What is the value of arccos(tan(-(5pi)/4)) ?

The value of arccos(tan(-(5pi)/4)) is pi

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arccos(tan(-(5pi)/4))

arccos (tan (- \frac{5 π}{4}))

π

Decimal Notation

180

Solution steps

arccos (tan (- \frac{5 π}{4}))

Rewrite using trig identities:

tan (- \frac{5 π}{4}) = - 1

tan (- \frac{5 π}{4})

Use the following property:

tan (- x) = - tan (x)

tan (- \frac{5 π}{4}) = - tan (\frac{5 π}{4})

= - tan (\frac{5 π}{4})

tan (\frac{5 π}{4}) = tan (\frac{π}{4})

tan (\frac{5 π}{4})

Rewrite

\frac{5 π}{4}

π + \frac{π}{4}

= tan (π + \frac{π}{4})

Apply the periodicity of

tan

tan (x + π) = tan (x)

tan (π + \frac{π}{4}) = 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

cos (arccos (3))

arccos (180)

\frac{π}{2} + cos (\frac{π}{2})

\frac{16}{sin ( 4 5 ^{\circ} )}

\frac{cos ( 2 π )}{2}

What is the value of arccos(tan(-(5pi)/4)) ?
The value of arccos(tan(-(5pi)/4)) is pi