What is the value of tan((19pi)/(12)) ?

The value of tan((19pi)/(12)) is -2-sqrt(3)

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tan((19pi)/(12))

Solution

tan (\frac{19 π}{12})

Solution

- 2 - 3

Decimal Notation

- 3.73205 \dots

Solution steps

tan (\frac{19 π}{12})

tan (\frac{19 π}{12}) = tan (\frac{7 π}{12})

tan (\frac{19 π}{12})

Rewrite

\frac{19 π}{12}

π + \frac{7 π}{12}

= tan (π + \frac{7 π}{12})

Apply the periodicity of

tan

tan (x + π) = tan (x)

tan (π + \frac{7 π}{12}) = tan (\frac{7 π}{12})

= tan (\frac{7 π}{12})

= tan (\frac{7 π}{12})

Rewrite using trig identities:

\frac{tan ( \frac{π}{3} ) + tan ( \frac{π}{4} )}{1 - tan ( \frac{π}{3} ) tan ( \frac{π}{4} )}

tan (\frac{7 π}{12})

Write

tan (\frac{7 π}{12})

tan (\frac{π}{3} + \frac{π}{4})

= tan (\frac{π}{3} + \frac{π}{4})

Use the Angle Sum identity:

tan (s + t) = \frac{tan ( s ) + tan ( t )}{1 - tan ( s ) tan ( t )}

= \frac{tan ( \frac{π}{3} ) + tan ( \frac{π}{4} )}{1 - tan ( \frac{π}{3} ) tan ( \frac{π}{4} )}

= \frac{tan ( \frac{π}{3} ) + tan ( \frac{π}{4} )}{1 - tan ( \frac{π}{3} ) tan ( \frac{π}{4} )}

Use the following trivial identity:

tan (\frac{π}{3}) = 3

tan (\frac{π}{3})

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}

= 3

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

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

Simplify

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

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

Multiply:

3 \cdot 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) = 4 + 23

(3 + 1) (1 + 3)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

(3 + 1) (1 + 3) = (3 + 1)^{1 + 1}

= (3 + 1)^{1 + 1}

Add the numbers:

1 + 1 = 2

= (3 + 1)^{2}

Apply Perfect Square Formula:

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

a = 3, b = 1

= (3)^{2} + 23 \cdot 1 + 1^{2}

Simplify

(3)^{2} + 23 \cdot 1 + 1^{2} : 4 + 23

(3)^{2} + 23 \cdot 1 + 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= (3)^{2} + 2 \cdot 1 \cdot 3 + 1

(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

23 \cdot 1 = 23

23 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 23

= 3 + 23 + 1

Add the numbers:

3 + 1 = 4

= 4 + 23

= 4 + 23

(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{4 + 2 3}{- 2}

Apply the fraction rule:

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

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

Cancel

\frac{4 + 2 3}{2} : 2 + 3

\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)

Distribute parentheses

= - (2) - (3)

Apply minus-plus rules

+ (- a) = - a

= - 2 - 3

= - 2 - 3

= - 2 - 3

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Frequently Asked Questions (FAQ)

What is the value of tan((19pi)/(12)) ?
The value of tan((19pi)/(12)) is -2-sqrt(3)