What is the value of tan(345) ?

The value of tan(345) is -2+sqrt(3)

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Popular >

tan(345)

Solution

tan (34 5^{\circ})

Solution

- 2 + 3

Decimal Notation

- 0.26794 \dots

Solution steps

tan (34 5^{\circ})

tan (34 5^{\circ}) = tan (16 5^{\circ})

tan (34 5^{\circ})

Rewrite

34 5^{\circ}

18 0^{\circ} + 16 5^{\circ}

= tan (18 0^{\circ} + 16 5^{\circ})

Apply the periodicity of

tan

tan (x + 18 0^{\circ}) = tan (x)

tan (18 0^{\circ} + 16 5^{\circ}) = tan (16 5^{\circ})

= tan (16 5^{\circ})

= tan (16 5^{\circ})

Rewrite using trig identities:

\frac{tan ( 10 5 ^{\circ} ) + tan ( 6 0 ^{\circ} )}{1 - tan ( 10 5 ^{\circ} ) tan ( 6 0 ^{\circ} )}

tan (16 5^{\circ})

Write

tan (16 5^{\circ})

tan (10 5^{\circ} + 6 0^{\circ})

= tan (10 5^{\circ} + 6 0^{\circ})

Use the Angle Sum identity:

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

= \frac{tan ( 10 5 ^{\circ} ) + tan ( 6 0 ^{\circ} )}{1 - tan ( 10 5 ^{\circ} ) tan ( 6 0 ^{\circ} )}

= \frac{tan ( 10 5 ^{\circ} ) + tan ( 6 0 ^{\circ} )}{1 - tan ( 10 5 ^{\circ} ) tan ( 6 0 ^{\circ} )}

Rewrite using trig identities:

tan (10 5^{\circ}) = - 2 - 3

tan (10 5^{\circ})

Rewrite using trig identities:

\frac{tan ( 6 0 ^{\circ} ) + tan ( 4 5 ^{\circ} )}{1 - tan ( 6 0 ^{\circ} ) tan ( 4 5 ^{\circ} )}

tan (10 5^{\circ})

Write

tan (10 5^{\circ})

tan (6 0^{\circ} + 4 5^{\circ})

= tan (6 0^{\circ} + 4 5^{\circ})

Use the Angle Sum identity:

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

= \frac{tan ( 6 0 ^{\circ} ) + tan ( 4 5 ^{\circ} )}{1 - tan ( 6 0 ^{\circ} ) tan ( 4 5 ^{\circ} )}

= \frac{tan ( 6 0 ^{\circ} ) + tan ( 4 5 ^{\circ} )}{1 - tan ( 6 0 ^{\circ} ) tan ( 4 5 ^{\circ} )}

Use the following trivial identity:

tan (6 0^{\circ}) = 3

tan (6 0^{\circ})

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 3

Use the following trivial identity:

tan (4 5^{\circ}) = 1

tan (4 5^{\circ})

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} 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 (6 0^{\circ}) = 3

tan (6 0^{\circ})

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} 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

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Frequently Asked Questions (FAQ)

What is the value of tan(345) ?
The value of tan(345) is -2+sqrt(3)