What is the value of (sin(pi)-cos(4pi))/(6pi) ?

The value of (sin(pi)-cos(4pi))/(6pi) is -1/(6pi)

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(sin(pi)-cos(4pi))/(6pi)

\frac{sin ( π ) - cos ( 4 π )}{6 π}

- \frac{1}{6 π}

Decimal Notation

- 0.05305 \dots

Solution steps

\frac{sin ( π ) - cos ( 4 π )}{6 π}

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

Rewrite using trig identities:

cos (4 π) = 1

cos (4 π)

cos (4 π) = cos (0)

cos (4 π)

Rewrite

4 π

2 π \cdot 2 + 0

= cos (2 π 2 + 0)

Apply the periodicity of

cos

cos (x + 2 π \cdot k) = cos (x)

cos (2 π \cdot 2 + 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{0 - 1}{6 π}

Simplify

= - \frac{1}{6 π}

cos (6 2^{\circ}) + cos (2 8^{\circ})

tan (13 8^{\circ})

- cos^{4} (0)

arctan (0.572)

\frac{6}{( 10 \cdot 1 0 ^{- 6} ) \cdot ( 20000 ) sin ( 4 0 ^{\circ} )}

What is the value of (sin(pi)-cos(4pi))/(6pi) ?
The value of (sin(pi)-cos(4pi))/(6pi) is -1/(6pi)