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tan(θ)>0\land cos(θ)<0

Solução

tan (θ) > 0 and cos (θ) < 0

Falso para todo θ \in R

Passos da solução

tan (θ) > 0 and cos (θ) < 0

tan (θ) > 0 : πn < θ < \frac{π}{2} + πn

tan (θ) > 0

tan (x) > a

então

arctan (a) + πn < x < \frac{π}{2} + πn

arctan (0) + πn < θ < \frac{π}{2} + πn

Simplificar

arctan (0) : 0

arctan (0)

Utilizar a seguinte identidade trivial:

arctan (0) = 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 + πn < θ < \frac{π}{2} + πn

Simplificar

πn < θ < \frac{π}{2} + πn

cos (θ) < 0 : \frac{π}{2} + 2 πn < θ < \frac{3 π}{2} + 2 πn

cos (θ) < 0

Para

cos (x) < a

, se

- 1 < a \leq 1

então

arccos (a) + 2 πn < x < 2 π - arccos (a) + 2 πn

arccos (0) + 2 πn < θ < 2 π - arccos (0) + 2 πn

Simplificar

arccos (0) : \frac{π}{2}

arccos (0)

Utilizar a seguinte identidade trivial:

arccos (0) = \frac{π}{2}

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}

= \frac{π}{2}

Simplificar

2 π - arccos (0) : \frac{3 π}{2}

2 π - arccos (0)

Utilizar a seguinte identidade trivial:

arccos (0) = \frac{π}{2}

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}

= 2 π - \frac{π}{2}

Simplificar

2 π - \frac{π}{2}

Converter para fração:

2 π = \frac{2 π 2}{2}

= \frac{2 π 2}{2} - \frac{π}{2}

Já que os denominadores são iguais, combinar as frações:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 π 2 - π}{2}

2 π 2 - π = 3 π

2 π 2 - π

Multiplicar os números:

2 \cdot 2 = 4

= 4 π - π

Somar elementos similares:

4 π - π = 3 π

= 3 π

= \frac{3 π}{2}

= \frac{3 π}{2}

\frac{π}{2} + 2 πn < θ < \frac{3 π}{2} + 2 πn

Combinar os intervalos

πn < θ < \frac{π}{2} + πn and \frac{π}{2} + 2 πn < θ < \frac{3 π}{2} + 2 πn

Junte intervalos que se sobrepoem

Falso para todo θ \in R

\frac{11 π}{2} \leq arctan (θ) \leq \frac{13 π}{9}

0 \leq tan^{2} (x) \leq 3

tan (θ) = 2 and cos (θ) < 0

tan (θ) = - \frac{4}{3} and sin (θ) < 0, sec (θ)

tan (x) 0 \leq x \leq \frac{π}{6}