What is the value of sin^2((5pi)/4) ?

The value of sin^2((5pi)/4) is 1/2

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sin^2((5pi)/4)

sin^{2} (\frac{5 π}{4})

\frac{1}{2}

Decimal Notation

0.5

Solution steps

sin^{2} (\frac{5 π}{4})

Rewrite using trig identities:

\frac{1 - cos ( \frac{π}{2} )}{2}

sin^{2} (\frac{5 π}{4})

Use the following identity:

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

Use the Double Angle identity

cos (2 θ) = 1 - 2 sin^{2} (θ)

Switch sides

2 sin^{2} (θ) - 1 = - cos (2 θ)

Add

1

to both sides

2 sin^{2} (θ) = 1 - cos (2 θ)

Divide both sides by

2

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

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

Simplify:

2 \cdot \frac{5 π}{4} = \frac{5 π}{2}

2 \cdot \frac{5 π}{4}

Multiply fractions:

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

= \frac{5 π 2}{4}

Multiply the numbers:

5 \cdot 2 = 10

= \frac{10 π}{4}

Cancel the common factor:

2

= \frac{5 π}{2}

= \frac{1 - cos ( \frac{5 π}{2} )}{2}

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

cos (\frac{5 π}{2})

Rewrite

\frac{5 π}{2}

2 π + \frac{π}{2}

= cos (2 π + \frac{π}{2})

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

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

= cos (\frac{π}{2})

= \frac{1 - cos ( \frac{π}{2} )}{2}

= \frac{1 - cos ( \frac{π}{2} )}{2}

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

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}

= 0

= \frac{1 - 0}{2}

Simplify

= \frac{1}{2}

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