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受欢迎的三角函数 >

-(sqrt(2))/2 <sin(x/2)<(sqrt(2))/2

解答

- \frac{2}{2} < sin (\frac{x}{2}) < \frac{2}{2}

解答

4 πn \leq x < \frac{π}{2} + 4 πn or \frac{3 π}{2} + 4 πn < x < \frac{5 π}{2} + 4 πn or \frac{7 π}{2} + 4 πn < x < 4 π + 4 πn

+2

间隔符号

[4 πn, \frac{π}{2} + 4 πn) \cup (\frac{3 π}{2} + 4 πn, \frac{5 π}{2} + 4 πn) \cup (\frac{7 π}{2} + 4 πn, 4 π + 4 πn)

十进制

4 πn \leq x < 1.57079 \dots + 4 πn or 4.71238 \dots + 4 πn < x < 7.85398 \dots + 4 πn or 10.99557 \dots + 4 πn < x < 12.56637 \dots + 4 πn

求解步骤

- \frac{2}{2} < sin (\frac{x}{2}) < \frac{2}{2}

若

a < u < b

，则

a < u and u < b

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

- \frac{2}{2} < sin (\frac{x}{2}) : - \frac{π}{2} + 4 πn < x < \frac{5 π}{2} + 4 πn

- \frac{2}{2} < sin (\frac{x}{2})

交换两边

sin (\frac{x}{2}) > - \frac{2}{2}

对于

sin (x) > a

，若

- 1 \leq a < 1

，则

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

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

若

a < u < b

，则

a < u and u < b

arcsin (- \frac{2}{2}) + 2 πn < \frac{x}{2} and \frac{x}{2} < π - arcsin (- \frac{2}{2}) + 2 πn

arcsin (- \frac{2}{2}) + 2 πn < \frac{x}{2} : x > - \frac{π}{2} + 4 πn

arcsin (- \frac{2}{2}) + 2 πn < \frac{x}{2}

交换两边

\frac{x}{2} > arcsin (- \frac{2}{2}) + 2 πn

化简

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

arcsin (- \frac{2}{2}) + 2 πn

arcsin (- \frac{2}{2}) = - \frac{π}{4}

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

利用以下特性:

arcsin (- x) = - arcsin (x)

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

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

使用以下普通恒等式:

arcsin (\frac{2}{2}) = \frac{π}{4}

arcsin (\frac{2}{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}{4}

= - \frac{π}{4}

= - \frac{π}{4} + 2 πn

\frac{x}{2} > - \frac{π}{4} + 2 πn

在两边乘以

2

\frac{x}{2} > - \frac{π}{4} + 2 πn

在两边乘以

2

\frac{2 x}{2} > - 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

化简

\frac{2 x}{2} > - 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

- 2 \cdot \frac{π}{4} + 2 \cdot 2 πn : - \frac{π}{2} + 4 πn

- 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

分式相乘:

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

= \frac{π 2}{4}

约分:

2

= \frac{π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= - \frac{π}{2} + 4 πn

x > - \frac{π}{2} + 4 πn

x > - \frac{π}{2} + 4 πn

x > - \frac{π}{2} + 4 πn

\frac{x}{2} < π - arcsin (- \frac{2}{2}) + 2 πn : x < \frac{5 π}{2} + 4 πn

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

化简

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

π - arcsin (- \frac{2}{2}) + 2 πn

arcsin (- \frac{2}{2}) = - \frac{π}{4}

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

利用以下特性:

arcsin (- x) = - arcsin (x)

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

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

使用以下普通恒等式:

arcsin (\frac{2}{2}) = \frac{π}{4}

arcsin (\frac{2}{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}{4}

= - \frac{π}{4}

= π - (- \frac{π}{4}) + 2 πn

使用法则

- (- a) = a

= π + \frac{π}{4} + 2 πn

\frac{x}{2} < π + \frac{π}{4} + 2 πn

在两边乘以

2

\frac{x}{2} < π + \frac{π}{4} + 2 πn

在两边乘以

2

\frac{2 x}{2} < 2 π + 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

化简

\frac{2 x}{2} < 2 π + 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 π + 2 \cdot \frac{π}{4} + 2 \cdot 2 πn : 2 π + \frac{π}{2} + 4 πn

2 π + 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

分式相乘:

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

= \frac{π 2}{4}

约分:

2

= \frac{π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= 2 π + \frac{π}{2} + 4 πn

x < 2 π + \frac{π}{2} + 4 πn

x < 2 π + \frac{π}{2} + 4 πn

化简

2 π + \frac{π}{2} : \frac{5 π}{2}

2 π + \frac{π}{2}

将项转换为分式:

2 π = \frac{2 π 2}{2}

= \frac{2 π 2}{2} + \frac{π}{2}

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{2 π 2 + π}{2}

2 π 2 + π = 5 π

2 π 2 + π

数字相乘:

2 \cdot 2 = 4

= 4 π + π

同类项相加:

4 π + π = 5 π

= 5 π

= \frac{5 π}{2}

x < \frac{5 π}{2} + 4 πn

x < \frac{5 π}{2} + 4 πn

合并区间

x > - \frac{π}{2} + 4 πn and x < \frac{5 π}{2} + 4 πn

合并重叠的区间

- \frac{π}{2} + 4 πn < x < \frac{5 π}{2} + 4 πn

sin (\frac{x}{2}) < \frac{2}{2} : - \frac{5 π}{2} + 4 πn < x < \frac{π}{2} + 4 πn

sin (\frac{x}{2}) < \frac{2}{2}

对于

sin (x) < a

，若

- 1 < a \leq 1

，则

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

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

若

a < u < b

，则

a < u and u < b

- π - arcsin (\frac{2}{2}) + 2 πn < \frac{x}{2} and \frac{x}{2} < arcsin (\frac{2}{2}) + 2 πn

- π - arcsin (\frac{2}{2}) + 2 πn < \frac{x}{2} : x > - \frac{5 π}{2} + 4 πn

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

交换两边

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

化简

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

- π - arcsin (\frac{2}{2}) + 2 πn

使用以下普通恒等式:

arcsin (\frac{2}{2}) = \frac{π}{4}

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}{4} + 2 πn

\frac{x}{2} > - π - \frac{π}{4} + 2 πn

在两边乘以

2

\frac{x}{2} > - π - \frac{π}{4} + 2 πn

在两边乘以

2

\frac{2 x}{2} > - 2 π - 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

化简

\frac{2 x}{2} > - 2 π - 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

- 2 π - 2 \cdot \frac{π}{4} + 2 \cdot 2 πn : - 2 π - \frac{π}{2} + 4 πn

- 2 π - 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

分式相乘:

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

= \frac{π 2}{4}

约分:

2

= \frac{π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= - 2 π - \frac{π}{2} + 4 πn

x > - 2 π - \frac{π}{2} + 4 πn

x > - 2 π - \frac{π}{2} + 4 πn

化简

- 2 π - \frac{π}{2} : - \frac{5 π}{2}

- 2 π - \frac{π}{2}

将项转换为分式:

2 π = \frac{2 π 2}{2}

= - \frac{2 π 2}{2} - \frac{π}{2}

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{- 2 π 2 - π}{2}

- 2 π 2 - π = - 5 π

- 2 π 2 - π

数字相乘:

2 \cdot 2 = 4

= - 4 π - π

同类项相加:

- 4 π - π = - 5 π

= - 5 π

= \frac{- 5 π}{2}

使用分式法则:

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

= - \frac{5 π}{2}

x > - \frac{5 π}{2} + 4 πn

x > - \frac{5 π}{2} + 4 πn

\frac{x}{2} < arcsin (\frac{2}{2}) + 2 πn : x < \frac{π}{2} + 4 πn

\frac{x}{2} < arcsin (\frac{2}{2}) + 2 πn

化简

arcsin (\frac{2}{2}) + 2 πn : \frac{π}{4} + 2 πn

arcsin (\frac{2}{2}) + 2 πn

使用以下普通恒等式:

arcsin (\frac{2}{2}) = \frac{π}{4}

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}{4} + 2 πn

\frac{x}{2} < \frac{π}{4} + 2 πn

在两边乘以

2

\frac{x}{2} < \frac{π}{4} + 2 πn

在两边乘以

2

\frac{2 x}{2} < 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

化简

\frac{2 x}{2} < 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{π}{4} + 2 \cdot 2 πn : \frac{π}{2} + 4 πn

2 \cdot \frac{π}{4} + 2 \cdot 2 πn

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

分式相乘:

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

= \frac{π 2}{4}

约分:

2

= \frac{π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{π}{2} + 4 πn

x < \frac{π}{2} + 4 πn

x < \frac{π}{2} + 4 πn

x < \frac{π}{2} + 4 πn

合并区间

x > - \frac{5 π}{2} + 4 πn and x < \frac{π}{2} + 4 πn

合并重叠的区间

- \frac{5 π}{2} + 4 πn < x < \frac{π}{2} + 4 πn

合并区间

- \frac{π}{2} + 4 πn < x < \frac{5 π}{2} + 4 πn and - \frac{5 π}{2} + 4 πn < x < \frac{π}{2} + 4 πn

合并重叠的区间

4 πn \leq x < \frac{π}{2} + 4 πn or \frac{3 π}{2} + 4 πn < x < \frac{5 π}{2} + 4 πn or \frac{7 π}{2} + 4 πn < x < 4 π + 4 πn

流行的例子

-sqrt(3)<= tan(x)<= ((sqrt(3)))/3

- 3 \leq tan (x) \leq \frac{( 3 )}{3}

0 < cos (θ) < 1

tan(θ)=-12/5 \land sin(θ)>0

tan (θ) = - \frac{12}{5} and sin (θ) > 0

-1<= arccos(x^2)<= 1

- 1 \leq arccos (x^{2}) \leq 1

1-cos(θ)0<= θ<= 2pi

1 - cos (θ) 0 \leq θ \leq 2 π