What is the value of cos(120)+cos(45) ?

The value of cos(120)+cos(45) is (-1+sqrt(2))/2

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cos(120)+cos(45)

cos (12 0^{\circ}) + cos (4 5^{\circ})

\frac{- 1 + 2}{2}

Decimal Notation

0.20710 \dots

Solution steps

cos (12 0^{\circ}) + cos (4 5^{\circ})

Use the following trivial identity:

cos (12 0^{\circ}) = - \frac{1}{2}

cos (12 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}

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

Simplify

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

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

Apply rule

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

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

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

cos (11 5^{\circ})

2 sin (2 π)

3 cos (3 0^{\circ})

π cos (π)

- cot (\frac{π}{2})

What is the value of cos(120)+cos(45) ?
The value of cos(120)+cos(45) is (-1+sqrt(2))/2