What is the value of sin^2((9pi)/8) ?

The value of sin^2((9pi)/8) is (2-sqrt(2))/4

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Popular Trigonometry >

sin^2((9pi)/8)

Solution

sin^{2} (\frac{9 π}{8})

Solution

\frac{2 - 2}{4}

Decimal

0.14644 \dots

Solution steps

sin^{2} (\frac{9 π}{8})

Rewrite using trig identities:

sin (\frac{9 π}{8}) = - \frac{2 - 2}{2}

sin (\frac{9 π}{8})

Rewrite using trig identities:

- \frac{1 - cos ( \frac{π}{4} )}{2}

sin (\frac{9 π}{8})

Write

sin (\frac{9 π}{8})

sin (\frac{\frac{9 π}{4}}{2})

= sin (\frac{\frac{9 π}{4}}{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{9 π}{4} )}{2}

cos (\frac{9 π}{4}) = cos (\frac{π}{4})

cos (\frac{9 π}{4})

Rewrite

\frac{9 π}{4}

2 π + \frac{π}{4}

= cos (2 π + \frac{π}{4})

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

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

= cos (\frac{π}{4})

= - \frac{1 - cos ( \frac{π}{4} )}{2}

= - \frac{1 - cos ( \frac{π}{4} )}{2}

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{π}{4})

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

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

Simplify

- \frac{1 - \frac{2}{2}}{2} : - \frac{2 - 2}{2}

- \frac{1 - \frac{2}{2}}{2}

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

\frac{1 - \frac{2}{2}}{2}

Join

1 - \frac{2}{2} : \frac{2 - 2}{2}

1 - \frac{2}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2 - 2}{2}

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

Apply the fraction rule:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 - 2}{4}

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

Simplify

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

\frac{2 - 2}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{2 - 2}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2 - 2}{2}

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

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

= (- \frac{2 - 2}{2})^{2}

Simplify

(- \frac{2 - 2}{2})^{2} : \frac{2 - 2}{4}

(- \frac{2 - 2}{2})^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- \frac{2 - 2}{2})^{2} = (\frac{2 - 2}{2})^{2}

= (\frac{2 - 2}{2})^{2}

Apply exponent rule:

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

= \frac{( 2 - 2 ) ^{2}}{2 ^{2}}

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

Apply radical rule:

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

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

Apply exponent rule:

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

= (2 - 2)^{\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

= 2 - 2

= \frac{2 - 2}{2 ^{2}}

2^{2} = 4

= \frac{2 - 2}{4}

= \frac{2 - 2}{4}

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Frequently Asked Questions (FAQ)

What is the value of sin^2((9pi)/8) ?
The value of sin^2((9pi)/8) is (2-sqrt(2))/4