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受欢迎的三角函数 >

cos(θ)= 3/4 \land cot(θ)<0

解答

cos (θ) = \frac{3}{4} and cot (θ) < 0

解答

对所有 θ \in R 为假

求解步骤

cos (θ) = \frac{3}{4} and cot (θ) < 0

cos (θ) = \frac{3}{4} : θ = 0.72273 \dots + 2 πn, θ = 2 π - 0.72273 \dots + 2 πn

cos (θ) = \frac{3}{4}

使用反三角函数性质

cos (θ) = \frac{3}{4}

cos (θ) = \frac{3}{4}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{3}{4}) + 2 πn, θ = 2 π - arccos (\frac{3}{4}) + 2 πn

θ = arccos (\frac{3}{4}) + 2 πn, θ = 2 π - arccos (\frac{3}{4}) + 2 πn

以小数形式表示解

θ = 0.72273 \dots + 2 πn, θ = 2 π - 0.72273 \dots + 2 πn

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

cot (θ) < 0

若

cot (x) < a

，则

arccot (a) + πn < x < π + πn

arccot (0) + πn < θ < π + πn

化简

arccot (0) : \frac{π}{2}

arccot (0)

使用以下普通恒等式:

arccot (0) = \frac{π}{2}

x - 3 - 1 - \frac{3}{3} 0 \frac{3}{3} 13 arccot (x) \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} arccot (x) 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ}

= \frac{π}{2}

\frac{π}{2} + πn < θ < π + πn

合并区间

(θ = 0.72273 \dots + 2 πn or θ = 2 π - 0.72273 \dots + 2 πn) and \frac{π}{2} + πn < θ < π + πn

合并重叠的区间

对所有 θ \in R 为假

流行的例子

sin(θ)>0\land cos(θ)>0

sin (θ) > 0 and cos (θ) > 0

sin(4θ)0<= θ<= pi

sin (4 θ) 0 \leq θ \leq π

[sin(pi)t]0<= t<= 2

[sin (π) t] 0 \leq t \leq 2

cosh(θ)= 7/3 \land θ<0,sinh(θ)

cosh (θ) = \frac{7}{3} and θ < 0, sinh (θ)

cos(x)<= sin^2(x)<= (sqrt(3))/2 sin(x)

cos (x) \leq sin^{2} (x) \leq \frac{3}{2} sin (x)