What is the value of (2pi+cos(pi))/(2sqrt(pi^2+sin(pi))) ?

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

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(2pi+cos(pi))/(2sqrt(pi^2+sin(pi)))

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

\frac{2 π - 1}{2 π}

Decimal Notation

0.84084 \dots

Solution steps

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

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

= \frac{2 π - 1}{2 π ^{2} + 0}

Simplify

\frac{2 π - 1}{2 π ^{2} + 0} : \frac{2 π - 1}{2 π}

\frac{2 π - 1}{2 π ^{2} + 0}

2 π^{2} + 0 = 2 π

2 π^{2} + 0

π^{2} + 0 = π

π^{2} + 0

π^{2} + 0 = π^{2}

= π^{2}

Apply radical rule:

assuming

a \geq 0

= π

= 2 π

= \frac{2 π - 1}{2 π}

= \frac{2 π - 1}{2 π}

cos (6 0^{\circ}) \cdot 20

arccos (\frac{6}{4})

cos (\frac{4}{π})

36 cos (2 8^{\circ})

cot (\frac{100 π}{101})

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