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

The value of arctan(cos(pi/2)) is 0

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arctan(cos(pi/2))

arctan (cos (\frac{π}{2}))

0

Solution steps

arctan (cos (\frac{π}{2}))

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

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}

= arctan (0)

= arctan (0)

Use the following trivial identity:

arctan (0) = 0

arctan (0)

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= 0

= 0

sin (arctan (3))

cos (11 0^{\circ}) cos (2 0^{\circ}) - sin (11 0^{\circ}) sin (2 0^{\circ})

sec (40 5^{\circ})

\frac{8.8}{sin ( 3 3 ^{\circ} )}

cos (arctan (- 3))