What is the value of sin^2(225) ?

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sin^2(225)

sin^{2} (22 5^{\circ})

\frac{1}{2}

Decimal Notation

0.5

Solution steps

sin^{2} (22 5^{\circ})

Rewrite using trig identities:

\frac{1 - cos ( 9 0 ^{\circ} )}{2}

sin^{2} (22 5^{\circ})

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 22 5 ^{\circ} )}{2}

Simplify

= \frac{1 - cos ( 45 0 ^{\circ} )}{2}

cos (45 0^{\circ}) = cos (9 0^{\circ})

cos (45 0^{\circ})

Rewrite

45 0^{\circ}

36 0^{\circ} + 9 0^{\circ}

= cos (36 0^{\circ} + 9 0^{\circ})

Apply the periodicity of

cos

cos (x + 36 0^{\circ}) = cos (x)

cos (36 0^{\circ} + 9 0^{\circ}) = cos (9 0^{\circ})

= cos (9 0^{\circ})

= \frac{1 - cos ( 9 0 ^{\circ} )}{2}

= \frac{1 - cos ( 9 0 ^{\circ} )}{2}

Use the following trivial identity:

cos (9 0^{\circ}) = 0

cos (9 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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}

15 sin (4 5^{\circ})

tan (81 0^{\circ})

sec (7 2^{\circ})

arccos (cos (\frac{π}{6}))

\frac{2 cos ^{2} ( 4 5 ^{\circ} ) + 3 csc ( 3 7 ^{\circ} )}{2 tan ( 4 5 ^{\circ} )}