What is the value of (cos(135))/(tan(225)) ?

The value of (cos(135))/(tan(225)) is -(sqrt(2))/2

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(cos(135))/(tan(225))

\frac{cos ( 13 5 ^{\circ} )}{tan ( 22 5 ^{\circ} )}

- \frac{2}{2}

Decimal Notation

- 0.70710 \dots

Solution steps

\frac{cos ( 13 5 ^{\circ} )}{tan ( 22 5 ^{\circ} )}

Use the following trivial identity:

cos (13 5^{\circ}) = - \frac{2}{2}

cos (13 5^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - \frac{2}{2}

Rewrite using trig identities:

tan (22 5^{\circ}) = 1

tan (22 5^{\circ})

tan (22 5^{\circ}) = tan (4 5^{\circ})

tan (22 5^{\circ})

Rewrite

22 5^{\circ}

18 0^{\circ} + 4 5^{\circ}

= tan (18 0^{\circ} + 4 5^{\circ})

Apply the periodicity of

tan

tan (x + 18 0^{\circ}) = tan (x)

tan (18 0^{\circ} + 4 5^{\circ}) = tan (4 5^{\circ})

= tan (4 5^{\circ})

= tan (4 5^{\circ})

Use the following trivial identity:

tan (4 5^{\circ}) = 1

tan (4 5^{\circ})

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

= 1

= \frac{- \frac{2}{2}}{1}

Simplify

= - \frac{2}{2}

50 (3 + 3 cos (5 3^{\circ}))

\frac{cos ( π )}{12}

cos (5.8)

arctan (\frac{- 16}{0})

arctan (1.3333)

What is the value of (cos(135))/(tan(225)) ?
The value of (cos(135))/(tan(225)) is -(sqrt(2))/2