Is tan(11/12 pi)=tan(-pi/(12)) ?

The answer to whether tan(11/12 pi)=tan(-pi/(12)) is True

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prove tan(11/12 pi)=tan(-pi/(12))

Solution

prove

tan (\frac{11}{12} π) = tan (- \frac{π}{12})

Solution

True

Solution steps

tan (\frac{11}{12} π) = tan (- \frac{π}{12})

Manipulating left side

tan (\frac{11}{12} π)

Simplify

tan (\frac{11}{12} π) : - 2 + 3

tan (\frac{11}{12} π)

Multiply

\frac{11}{12} π : \frac{11 π}{12}

\frac{11}{12} π

Multiply fractions:

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

= \frac{11 π}{12}

= tan (\frac{11 π}{12})

Rewrite using trig identities:

\frac{tan ( \frac{7 π}{12} ) + tan ( \frac{π}{3} )}{1 - tan ( \frac{7 π}{12} ) tan ( \frac{π}{3} )}

tan (\frac{11 π}{12})

Write

tan (\frac{11 π}{12})

tan (\frac{7 π}{12} + \frac{π}{3})

= tan (\frac{7 π}{12} + \frac{π}{3})

Use the Angle Sum identity:

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

= \frac{tan ( \frac{7 π}{12} ) + tan ( \frac{π}{3} )}{1 - tan ( \frac{7 π}{12} ) tan ( \frac{π}{3} )}

= \frac{tan ( \frac{7 π}{12} ) + tan ( \frac{π}{3} )}{1 - tan ( \frac{7 π}{12} ) tan ( \frac{π}{3} )}

Rewrite using trig identities:

tan (\frac{7 π}{12}) = - 2 - 3

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

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

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

Simplify

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

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

Add similar elements:

- 3 + 3 = 0

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

Apply the fraction rule:

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

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

Expand

1 - (- 2 - 3) 3 : 4 + 23

1 - (- 2 - 3) 3

= 1 - 3 (- 2 - 3)

Expand

- 3 (- 2 - 3) : 23 + 3

- 3 (- 2 - 3)

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = - 2, c = 3

= - 3 (- 2) - (- 3) 3

Apply minus-plus rules

- (- a) = a

= 23 + 33

Apply radical rule:

a a = a

33 = 3

= 23 + 3

= 1 + 23 + 3

Add the numbers:

1 + 3 = 4

= 4 + 23

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

Cancel

\frac{2}{4 + 2 3} : \frac{1}{( 2 + 3 )}

\frac{2}{4 + 2 3}

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 ( 2 + 3 )}

Divide the numbers:

\frac{2}{2} = 1

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

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

Remove parentheses:

(a) = a

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

Rationalize

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

- \frac{1}{2 + 3}

Multiply by the conjugate

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

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

1 \cdot (2 - 3) = 2 - 3

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

(2 + 3) (2 - 3)

Apply Difference of Two Squares Formula:

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

a = 2, b = 3

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

Simplify

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

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

2^{2} = 4

2^{2}

2^{2} = 4

= 4

(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

= 4 - 3

Subtract the numbers:

4 - 3 = 1

= 1

= 1

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

Apply rule

\frac{a}{1} = a

= - (2 - 3)

Distribute parentheses

= - (2) - (- 3)

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 2 + 3

= - 2 + 3

= - 2 + 3

= - 2 + 3

Manipulating right side

tan (- \frac{π}{12})

Use the negative angle identity:

tan (- x) = - tan (x)

= - tan (\frac{π}{12})

Simplify

- tan (\frac{π}{12}) : - 2 + 3

- tan (\frac{π}{12})

tan (\frac{π}{12}) = 2 - 3

tan (\frac{π}{12})

Rewrite using trig identities:

\frac{tan ( \frac{π}{4} ) - tan ( \frac{π}{6} )}{1 + tan ( \frac{π}{4} ) tan ( \frac{π}{6} )}

tan (\frac{π}{12})

Write

tan (\frac{π}{12})

tan (\frac{π}{4} - \frac{π}{6})

= tan (\frac{π}{4} - \frac{π}{6})

Use the Angle Difference identity:

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

= \frac{tan ( \frac{π}{4} ) - tan ( \frac{π}{6} )}{1 + tan ( \frac{π}{4} ) tan ( \frac{π}{6} )}

= \frac{tan ( \frac{π}{4} ) - tan ( \frac{π}{6} )}{1 + tan ( \frac{π}{4} ) tan ( \frac{π}{6} )}

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

Use the following trivial identity:

tan (\frac{π}{6}) = \frac{3}{3}

tan (\frac{π}{6})

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}

= \frac{3}{3}

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

Simplify

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

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

Multiply:

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

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

Join

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

1 + \frac{3}{3}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3 + 3}{3}

Factor

3 + 3 : 3 (3 + 1)

3 + 3

3 = 33

= 33 + 3

Factor out common term

3

= 3 (3 + 1)

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

Cancel

\frac{3 ( 3 + 1 )}{3} : \frac{3 + 1}{3}

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

Apply radical rule:

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

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

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

Apply exponent rule:

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

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

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

Subtract the numbers:

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

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

Apply radical rule:

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

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

= \frac{3 + 1}{3}

= \frac{3 + 1}{3}

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

Join

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

1 - \frac{3}{3}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3 - 3}{3}

Factor

3 - 3 : 3 (3 - 1)

3 - 3

3 = 33

= 33 - 3

Factor out common term

3

= 3 (3 - 1)

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

Cancel

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

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

Apply radical rule:

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

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

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

Apply exponent rule:

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

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

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

Subtract the numbers:

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

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

Apply radical rule:

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

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

= \frac{3 - 1}{3}

= \frac{3 - 1}{3}

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

Divide fractions:

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

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

Cancel the common factor:

3

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

Rationalize

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

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

Multiply by the conjugate

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

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

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

(3 - 1) (3 - 1)

Apply exponent rule:

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

(3 - 1) (3 - 1) = (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

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

(3 + 1) (3 - 1)

Apply Difference of Two Squares Formula:

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

a = 3, b = 1

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= (3)^{2} - 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

= 3 - 1

Subtract the numbers:

3 - 1 = 2

= 2

= 2

= \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)

Simplify

- (2 - 3)

Distribute parentheses

= - (2) - (- 3)

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 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(11/12 pi)=tan(-pi/(12)) ?
The answer to whether tan(11/12 pi)=tan(-pi/(12)) is True