What is the value of cos(60+225) ?

The value of cos(60+225) is (sqrt(2)(sqrt(3)-1))/4

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cos(60+225)

Solution

cos (6 0^{\circ} + 22 5^{\circ})

Solution

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

Decimal Notation

0.25881 \dots

Solution steps

cos (6 0^{\circ} + 22 5^{\circ})

Simplify

= cos (28 5^{\circ})

Rewrite using trig identities:

cos (15 0^{\circ}) cos (13 5^{\circ}) - sin (15 0^{\circ}) sin (13 5^{\circ})

cos (28 5^{\circ})

Write

cos (28 5^{\circ})

cos (15 0^{\circ} + 13 5^{\circ})

= cos (15 0^{\circ} + 13 5^{\circ})

Use the Angle Sum identity:

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

= cos (15 0^{\circ}) cos (13 5^{\circ}) - sin (15 0^{\circ}) sin (13 5^{\circ})

= cos (15 0^{\circ}) cos (13 5^{\circ}) - sin (15 0^{\circ}) sin (13 5^{\circ})

Use the following trivial identity:

cos (15 0^{\circ}) = - \frac{3}{2}

cos (15 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{3}{2}

Use the following trivial identity:

cos (13 5^{\circ}) = - \frac{2}{2}

cos (13 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 (15 0^{\circ}) = \frac{1}{2}

sin (15 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{1}{2}

Use the following trivial identity:

sin (13 5^{\circ}) = \frac{2}{2}

sin (13 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{3}{2}) (- \frac{2}{2}) - \frac{1}{2} \cdot \frac{2}{2}

Simplify

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

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

Remove parentheses:

(- a) = - a, - (- a) = a

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

Factor out common term

\frac{2}{2}

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

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

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

Apply rule

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

= \frac{3 - 1}{2}

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

Multiply fractions:

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

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

Multiply the numbers:

2 \cdot 2 = 4

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

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

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Frequently Asked Questions (FAQ)

What is the value of cos(60+225) ?
The value of cos(60+225) is (sqrt(2)(sqrt(3)-1))/4