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

The value of pi/2 cos(2pi) is pi/2

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pi/2 cos(2pi)

\frac{π}{2} cos (2 π)

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\frac{π}{2} cos (2 π)

Rewrite using trig identities:

cos (2 π) = 1

cos (2 π)

cos (2 π) = cos (0)

cos (2 π)

Rewrite

2 π

2 π + 0

= cos (2 π + 0)

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

cos (2 π + 0) = cos (0)

= cos (0)

= cos (0)

Use the following trivial identity:

cos (0) = 1

cos (0)

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

= 1

= \frac{π}{2} \cdot 1

Simplify

= \frac{π}{2}

\frac{1}{1 - sin ( \frac{π}{6} )}

cot (arcsin (\frac{1}{4}))

\frac{sin ( 24 0 ^{\circ} )}{cos ( 12 0 ^{\circ} )}

5 + 4 cos (\frac{π}{2})

(sin (3 0^{\circ})) (tan (4 5^{\circ})) + (tan (3 0^{\circ})) (sin (6 0^{\circ}))