What is the value of 4cos((5pi)/3) ?

The value of 4cos((5pi)/3) is 2

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4cos((5pi)/3)

4 cos (\frac{5 π}{3})

2

Solution steps

4 cos (\frac{5 π}{3})

Rewrite using trig identities:

cos (\frac{5 π}{3}) = \frac{1}{2}

cos (\frac{5 π}{3})

Rewrite using trig identities:

cos (π) cos (\frac{2 π}{3}) - sin (π) sin (\frac{2 π}{3})

cos (\frac{5 π}{3})

Write

cos (\frac{5 π}{3})

cos (π + \frac{2 π}{3})

= cos (π + \frac{2 π}{3})

Use the Angle Sum identity:

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

= cos (π) cos (\frac{2 π}{3}) - sin (π) sin (\frac{2 π}{3})

= cos (π) cos (\frac{2 π}{3}) - sin (π) sin (\frac{2 π}{3})

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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}

= (- 1)

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}

Use the following trivial identity:

sin (π) = 0

sin (π)

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

= 0

Use the following trivial identity:

sin (\frac{2 π}{3}) = \frac{3}{2}

sin (\frac{2 π}{3})

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

= (- 1) (- \frac{1}{2}) - 0 \cdot \frac{3}{2}

Simplify

= \frac{1}{2}

= 4 \cdot \frac{1}{2}

Simplify

4 \cdot \frac{1}{2} : 2

4 \cdot \frac{1}{2}

Convert element to fraction:

4 = \frac{4}{1}

= \frac{4}{1} \cdot \frac{1}{2}

Cross-cancel common factor:

2

= \frac{2}{1}

Apply the fraction rule:

\frac{a}{1} = a

= 2

= 2

sec (1) tan (1)

arccos (\frac{1}{6})

e^{a r c t a n (1)}

sin (\frac{5 π}{4}) + cos (\frac{5 π}{4})

tan (- \frac{11 π}{4})