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sin(A)=(-4)/5 \land cos(A)>0,cos(A)

Solución

sin (A) = \frac{- 4}{5} and cos (A) > 0, cos (A)

A = - 0.92729 \dots + 2 πn

Notación de intervalos

A = - 0.92729 \dots + 2 πn

Pasos de solución

sin (A) = \frac{- 4}{5} and cos (A) > 0

sin (A) = \frac{- 4}{5} : A = - 0.92729 \dots + 2 πn, A = π + 0.92729 \dots + 2 πn

sin (A) = \frac{- 4}{5}

Aplicar las propiedades de las fracciones:

\frac{- a}{b} = - \frac{a}{b}

sin (A) = - \frac{4}{5}

Aplicar propiedades trigonométricas inversas

sin (A) = - \frac{4}{5}

Soluciones generales para

sin (A) = - \frac{4}{5}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

A = arcsin (- \frac{4}{5}) + 2 πn, A = π + arcsin (\frac{4}{5}) + 2 πn

A = arcsin (- \frac{4}{5}) + 2 πn, A = π + arcsin (\frac{4}{5}) + 2 πn

Mostrar soluciones en forma decimal

A = - 0.92729 \dots + 2 πn, A = π + 0.92729 \dots + 2 πn

cos (A) > 0 : - \frac{π}{2} + 2 πn < A < \frac{π}{2} + 2 πn

cos (A) > 0

Para

cos (x) > a

, si

- 1 \leq a < 1

entonces

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

- arccos (0) + 2 πn < A < arccos (0) + 2 πn

Simplificar

- arccos (0) : - \frac{π}{2}

- arccos (0)

Utilizar la siguiente identidad 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

arccos (0) : \frac{π}{2}

arccos (0)

Utilizar la siguiente identidad 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}

- \frac{π}{2} + 2 πn < A < \frac{π}{2} + 2 πn

Combinar los rangos

(A = π + 0.92729 \dots + 2 πn or A = - 0.92729 \dots + 2 πn) and - \frac{π}{2} + 2 πn < A < \frac{π}{2} + 2 πn

Mezclar intervalos sobrepuestos

A = - 0.92729 \dots + 2 πn

sin (θ) = \frac{2}{5} and sec (θ) > 0

csc (θ) < 0 and cos (θ) > 0

sin (2 x) 0 \leq \frac{π}{2} and 0 - \frac{π}{2} < 0

[sin (π) t] 2 \leq t \leq 4

3 - 3 (0)^{2} \leq g (0) \leq 3 cos (0)