What is the value of arccos(tan(-pi/4)) ?

The value of arccos(tan(-pi/4)) is pi

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arccos(tan(-pi/4))

arccos (tan (- \frac{π}{4}))

π

Decimal Notation

180

Solution steps

arccos (tan (- \frac{π}{4}))

Rewrite using trig identities:

tan (- \frac{π}{4}) = - 1

tan (- \frac{π}{4})

Use the following property:

tan (- x) = - tan (x)

tan (- \frac{π}{4}) = - tan (\frac{π}{4})

= - tan (\frac{π}{4})

Use the following trivial identity:

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

= - 1

= arccos (- 1)

Use the following trivial identity:

arccos (- 1) = π

arccos (- 1)

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= π

= π

\frac{sin ( π )}{π}

\frac{tan ( 6 0 ^{\circ} ) - tan ( 3 0 ^{\circ} )}{1 + tan ( 6 0 ^{\circ} ) tan ( 3 0 ^{\circ} )}

sin (arctan (- 8))

cot (\frac{11 π}{4})

cos (\frac{3 π}{10})