What is the value of sin((8pi)/5) ?

The value of sin((8pi)/5) is -(sqrt(2)sqrt(5+\sqrt{5)})/4

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Popular >

sin((8pi)/5)

Solution

sin (\frac{8 π}{5})

Solution

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

Decimal Notation

- 0.95105 \dots

Solution steps

sin (\frac{8 π}{5})

Rewrite using trig identities:

cos (\frac{11 π}{10})

sin (\frac{8 π}{5})

Use the following identity:

sin (x) = cos (\frac{π}{2} - x)

= cos (\frac{π}{2} - \frac{8 π}{5})

Simplify:

\frac{π}{2} - \frac{8 π}{5} = - \frac{11 π}{10}

\frac{π}{2} - \frac{8 π}{5}

Least Common Multiplier of

2, 5 : 10

2, 5

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

5 : 5

5

5

is a prime number, therefore no factorization is possible

= 5

Multiply each factor the greatest number of times it occurs in either

2

5

= 2 \cdot 5

Multiply the numbers:

2 \cdot 5 = 10

= 10

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

10

For

\frac{π}{2} :

multiply the denominator and numerator by

5

\frac{π}{2} = \frac{π 5}{2 \cdot 5} = \frac{π 5}{10}

For

\frac{8 π}{5} :

multiply the denominator and numerator by

2

\frac{8 π}{5} = \frac{8 π 2}{5 \cdot 2} = \frac{16 π}{10}

= \frac{π 5}{10} - \frac{16 π}{10}

Since the denominators are equal, combine the fractions:

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

= \frac{π 5 - 16 π}{10}

Add similar elements:

5 π - 16 π = - 11 π

= \frac{- 11 π}{10}

Apply the fraction rule:

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

= - \frac{11 π}{10}

= cos (- \frac{11 π}{10})

Use the following property:

cos (- x) = cos (x)

cos (- \frac{11 π}{10}) = cos (\frac{11 π}{10})

= cos (\frac{11 π}{10})

= cos (\frac{11 π}{10})

Rewrite using trig identities:

- \frac{1 + cos ( \frac{π}{5} )}{2}

cos (\frac{11 π}{10})

Write

cos (\frac{11 π}{10})

cos (\frac{\frac{11 π}{5}}{2})

= cos (\frac{\frac{11 π}{5}}{2})

Use the Half Angle identity:

cos (\frac{θ}{2}) = - \frac{1 + cos ( θ )}{2}

Use the Double Angle identity

cos (2 θ) = 2 cos^{2} (θ) - 1

Substitute

θ

with

\frac{θ}{2}

cos (θ) = 2 cos^{2} (\frac{θ}{2}) - 1

Switch sides

2 cos^{2} (\frac{θ}{2}) = 1 + cos (θ)

Divide both sides by

2

cos^{2} (\frac{θ}{2}) = \frac{( 1 + cos ( θ ) )}{2}

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

range [0, \frac{π}{2}] [\frac{π}{2}, π] [π, \frac{3 π}{2}] [\frac{3 π}{2}, 2 π] quadrant I II III I V sin positive positive negative negative cos positive negative negative positive

cos (\frac{θ}{2}) = - \frac{( 1 + cos ( θ ) )}{2}

= - \frac{1 + cos ( \frac{11 π}{5} )}{2}

cos (\frac{11 π}{5}) = cos (\frac{π}{5})

cos (\frac{11 π}{5})

Rewrite

\frac{11 π}{5}

2 π + \frac{π}{5}

= cos (2 π + \frac{π}{5})

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

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

= cos (\frac{π}{5})

= - \frac{1 + cos ( \frac{π}{5} )}{2}

= - \frac{1 + cos ( \frac{π}{5} )}{2}

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}

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

Simplify

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

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

\frac{1 + \frac{5 + 1}{4}}{2} = \frac{5 + 5}{8}

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

Join

1 + \frac{5 + 1}{4} : \frac{5 + 5}{4}

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

Convert element to fraction:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} + \frac{5 + 1}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 4 + 5 + 1}{4}

1 \cdot 4 + 5 + 1 = 5 + 5

1 \cdot 4 + 5 + 1

Multiply the numbers:

1 \cdot 4 = 4

= 4 + 5 + 1

Add the numbers:

4 + 1 = 5

= 5 + 5

= \frac{5 + 5}{4}

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

Apply the fraction rule:

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

= \frac{5 + 5}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{5 + 5}{8}

= - \frac{5 + 5}{8}

Simplify

\frac{5 + 5}{8} : \frac{5 + 5}{2 2}

\frac{5 + 5}{8}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{5 + 5}{8}

8 = 22

8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

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

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

222 = 4

222

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}

Popular Examples

cot((sqrt(3))/3)

cot (\frac{3}{3})

cos(4/3)

cos (\frac{4}{3})

-2cos((3pi)/4)

- 2 cos (\frac{3 π}{4})

tan(2arccos(-3/5))

tan (2 arccos (- \frac{3}{5}))

arcsin(9/11)

arcsin (\frac{9}{11})

Frequently Asked Questions (FAQ)

What is the value of sin((8pi)/5) ?
The value of sin((8pi)/5) is -(sqrt(2)sqrt(5+\sqrt{5)})/4