - 1 \cdot 1 + 1 \cdot 3

Add similar elements:

1 \cdot 3 + 1 \cdot 3 = 23

= 23 - 33 - 1 \cdot 1

Apply radical rule:

a a = a

33 = 3

= 23 - 3 - 1 \cdot 1

Multiply the numbers:

1 \cdot 1 = 1

= 23 - 3 - 1

Subtract the numbers:

- 3 - 1 = - 4

= 23 - 4

= 23 - 4

(1 + 3) (1 - 3) = - 2

(1 + 3) (1 - 3)

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = 3

= 1^{2} - (3)^{2}

Simplify

1^{2} - (3)^{2} : - 2

1^{2} - (3)^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - (3)^{2}

(3)^{2} = 3

(3)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

= 1 - 3

Subtract the numbers:

1 - 3 = - 2

= - 2

= - 2

= \frac{2 3 - 4}{- 2}

Apply the fraction rule:

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

23 - 4 = - (4 - 23)

= \frac{4 - 2 3}{2}

Factor

4 - 23 : 2 (2 - 3)

4 - 23

Rewrite as

= 2 \cdot 2 - 23

Factor out common term

2

= 2 (2 - 3)

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

Divide the numbers:

\frac{2}{2} = 1

= 2 - 3

= 2 - 3

= 2 - 3

= 2 - 3

= 2 - 3

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is tan((4pi)/3-pi/4)=2-sqrt(3) ?
The answer to whether tan((4pi)/3-pi/4)=2-sqrt(3) is True