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受欢迎的三角函数 >

-1/2 <= sin(x)<= (sqrt(3))/2

解答

- \frac{1}{2} \leq sin (x) \leq \frac{3}{2}

解答

2 πn \leq x \leq \frac{π}{3} + 2 πn or \frac{2 π}{3} + 2 πn \leq x \leq \frac{7 π}{6} + 2 πn or \frac{11 π}{6} + 2 πn \leq x < 2 π + 2 πn

+2

间隔符号

[2 πn, \frac{π}{3} + 2 πn] \cup [\frac{2 π}{3} + 2 πn, \frac{7 π}{6} + 2 πn] \cup [\frac{11 π}{6} + 2 πn, 2 π + 2 πn)

十进制

2 πn \leq x \leq 1.04719 \dots + 2 πn or 2.09439 \dots + 2 πn \leq x \leq 3.66519 \dots + 2 πn or 5.75958 \dots + 2 πn \leq x < 6.28318 \dots + 2 πn

求解步骤

- \frac{1}{2} \leq sin (x) \leq \frac{3}{2}

若

a \leq u \leq b

，则

a \leq u and u \leq b

- \frac{1}{2} \leq sin (x) and sin (x) \leq \frac{3}{2}

- \frac{1}{2} \leq sin (x) : - \frac{π}{6} + 2 πn \leq x \leq \frac{7 π}{6} + 2 πn

- \frac{1}{2} \leq sin (x)

交换两边

sin (x) \geq - \frac{1}{2}

对于

sin (x) \geq a

，若

- 1 < a < 1

，则

arcsin (a) + 2 πn \leq x \leq π - arcsin (a) + 2 πn

arcsin (- \frac{1}{2}) + 2 πn \leq x \leq π - arcsin (- \frac{1}{2}) + 2 πn

化简

arcsin (- \frac{1}{2}) : - \frac{π}{6}

arcsin (- \frac{1}{2})

利用以下特性:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{2}) = - arcsin (\frac{1}{2})

= - arcsin (\frac{1}{2})

使用以下普通恒等式:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{6}

= - \frac{π}{6}

化简

π - arcsin (- \frac{1}{2}) : \frac{7 π}{6}

π - arcsin (- \frac{1}{2})

arcsin (- \frac{1}{2}) = - \frac{π}{6}

arcsin (- \frac{1}{2})

利用以下特性:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{2}) = - arcsin (\frac{1}{2})

= - arcsin (\frac{1}{2})

使用以下普通恒等式:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{6}

= - \frac{π}{6}

= π - (- \frac{π}{6})

化简

π - (- \frac{π}{6})

使用法则

- (- a) = a

= π + \frac{π}{6}

将项转换为分式:

π = \frac{π 6}{6}

= \frac{π 6}{6} + \frac{π}{6}

因为分母相等，所以合并分式:

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

= \frac{π 6 + π}{6}

同类项相加:

6 π + π = 7 π

= \frac{7 π}{6}

= \frac{7 π}{6}

- \frac{π}{6} + 2 πn \leq x \leq \frac{7 π}{6} + 2 πn

sin (x) \leq \frac{3}{2} : - \frac{4 π}{3} + 2 πn \leq x \leq \frac{π}{3} + 2 πn

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

对于

sin (x) \leq a

，若

- 1 < a < 1

，则

- π - arcsin (a) + 2 πn \leq x \leq arcsin (a) + 2 πn

- π - arcsin (\frac{3}{2}) + 2 πn \leq x \leq arcsin (\frac{3}{2}) + 2 πn

化简

- π - arcsin (\frac{3}{2}) : - \frac{4 π}{3}

- π - arcsin (\frac{3}{2})

使用以下普通恒等式:

arcsin (\frac{3}{2}) = \frac{π}{3}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= - π - \frac{π}{3}

化简

- π - \frac{π}{3}

将项转换为分式:

π = \frac{π 3}{3}

= - \frac{π 3}{3} - \frac{π}{3}

因为分母相等，所以合并分式:

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

= \frac{- π 3 - π}{3}

同类项相加:

- 3 π - π = - 4 π

= \frac{- 4 π}{3}

使用分式法则:

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

= - \frac{4 π}{3}

= - \frac{4 π}{3}

化简

arcsin (\frac{3}{2}) : \frac{π}{3}

arcsin (\frac{3}{2})

使用以下普通恒等式:

arcsin (\frac{3}{2}) = \frac{π}{3}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{3}

- \frac{4 π}{3} + 2 πn \leq x \leq \frac{π}{3} + 2 πn

合并区间

- \frac{π}{6} + 2 πn \leq x \leq \frac{7 π}{6} + 2 πn and - \frac{4 π}{3} + 2 πn \leq x \leq \frac{π}{3} + 2 πn

合并重叠的区间

2 πn \leq x \leq \frac{π}{3} + 2 πn or \frac{2 π}{3} + 2 πn \leq x \leq \frac{7 π}{6} + 2 πn or \frac{11 π}{6} + 2 πn \leq x < 2 π + 2 πn

流行的例子

sec(θ)= 13/5 \land sin(θ)<0

sec (θ) = \frac{13}{5} and sin (θ) < 0

cot(θ)= 2/3 \land csc(θ)<0

cot (θ) = \frac{2}{3} and csc (θ) < 0

-(sqrt(3))/2 <cos(x)<(sqrt(3))/2

- \frac{3}{2} < cos (x) < \frac{3}{2}

0<=-cos(b)0.6428<= 180

0 \leq - cos (b) 0.6428 \leq 180

-sqrt(3)<tan(θ)<1

- 3 < tan (θ) < 1