What is the value of cot((6pi)/5) ?

The value of cot((6pi)/5) is ((3sqrt(10)+5sqrt(2))sqrt(5-\sqrt{5)})/(20)

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cot((6pi)/5)

Solution

cot (\frac{6 π}{5})

Solution

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

Decimal Notation

1.37638 \dots

Solution steps

cot (\frac{6 π}{5})

cot (\frac{6 π}{5}) = cot (\frac{π}{5})

cot (\frac{6 π}{5})

Rewrite

\frac{6 π}{5}

π + \frac{π}{5}

= cot (π + \frac{π}{5})

Apply the periodicity of

cot

cot (x + π) = cot (x)

cot (π + \frac{π}{5}) = cot (\frac{π}{5})

= cot (\frac{π}{5})

= cot (\frac{π}{5})

Rewrite using trig identities:

\frac{cos ( \frac{π}{5} )}{sin ( \frac{π}{5} )}

cot (\frac{π}{5})

Use the basic trigonometric identity:

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

= \frac{cos ( \frac{π}{5} )}{sin ( \frac{π}{5} )}

= \frac{cos ( \frac{π}{5} )}{sin ( \frac{π}{5} )}

Rewrite using trig identities:

cos (\frac{π}{5}) = \frac{5 + 1}{4}

cos (\frac{π}{5})

Show that:

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

Use the following product to sum identity:

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

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

Show that:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

Use the Double Angle identity:

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

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Divide both sides by

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Use the following identity:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Divide both sides by

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

Divide both sides by

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

Substitute

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

\frac{1}{2} = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

sin (\frac{3 π}{10}) = cos (\frac{π}{2} - \frac{3 π}{10})

\frac{1}{2} = cos (\frac{π}{2} - \frac{3 π}{10}) - sin (\frac{π}{10})

\frac{1}{2} = cos (\frac{π}{5}) - sin (\frac{π}{10})

Show that:

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

Use the factorization rule:

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

a = cos (\frac{π}{5}) + sin (\frac{π}{10})

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = ((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) ((cos (\frac{π}{5}) + sin (\frac{π}{10})) - (cos (\frac{π}{5}) - sin (\frac{π}{10})))

Refine

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 2 (2 cos (\frac{π}{5}) sin (\frac{π}{10}))

Show that:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

Use the Double Angle identity:

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

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Divide both sides by

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Use the following identity:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Divide both sides by

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

Divide both sides by

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

Substitute

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 1

Substitute

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (\frac{1}{2})^{2} = 1

Refine

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} = 1

Add

\frac{1}{4}

to both sides

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

Refine

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} = \frac{5}{4}

Take the square root of both sides

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \pm \frac{5}{4}

cos (\frac{π}{5})

cannot be negative

sin (\frac{π}{10})

cannot be negative

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

Add the following equations

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{2}

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

Refine

cos (\frac{π}{5}) = \frac{5 + 1}{4}

= \frac{5 + 1}{4}

Rewrite using trig identities:

sin (\frac{π}{5}) = \frac{2 5 - 5}{4}

sin (\frac{π}{5})

Show that:

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

Use the following product to sum identity:

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

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

Show that:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

Use the Double Angle identity:

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

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Divide both sides by

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Use the following identity:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Divide both sides by

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

Divide both sides by

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

Substitute

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

\frac{1}{2} = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

sin (\frac{3 π}{10}) = cos (\frac{π}{2} - \frac{3 π}{10})

\frac{1}{2} = cos (\frac{π}{2} - \frac{3 π}{10}) - sin (\frac{π}{10})

\frac{1}{2} = cos (\frac{π}{5}) - sin (\frac{π}{10})

Show that:

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

Use the factorization rule:

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

a = cos (\frac{π}{5}) + sin (\frac{π}{10})

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = ((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) ((cos (\frac{π}{5}) + sin (\frac{π}{10})) - (cos (\frac{π}{5}) - sin (\frac{π}{10})))

Refine

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 2 (2 cos (\frac{π}{5}) sin (\frac{π}{10}))

Show that:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

Use the Double Angle identity:

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

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Divide both sides by

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Use the following identity:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

Divide both sides by

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

Divide both sides by

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

Substitute

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 1

Substitute

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (\frac{1}{2})^{2} = 1

Refine

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} = 1

Add

\frac{1}{4}

to both sides

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

Refine

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} = \frac{5}{4}

Take the square root of both sides

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \pm \frac{5}{4}

cos (\frac{π}{5})

cannot be negative

sin (\frac{π}{10})

cannot be negative

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

Add the following equations

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{2}

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

Refine

cos (\frac{π}{5}) = \frac{5 + 1}{4}

Square both sides

(cos (\frac{π}{5}))^{2} = (\frac{5 + 1}{4})^{2}

Use the following identity:

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

sin^{2} (\frac{π}{5}) = 1 - cos^{2} (\frac{π}{5})

Substitute

cos (\frac{π}{5}) = \frac{5 + 1}{4}

sin^{2} (\frac{π}{5}) = 1 - (\frac{5 + 1}{4})^{2}

Refine

sin^{2} (\frac{π}{5}) = \frac{5 - 5}{8}

Take the square root of both sides

sin (\frac{π}{5}) = \pm \frac{5 - 5}{8}

sin (\frac{π}{5})

cannot be negative

sin (\frac{π}{5}) = \frac{5 - 5}{8}

Refine

sin (\frac{π}{5}) = \frac{\frac{5 - 5}{2}}{2}

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

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

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

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

\frac{5 - 5}{2}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{5 - 5}{2}

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

Apply the fraction rule:

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

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

Rationalize

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

\frac{5 - 5}{2 2}

Multiply by the conjugate

\frac{2}{2}

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

2 \cdot 22 = 4

2 \cdot 22

Apply exponent rule:

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

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

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

Add similar elements:

\frac{1}{2} + \frac{1}{2} = 2 \cdot \frac{1}{2}

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

2 \cdot \frac{1}{2} = 1

2 \cdot \frac{1}{2}

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 2^{1 + 1}

Add the numbers:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

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

= \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}

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Frequently Asked Questions (FAQ)

What is the value of cot((6pi)/5) ?
The value of cot((6pi)/5) is ((3sqrt(10)+5sqrt(2))sqrt(5-\sqrt{5)})/(20)