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Popular Trigonometry >

tan(-pi/4)<= tan(a/2)<= tan(pi/4)

Solution

tan (- \frac{π}{4}) \leq tan (\frac{a}{2}) \leq tan (\frac{π}{4})

Solution

2 πn \leq a \leq \frac{π}{2} + 2 πn or \frac{3 π}{2} + 2 πn \leq a < 2 π + 2 πn

Interval Notation

[2 πn, \frac{π}{2} + 2 πn] \cup [\frac{3 π}{2} + 2 πn, 2 π + 2 πn)

Decimal

2 πn \leq a \leq 1.57079 \dots + 2 πn or 4.71238 \dots + 2 πn \leq a < 6.28318 \dots + 2 πn

Solution steps

tan (- \frac{π}{4}) \leq tan (\frac{a}{2}) \leq tan (\frac{π}{4})

a \leq u \leq b

then

a \leq u and u \leq b

tan (- \frac{π}{4}) \leq tan (\frac{a}{2}) and tan (\frac{a}{2}) \leq tan (\frac{π}{4})

tan (- \frac{π}{4}) \leq tan (\frac{a}{2}) : - \frac{π}{2} + 2 πn \leq a < π + 2 πn

tan (- \frac{π}{4}) \leq tan (\frac{a}{2})

Switch sides

tan (\frac{a}{2}) \geq tan (- \frac{π}{4})

Simplify

tan (- \frac{π}{4}) : - 1

tan (- \frac{π}{4})

Use the following property:

tan (- x) = - tan (x)

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

= - tan (\frac{π}{4})

Use the following trivial identity:

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

= - 1

tan (x) \geq a

then

arctan (a) + πn \leq x < \frac{π}{2} + πn

arctan (- 1) + πn \leq \frac{a}{2} < \frac{π}{2} + πn

a \leq u < b

then

a \leq u and u < b

arctan (- 1) + πn \leq \frac{a}{2} and \frac{a}{2} < \frac{π}{2} + πn

arctan (- 1) + πn \leq \frac{a}{2} : a \geq - \frac{π}{2} + 2 πn

arctan (- 1) + πn \leq \frac{a}{2}

Switch sides

\frac{a}{2} \geq arctan (- 1) + πn

Simplify

arctan (- 1) + πn : - \frac{π}{4} + πn

arctan (- 1) + πn

arctan (- 1) = - \frac{π}{4}

arctan (- 1)

Use the following property:

arctan (- x) = - arctan (x)

arctan (- 1) = - arctan (1)

= - arctan (1)

Use the following trivial identity:

arctan (1) = \frac{π}{4}

arctan (1)

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{4}

= - \frac{π}{4}

= - \frac{π}{4} + πn

\frac{a}{2} \geq - \frac{π}{4} + πn

Multiply both sides by

2

\frac{a}{2} \geq - \frac{π}{4} + πn

Multiply both sides by

2

\frac{2 a}{2} \geq - 2 \cdot \frac{π}{4} + 2 πn

Simplify

\frac{2 a}{2} \geq - 2 \cdot \frac{π}{4} + 2 πn

Simplify

\frac{2 a}{2} : a

\frac{2 a}{2}

Divide the numbers:

\frac{2}{2} = 1

= a

Simplify

- 2 \cdot \frac{π}{4} + 2 πn : - \frac{π}{2} + 2 πn

- 2 \cdot \frac{π}{4} + 2 πn

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

Multiply fractions:

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

= \frac{π 2}{4}

Cancel the common factor:

2

= \frac{π}{2}

= - \frac{π}{2} + 2 πn

a \geq - \frac{π}{2} + 2 πn

a \geq - \frac{π}{2} + 2 πn

a \geq - \frac{π}{2} + 2 πn

\frac{a}{2} < \frac{π}{2} + πn : a < π + 2 πn

\frac{a}{2} < \frac{π}{2} + πn

Multiply both sides by

2

\frac{a}{2} < \frac{π}{2} + πn

Multiply both sides by

2

\frac{2 a}{2} < 2 \cdot \frac{π}{2} + 2 πn

Simplify

\frac{2 a}{2} < 2 \cdot \frac{π}{2} + 2 πn

Simplify

\frac{2 a}{2} : a

\frac{2 a}{2}

Divide the numbers:

\frac{2}{2} = 1

= a

Simplify

2 \cdot \frac{π}{2} + 2 πn : π + 2 πn

2 \cdot \frac{π}{2} + 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

Multiply fractions:

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

= \frac{π 2}{2}

Cancel the common factor:

2

= π

= π + 2 πn

a < π + 2 πn

a < π + 2 πn

a < π + 2 πn

Combine the intervals

a \geq - \frac{π}{2} + 2 πn and a < π + 2 πn

Merge Overlapping Intervals

- \frac{π}{2} + 2 πn \leq a < π + 2 πn

tan (\frac{a}{2}) \leq tan (\frac{π}{4}) : - π + 2 πn < a \leq \frac{π}{2} + 2 πn

tan (\frac{a}{2}) \leq tan (\frac{π}{4})

Use the following trivial identity:

tan (\frac{π}{4}) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

tan (\frac{a}{2}) \leq 1

tan (x) \leq a

then

- \frac{π}{2} + πn < x \leq arctan (a) + πn

- \frac{π}{2} + πn < \frac{a}{2} \leq arctan (1) + πn

a < u \leq b

then

a < u and u \leq b

- \frac{π}{2} + πn < \frac{a}{2} and \frac{a}{2} \leq arctan (1) + πn

- \frac{π}{2} + πn < \frac{a}{2} : a > - π + 2 πn

- \frac{π}{2} + πn < \frac{a}{2}

Switch sides

\frac{a}{2} > - \frac{π}{2} + πn

Multiply both sides by

2

\frac{a}{2} > - \frac{π}{2} + πn

Multiply both sides by

2

\frac{2 a}{2} > - 2 \cdot \frac{π}{2} + 2 πn

Simplify

\frac{2 a}{2} > - 2 \cdot \frac{π}{2} + 2 πn

Simplify

\frac{2 a}{2} : a

\frac{2 a}{2}

Divide the numbers:

\frac{2}{2} = 1

= a

Simplify

- 2 \cdot \frac{π}{2} + 2 πn : - π + 2 πn

- 2 \cdot \frac{π}{2} + 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

Multiply fractions:

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

= \frac{π 2}{2}

Cancel the common factor:

2

= π

= - π + 2 πn

a > - π + 2 πn

a > - π + 2 πn

a > - π + 2 πn

\frac{a}{2} \leq arctan (1) + πn : a \leq \frac{π}{2} + 2 πn

\frac{a}{2} \leq arctan (1) + πn

Simplify

arctan (1) + πn : \frac{π}{4} + πn

arctan (1) + πn

Use the following trivial identity:

arctan (1) = \frac{π}{4}

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{4} + πn

\frac{a}{2} \leq \frac{π}{4} + πn

Multiply both sides by

2

\frac{a}{2} \leq \frac{π}{4} + πn

Multiply both sides by

2

\frac{2 a}{2} \leq 2 \cdot \frac{π}{4} + 2 πn

Simplify

\frac{2 a}{2} \leq 2 \cdot \frac{π}{4} + 2 πn

Simplify

\frac{2 a}{2} : a

\frac{2 a}{2}

Divide the numbers:

\frac{2}{2} = 1

= a

Simplify

2 \cdot \frac{π}{4} + 2 πn : \frac{π}{2} + 2 πn

2 \cdot \frac{π}{4} + 2 πn

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

Multiply fractions:

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

= \frac{π 2}{4}

Cancel the common factor:

2

= \frac{π}{2}

= \frac{π}{2} + 2 πn

a \leq \frac{π}{2} + 2 πn

a \leq \frac{π}{2} + 2 πn

a \leq \frac{π}{2} + 2 πn

Combine the intervals

a > - π + 2 πn and a \leq \frac{π}{2} + 2 πn

Merge Overlapping Intervals

- π + 2 πn < a \leq \frac{π}{2} + 2 πn

Combine the intervals

- \frac{π}{2} + 2 πn \leq a < π + 2 πn and - π + 2 πn < a \leq \frac{π}{2} + 2 πn

Merge Overlapping Intervals

2 πn \leq a \leq \frac{π}{2} + 2 πn or \frac{3 π}{2} + 2 πn \leq a < 2 π + 2 πn

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