What is the value of 2sin(1.1pi) ?

The value of 2sin(1.1pi) is -sqrt(2)sqrt(-(\sqrt{5)+1)/4+1}

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Popular >

2sin(1.1pi)

Solution

2 sin (1.1 π)

Solution

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

Decimal Notation

- 0.61803 \dots

Solution steps

2 sin (1.1 π)

= 2 sin (\frac{11}{10} π)

Simplify:

\frac{11}{10} π = \frac{11 π}{10}

\frac{11}{10} π

Multiply fractions:

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

= \frac{11 π}{10}

= 2 sin (\frac{11 π}{10})

Rewrite using trig identities:

sin (\frac{11 π}{10}) = - \frac{1 - cos ( \frac{π}{5} )}{2}

sin (\frac{11 π}{10})

Write

sin (\frac{11 π}{10})

sin (\frac{\frac{11 π}{5}}{2})

= sin (\frac{\frac{11 π}{5}}{2})

Use the Half Angle identity:

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

Use the Double Angle identity

cos (2 θ) = 1 - 2 sin^{2} (θ)

Substitute

θ

with

\frac{θ}{2}

cos (θ) = 1 - 2 sin^{2} (\frac{θ}{2})

Switch sides

2 sin^{2} (\frac{θ}{2}) = 1 - cos (θ)

Divide both sides by

2

sin^{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

sin (\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}

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

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

2 - \frac{1 - cos ( \frac{π}{5} )}{2}

Remove parentheses:

(- a) = - a

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

Apply exponent rule:

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

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

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

Multiply

2^{2} \cdot \frac{1 - cos ( \frac{π}{5} )}{2} : 2 (- cos (\frac{π}{5}) + 1)

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

Multiply fractions:

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

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

Cancel the common factor:

2

= 2 (- cos (\frac{π}{5}) + 1)

= - 2 (- cos (\frac{π}{5}) + 1)

Apply radical rule:

assuming

a \geq 0, b \geq 0

= - 2 - cos (\frac{π}{5}) + 1

= - 2 - cos (\frac{π}{5}) + 1

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}

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

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Frequently Asked Questions (FAQ)

What is the value of 2sin(1.1pi) ?
The value of 2sin(1.1pi) is -sqrt(2)sqrt(-(\sqrt{5)+1)/4+1}