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sin(θ)cos(θ)=0.222\land tan(θ)<0

Solução

sin (θ) cos (θ) = 0.222 and tan (θ) < 0

Falso para todo θ \in R

Passos da solução

sin (θ) cos (θ) = 0.222 and tan (θ) < 0

sin (θ) cos (θ) = 0.222 : θ = \frac{0.46005 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.46005 \dots}{2} + πn

sin (θ) cos (θ) = 0.222

Reeecreva usando identidades trigonométricas

sin (θ) cos (θ)

Utilizar a identidade trigonométrica do arco duplo:

2 sin (x) cos (x) = sin (2 x)

sin (x) cos (x) = \frac{sin ( 2 x )}{2}

= \frac{sin ( 2 θ )}{2}

\frac{sin ( 2 θ )}{2} = 0.222

Multiplicar ambos os lados por

2

\frac{sin ( 2 θ )}{2} = 0.222

Multiplicar ambos os lados por

2

\frac{2 sin ( 2 θ )}{2} = 0.222 \cdot 2

Simplificar

sin (2 θ) = 0.444

sin (2 θ) = 0.444

Aplique as propriedades trigonométricas inversas

sin (2 θ) = 0.444

Soluções gerais para

sin (2 θ) = 0.444

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

2 θ = arcsin (0.444) + 2 πn, 2 θ = π - arcsin (0.444) + 2 πn

2 θ = arcsin (0.444) + 2 πn, 2 θ = π - arcsin (0.444) + 2 πn

Resolver

2 θ = arcsin (0.444) + 2 πn : θ = \frac{arcsin ( 0.444 )}{2} + πn

2 θ = arcsin (0.444) + 2 πn

Dividir ambos os lados por

2

2 θ = arcsin (0.444) + 2 πn

Dividir ambos os lados por

2

\frac{2 θ}{2} = \frac{arcsin ( 0.444 )}{2} + \frac{2 πn}{2}

Simplificar

θ = \frac{arcsin ( 0.444 )}{2} + πn

θ = \frac{arcsin ( 0.444 )}{2} + πn

Resolver

2 θ = π - arcsin (0.444) + 2 πn : θ = \frac{π}{2} - \frac{arcsin ( 0.444 )}{2} + πn

2 θ = π - arcsin (0.444) + 2 πn

Dividir ambos os lados por

2

2 θ = π - arcsin (0.444) + 2 πn

Dividir ambos os lados por

2

\frac{2 θ}{2} = \frac{π}{2} - \frac{arcsin ( 0.444 )}{2} + \frac{2 πn}{2}

Simplificar

θ = \frac{π}{2} - \frac{arcsin ( 0.444 )}{2} + πn

θ = \frac{π}{2} - \frac{arcsin ( 0.444 )}{2} + πn

θ = \frac{arcsin ( 0.444 )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( 0.444 )}{2} + πn

Mostrar soluções na forma decimal

θ = \frac{0.46005 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.46005 \dots}{2} + πn

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

tan (θ) < 0

tan (x) < a

então

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

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

- \frac{π}{2} + πn < θ < 0 + πn

Simplificar

- \frac{π}{2} + πn < θ < πn

Combinar os intervalos

(θ = \frac{0.46005 \dots}{2} + πn or θ = \frac{π}{2} - \frac{0.46005 \dots}{2} + πn) and - \frac{π}{2} + πn < θ < πn

Junte intervalos que se sobrepoem

Falso para todo θ \in R

\frac{( sin ( 56. 2 ^{\circ} ) \cdot 19 ) ^{2}}{( 2 \cdot 9.8 )}

e^{- 2 π} cos (2)

arctan (\frac{- 6}{- 4})

sin (2 0^{\circ}) + sin (4 0^{\circ}) + sin (8 0^{\circ})

200 (e^{- 2 π} cos (2))