What is the value of tan(54) ?

The value of tan(54) is ((3sqrt(10)+5sqrt(2))sqrt(5-\sqrt{5)})/(20)

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Popular >

tan(54)

Solution

tan (5 4^{\circ})

Solution

\frac{( 3 10 + 5 2 ) 5 - 5}{20}

Decimal Notation

1.37638 \dots

Solution steps

tan (5 4^{\circ})

Rewrite using trig identities:

\frac{sin ( 5 4 ^{\circ} )}{cos ( 5 4 ^{\circ} )}

tan (5 4^{\circ})

Use the basic trigonometric identity:

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

= \frac{sin ( 5 4 ^{\circ} )}{cos ( 5 4 ^{\circ} )}

= \frac{sin ( 5 4 ^{\circ} )}{cos ( 5 4 ^{\circ} )}

Rewrite using trig identities:

sin (5 4^{\circ}) = \frac{5 + 1}{4}

sin (5 4^{\circ})

Rewrite using trig identities:

cos (3 6^{\circ})

sin (5 4^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

= cos (9 0^{\circ} - 5 4^{\circ})

Simplify

= cos (3 6^{\circ})

= cos (3 6^{\circ})

Rewrite using trig identities:

\frac{5 + 1}{4}

cos (3 6^{\circ})

Show that:

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

Use the following product to sum identity:

2 sin (x) cos (y) = sin (x + y) - sin (x - y)

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = sin (5 4^{\circ}) - sin (1 8^{\circ})

Show that:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

Divide both sides by

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

Substitute

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

\frac{1}{2} = sin (5 4^{\circ}) - sin (1 8^{\circ})

sin (5 4^{\circ}) = cos (9 0^{\circ} - 5 4^{\circ})

\frac{1}{2} = cos (9 0^{\circ} - 5 4^{\circ}) - sin (1 8^{\circ})

\frac{1}{2} = cos (3 6^{\circ}) - sin (1 8^{\circ})

Show that:

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

Use the factorization rule:

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

a = cos (3 6^{\circ}) + sin (1 8^{\circ})

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = ((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) ((cos (3 6^{\circ}) + sin (1 8^{\circ})) - (cos (3 6^{\circ}) - sin (1 8^{\circ})))

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 2 (2 cos (3 6^{\circ}) sin (1 8^{\circ}))

Show that:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

Divide both sides by

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

Substitute

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 1

Substitute

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (\frac{1}{2})^{2} = 1

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} = 1

Add

\frac{1}{4}

to both sides

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} = \frac{5}{4}

Take the square root of both sides

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \pm \frac{5}{4}

cos (3 6^{\circ})

cannot be negative

sin (1 8^{\circ})

cannot be negative

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

Add the following equations

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{2}

((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) = (\frac{5}{2} + \frac{1}{2})

Refine

cos (3 6^{\circ}) = \frac{5 + 1}{4}

= \frac{5 + 1}{4}

= \frac{5 + 1}{4}

Rewrite using trig identities:

cos (5 4^{\circ}) = \frac{2 5 - 5}{4}

cos (5 4^{\circ})

Rewrite using trig identities:

sin (3 6^{\circ})

cos (5 4^{\circ})

Use the following identity:

cos (x) = sin (9 0^{\circ} - x)

= sin (9 0^{\circ} - 5 4^{\circ})

Simplify

= sin (3 6^{\circ})

= sin (3 6^{\circ})

Rewrite using trig identities:

\frac{2 5 - 5}{4}

sin (3 6^{\circ})

Show that:

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

Use the following product to sum identity:

2 sin (x) cos (y) = sin (x + y) - sin (x - y)

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = sin (5 4^{\circ}) - sin (1 8^{\circ})

Show that:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

Divide both sides by

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

Substitute

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

\frac{1}{2} = sin (5 4^{\circ}) - sin (1 8^{\circ})

sin (5 4^{\circ}) = cos (9 0^{\circ} - 5 4^{\circ})

\frac{1}{2} = cos (9 0^{\circ} - 5 4^{\circ}) - sin (1 8^{\circ})

\frac{1}{2} = cos (3 6^{\circ}) - sin (1 8^{\circ})

Show that:

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

Use the factorization rule:

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

a = cos (3 6^{\circ}) + sin (1 8^{\circ})

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = ((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) ((cos (3 6^{\circ}) + sin (1 8^{\circ})) - (cos (3 6^{\circ}) - sin (1 8^{\circ})))

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 2 (2 cos (3 6^{\circ}) sin (1 8^{\circ}))

Show that:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

Divide both sides by

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

Substitute

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 1

Substitute

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (\frac{1}{2})^{2} = 1

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} = 1

Add

\frac{1}{4}

to both sides

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} = \frac{5}{4}

Take the square root of both sides

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \pm \frac{5}{4}

cos (3 6^{\circ})

cannot be negative

sin (1 8^{\circ})

cannot be negative

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

Add the following equations

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{2}

((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) = (\frac{5}{2} + \frac{1}{2})

Refine

cos (3 6^{\circ}) = \frac{5 + 1}{4}

Square both sides

(cos (3 6^{\circ}))^{2} = (\frac{5 + 1}{4})^{2}

Use the following identity:

sin^{2} (x) = 1 - cos^{2} (x)

sin^{2} (3 6^{\circ}) = 1 - cos^{2} (3 6^{\circ})

Substitute

cos (3 6^{\circ}) = \frac{5 + 1}{4}

sin^{2} (3 6^{\circ}) = 1 - (\frac{5 + 1}{4})^{2}

Refine

sin^{2} (3 6^{\circ}) = \frac{5 - 5}{8}

Take the square root of both sides

sin (3 6^{\circ}) = \pm \frac{5 - 5}{8}

sin (3 6^{\circ})

cannot be negative

sin (3 6^{\circ}) = \frac{5 - 5}{8}

Refine

sin (3 6^{\circ}) = \frac{\frac{5 - 5}{2}}{2}

= \frac{\frac{5 - 5}{2}}{2}

Simplify

= \frac{2 5 - 5}{4}

= \frac{2 5 - 5}{4}

= \frac{\frac{5 + 1}{4}}{\frac{2 5 - 5}{4}}

Simplify

\frac{\frac{5 + 1}{4}}{\frac{2 5 - 5}{4}} : \frac{( 3 10 + 5 2 ) 5 - 5}{20}

\frac{\frac{5 + 1}{4}}{\frac{2 5 - 5}{4}}

Divide fractions:

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

= \frac{( 5 + 1 ) \cdot 4}{4 2 5 - 5}

Cancel the common factor:

4

= \frac{5 + 1}{2 5 - 5}

Rationalize

\frac{5 + 1}{2 5 - 5} : \frac{( 3 10 + 5 2 ) 5 - 5}{20}

\frac{5 + 1}{2 5 - 5}

Multiply by the conjugate

\frac{2}{2}

= \frac{( 5 + 1 ) 2}{2 5 - 5 2}

2 5 - 5 2 = 2 5 - 5

2 5 - 5 2

Apply radical rule:

a a = a

22 = 2

= 2 5 - 5

= \frac{2 ( 5 + 1 )}{2 5 - 5}

Multiply by the conjugate

\frac{5 - 5}{5 - 5}

= \frac{2 ( 5 + 1 ) 5 - 5}{2 5 - 5 5 - 5}

2 5 - 5 5 - 5 = 10 - 25

2 5 - 5 5 - 5

Apply radical rule:

a a = a

5 - 5 5 - 5 = 5 - 5

= 2 (5 - 5)

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 5, c = 5

= 2 \cdot 5 - 25

Multiply the numbers:

2 \cdot 5 = 10

= 10 - 25

= \frac{2 ( 5 + 1 ) 5 - 5}{10 - 2 5}

Factor out common term

- 2 : - 2 (5 - 5)

- 25 + 10

Rewrite

10

2 \cdot 5

= - 25 + 2 \cdot 5

Factor out common term

- 2

= - 2 (5 - 5)

= - \frac{2 ( 5 + 1 ) 5 - 5}{2 ( 5 - 5 )}

Cancel

- \frac{2 ( 5 + 1 ) 5 - 5}{2 ( 5 - 5 )} : \frac{2 ( 5 + 1 ) 5 - 5}{2 ( 5 - 5 )}

- \frac{2 ( 5 + 1 ) 5 - 5}{2 ( 5 - 5 )}

5 - 5 = - (5 - 5)

= - \frac{2 ( 1 + 5 ) 5 - 5}{- 2 ( 5 - 5 )}

Refine

= \frac{2 ( 5 + 1 ) 5 - 5}{2 ( 5 - 5 )}

= \frac{2 ( 5 + 1 ) 5 - 5}{2 ( 5 - 5 )}

Multiply by the conjugate

\frac{5 + 5}{5 + 5}

= \frac{2 ( 5 + 1 ) 5 - 5 ( 5 + 5 )}{2 ( 5 - 5 ) ( 5 + 5 )}

2 (5 + 1) 5 - 5 (5 + 5) = 610 5 - 5 + 102 5 - 5

2 (5 + 1) 5 - 5 (5 + 5)

= 2 (5 + 1) (5 + 5) 5 - 5

Expand

(5 + 1) (5 + 5) : 65 + 10

(5 + 1) (5 + 5)

Apply FOIL method:

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

a = 5, b = 1, c = 5, d = 5

= 5 \cdot 5 + 55 + 1 \cdot 5 + 1 \cdot 5

= 55 + 55 + 1 \cdot 5 + 1 \cdot 5

Simplify

55 + 55 + 1 \cdot 5 + 1 \cdot 5 : 65 + 10

55 + 55 + 1 \cdot 5 + 1 \cdot 5

Add similar elements:

55 + 1 \cdot 5 = 65

= 65 + 55 + 1 \cdot 5

Apply radical rule:

a a = a

55 = 5

= 65 + 5 + 1 \cdot 5

Multiply the numbers:

1 \cdot 5 = 5

= 65 + 5 + 5

Add the numbers:

5 + 5 = 10

= 65 + 10

= 65 + 10

= 2 5 - 5 (65 + 10)

Expand

2 5 - 5 (65 + 10) : 610 5 - 5 + 102 5 - 5

2 5 - 5 (65 + 10)

Apply the distributive law:

a (b + c) = ab + a c

a = 2 5 - 5, b = 65, c = 10

= 2 5 - 5 \cdot 65 + 2 5 - 5 \cdot 10

= 625 5 - 5 + 102 5 - 5

625 5 - 5 = 610 5 - 5

625 5 - 5

Apply radical rule:

a b = a \cdot b

25 5 - 5 = 2 \cdot 5 (5 - 5)

= 6 2 \cdot 5 (5 - 5)

Multiply the numbers:

2 \cdot 5 = 10

= 6 10 (5 - 5)

Apply radical rule:

assuming

a \geq 0, b \geq 0

10 (5 - 5) = 10 5 - 5

= 610 5 - 5

= 610 5 - 5 + 102 5 - 5

= 610 5 - 5 + 102 5 - 5

2 (5 - 5) (5 + 5) = 40

2 (5 - 5) (5 + 5)

Expand

(5 - 5) (5 + 5) : 20

(5 - 5) (5 + 5)

Apply Difference of Two Squares Formula:

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

a = 5, b = 5

= 5^{2} - (5)^{2}

Simplify

5^{2} - (5)^{2} : 20

5^{2} - (5)^{2}

5^{2} = 25

5^{2}

5^{2} = 25

= 25

(5)^{2} = 5

(5)^{2}

Apply radical rule:

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

= (5^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 5^{\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

= 5

= 25 - 5

Subtract the numbers:

25 - 5 = 20

= 20

= 20

= 2 \cdot 20

Expand

2 \cdot 20 : 40

2 \cdot 20

Distribute parentheses

= 2 \cdot 20

Multiply the numbers:

2 \cdot 20 = 40

= 40

= 40

= \frac{6 10 5 - 5 + 10 2 5 - 5}{40}

Factor

610 5 - 5 + 102 5 - 5 : 2 5 - 5 (310 + 52)

610 5 - 5 + 102 5 - 5

Rewrite as

= 3 \cdot 2 5 - 5 10 + 5 \cdot 2 5 - 5 2

Factor out common term

2 5 - 5

= 2 5 - 5 (310 + 52)

= \frac{2 5 - 5 ( 3 10 + 5 2 )}{40}

Cancel the common factor:

2

= \frac{( 3 10 + 5 2 ) 5 - 5}{20}

= \frac{( 3 10 + 5 2 ) 5 - 5}{20}

= \frac{( 3 10 + 5 2 ) 5 - 5}{20}

Popular Examples

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cos (157. 5^{\circ})

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tan (arccos (- \frac{4}{5}))

sin((-pi)/3)

sin (\frac{- π}{3})

sin((-pi)/6)

sin (\frac{- π}{6})

(sin(pi))/2

\frac{sin ( π )}{2}

Frequently Asked Questions (FAQ)

What is the value of tan(54) ?
The value of tan(54) is ((3sqrt(10)+5sqrt(2))sqrt(5-\sqrt{5)})/(20)