What is the value of 9cos((2pi)/3) ?

The value of 9cos((2pi)/3) is -9/2

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9cos((2pi)/3)

9 cos (\frac{2 π}{3})

- \frac{9}{2}

Decimal Notation

- 4.5

Solution steps

9 cos (\frac{2 π}{3})

Use the following trivial identity:

cos (\frac{2 π}{3}) = - \frac{1}{2}

cos (\frac{2 π}{3})

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}

= - \frac{1}{2}

= 9 (- \frac{1}{2})

Simplify

9 (- \frac{1}{2}) : - \frac{9}{2}

9 (- \frac{1}{2})

Remove parentheses:

(- a) = - a

= - 9 \cdot \frac{1}{2}

Multiply fractions:

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

= - \frac{1 \cdot 9}{2}

Multiply the numbers:

1 \cdot 9 = 9

= - \frac{9}{2}

= - \frac{9}{2}

1 + cos (- π)

1 + 2 sin (0)

arccos (- \frac{15}{17})

ln (tan (\frac{π}{3}) + sec (\frac{π}{3}))

arccos (0.8333)