What is the value of (pisin(pi)+cos(pi)-1)/((pi)^2) ?

The value of (pisin(pi)+cos(pi)-1)/((pi)^2) is -2/(pi^2)

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(pisin(pi)+cos(pi)-1)/((pi)^2)

\frac{π sin ( π ) + cos ( π ) - 1}{( π ) ^{2}}

- \frac{2}{π ^{2}}

Decimal Notation

- 0.20264 \dots

Solution steps

\frac{π sin ( π ) + cos ( π ) - 1}{( π ) ^{2}}

= \frac{π sin ( π ) + cos ( π ) - 1}{π ^{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:

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)

= \frac{π 0 - 1 - 1}{π ^{2}}

Simplify

= - \frac{2}{π ^{2}}

arccos (\frac{29}{36})

- 3 cos (2 (\frac{π}{3})) + 1

2 cos (6 0^{\circ}) sin (3 0^{\circ})

\frac{1}{cos ( \frac{1}{4} )}

9 cos (1 2^{\circ})

What is the value of (pisin(pi)+cos(pi)-1)/((pi)^2) ?
The value of (pisin(pi)+cos(pi)-1)/((pi)^2) is -2/(pi^2)