What is the value of cos(465) ?

The value of cos(465) is (sqrt(2)(1-sqrt(3)))/4

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Popular >

cos(465)

Solution

cos (46 5^{\circ})

Solution

\frac{2 ( 1 - 3 )}{4}

Decimal

- 0.25881 \dots

Solution steps

cos (46 5^{\circ})

cos (46 5^{\circ}) = cos (10 5^{\circ})

cos (46 5^{\circ})

Rewrite

46 5^{\circ}

36 0^{\circ} + 10 5^{\circ}

= cos (36 0^{\circ} + 10 5^{\circ})

Apply the periodicity of

cos

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

cos (36 0^{\circ} + 10 5^{\circ}) = cos (10 5^{\circ})

= cos (10 5^{\circ})

= cos (10 5^{\circ})

Rewrite using trig identities:

cos (6 0^{\circ}) cos (4 5^{\circ}) - sin (6 0^{\circ}) sin (4 5^{\circ})

cos (10 5^{\circ})

Write

cos (10 5^{\circ})

cos (6 0^{\circ} + 4 5^{\circ})

= cos (6 0^{\circ} + 4 5^{\circ})

Use the Angle Sum identity:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

= cos (6 0^{\circ}) cos (4 5^{\circ}) - sin (6 0^{\circ}) sin (4 5^{\circ})

= cos (6 0^{\circ}) cos (4 5^{\circ}) - sin (6 0^{\circ}) sin (4 5^{\circ})

Use the following trivial identity:

cos (6 0^{\circ}) = \frac{1}{2}

cos (6 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}

= \frac{1}{2}

Use the following trivial identity:

cos (4 5^{\circ}) = \frac{2}{2}

cos (4 5^{\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}

= \frac{2}{2}

Use the following trivial identity:

sin (6 0^{\circ}) = \frac{3}{2}

sin (6 0^{\circ})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

Use the following trivial identity:

sin (4 5^{\circ}) = \frac{2}{2}

sin (4 5^{\circ})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2}

Simplify

\frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2} : \frac{2 ( 1 - 3 )}{4}

\frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2}

Factor out common term

\frac{2}{2}

= \frac{2}{2} (\frac{1}{2} - \frac{3}{2})

\frac{1}{2} - \frac{3}{2} = \frac{1 - 3}{2}

\frac{1}{2} - \frac{3}{2}

Apply rule

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

= \frac{1 - 3}{2}

= \frac{2}{2} \cdot \frac{1 - 3}{2}

Multiply fractions:

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

= \frac{( 1 - 3 ) 2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 ( 1 - 3 )}{4}

= \frac{2 ( 1 - 3 )}{4}

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Frequently Asked Questions (FAQ)

What is the value of cos(465) ?
The value of cos(465) is (sqrt(2)(1-sqrt(3)))/4