What is the value of 6cos(480) ?

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6cos(480)

6 cos (48 0^{\circ})

- 3

Solution steps

6 cos (48 0^{\circ})

Rewrite using trig identities:

cos (48 0^{\circ}) = - \frac{1}{2}

cos (48 0^{\circ})

cos (48 0^{\circ}) = cos (12 0^{\circ})

cos (48 0^{\circ})

Rewrite

48 0^{\circ}

36 0^{\circ} + 12 0^{\circ}

= cos (36 0^{\circ} + 12 0^{\circ})

Apply the periodicity of

cos

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

cos (36 0^{\circ} + 12 0^{\circ}) = cos (12 0^{\circ})

= cos (12 0^{\circ})

= cos (12 0^{\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}

= - \frac{1}{2}

= 6 (- \frac{1}{2})

Simplify

6 (- \frac{1}{2}) : - 3

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

Remove parentheses:

(- a) = - a

= - 6 \cdot \frac{1}{2}

Multiply fractions:

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

= - \frac{1 \cdot 6}{2}

Multiply the numbers:

1 \cdot 6 = 6

= - \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= - 3

= - 3

2 - sin (π)

\frac{( 7.51 ) ^{2} \cdot sin ( 2 \cdot 20 )}{12.08}

- 3 cos (\frac{3 π}{4})

\frac{986}{tan ( 3 6 ^{\circ} )}

(cos (0) π)^{1} ln (cos (0) π)