What is the value of arctan(sin(pi)) ?

The value of arctan(sin(pi)) is 0

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arctan(sin(pi))

arctan (sin (π))

0

Solution steps

arctan (sin (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}

= 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

2 π + sin (2 π)

\frac{tan ( 7 ^{\circ} )}{1 - tan ^{2} ( 7 ^{\circ} )}

2 sin (27 0^{\circ})

\frac{tan ( 3 5 ^{\circ} ) + tan ( 1 0 ^{\circ} )}{1 - tan ( 3 5 ^{\circ} ) tan ( 1 0 ^{\circ} )}

sin^{2} (7)