What is the value of 4cos(2((3pi)/2)) ?

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

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4cos(2((3pi)/2))

4 cos (2 (\frac{3 π}{2}))

- 4

Solution steps

4 cos (2 (\frac{3 π}{2}))

= 4 cos (2 \cdot \frac{3 π}{2})

Simplify:

2 \cdot \frac{3 π}{2} = 3 π

2 \cdot \frac{3 π}{2}

Multiply fractions:

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

= \frac{3 π 2}{2}

Cancel the common factor:

2

= 3 π

= 4 cos (3 π)

cos (3 π) = cos (π)

cos (3 π)

Rewrite

3 π

2 π + π

= cos (2 π + π)

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

cos (2 π + π) = cos (π)

= cos (π)

= 4 cos (π)

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)

= 4 (- 1)

Simplify

= - 4

sin (π^{0})

- \frac{1}{2} cos (\frac{π}{2})

cos (\frac{35 π}{12})

sec^{2} (1 5^{\circ})

80 \cdot sin (2 0^{\circ})