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受欢迎的三角函数 >

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

解答

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

解答

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

+2

间隔符号

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

十进制

2 πn \leq x < 0.78539 \dots + 2 πn or 2.35619 \dots + 2 πn < x < 3.92699 \dots + 2 πn or 5.49778 \dots + 2 πn < x < 6.28318 \dots + 2 πn

求解步骤

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

若

a < u < b

，则

a < u and u < b

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

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

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

交换两边

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

化简

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}

化简

π - arcsin (- \frac{2}{2}) : \frac{5 π}{4}

π - arcsin (- \frac{2}{2})

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})

化简

π - (- \frac{π}{4})

使用法则

- (- a) = a

= π + \frac{π}{4}

将项转换为分式:

π = \frac{π 4}{4}

= \frac{π 4}{4} + \frac{π}{4}

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 4 + π}{4}

同类项相加:

4 π + π = 5 π

= \frac{5 π}{4}

= \frac{5 π}{4}

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

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

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

化简

- π - arcsin (\frac{2}{2}) : - \frac{5 π}{4}

- π - arcsin (\frac{2}{2})

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}

化简

- π - \frac{π}{4}

将项转换为分式:

π = \frac{π 4}{4}

= - \frac{π 4}{4} - \frac{π}{4}

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 4 - π}{4}

同类项相加:

- 4 π - π = - 5 π

= \frac{- 5 π}{4}

使用分式法则:

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

= - \frac{5 π}{4}

= - \frac{5 π}{4}

化简

arcsin (\frac{2}{2}) : \frac{π}{4}

arcsin (\frac{2}{2})

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}

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

合并区间

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

合并重叠的区间

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

流行的例子

-sqrt(2)<= sin(θ)+cos(θ)<= sqrt(2)

- 2 \leq sin (θ) + cos (θ) \leq 2

cos(θ)= 13/5 \land 180<θ<270,tan(2θ)

cos (θ) = \frac{13}{5} and 180 < θ < 270, tan (2 θ)

sin(θ)=-1/3 \land tan(θ)>0

sin (θ) = - \frac{1}{3} and tan (θ) > 0

tan(θ)= 1/3 \land sin(θ)>0

tan (θ) = \frac{1}{3} and sin (θ) > 0

cot(θ)>0\land cos(θ)>0

cot (θ) > 0 and cos (θ) > 0