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受欢迎的三角函数 >

sin(θ)+cos(θ)+1>0

解答

sin (θ) + cos (θ) + 1 > 0

解答

- \frac{π}{2} + 2 πn < θ < π + 2 πn

+2

间隔符号

(- \frac{π}{2} + 2 πn, π + 2 πn)

十进制

- 1.57079 \dots + 2 πn < θ < 3.14159 \dots + 2 πn

求解步骤

sin (θ) + cos (θ) + 1 > 0

利用以下特性:

cos (x) + sin (x) = 2 sin (\frac{π}{4} + x)

1 + 2 sin (\frac{π}{4} + θ) > 0

将

1

到右边

1 + 2 sin (\frac{π}{4} + θ) > 0

两边减去

1

1 + 2 sin (\frac{π}{4} + θ) - 1 > 0 - 1

化简

2 sin (\frac{π}{4} + θ) > - 1

2 sin (\frac{π}{4} + θ) > - 1

两边除以

2

2 sin (\frac{π}{4} + θ) > - 1

两边除以

2

\frac{2 sin ( \frac{π}{4} + θ )}{2} > \frac{- 1}{2}

化简

\frac{2 sin ( \frac{π}{4} + θ )}{2} > \frac{- 1}{2}

化简

\frac{2 sin ( \frac{π}{4} + θ )}{2} : sin (\frac{π}{4} + θ)

\frac{2 sin ( \frac{π}{4} + θ )}{2}

约分:

2

= sin (\frac{π}{4} + θ)

化简

\frac{- 1}{2} : - \frac{2}{2}

\frac{- 1}{2}

使用分式法则:

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

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

sin (\frac{π}{4} + θ) > - \frac{2}{2}

sin (\frac{π}{4} + θ) > - \frac{2}{2}

sin (\frac{π}{4} + θ) > - \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{π}{4} + θ) < π - arcsin (- \frac{2}{2}) + 2 πn

若

a < u < b

，则

a < u and u < b

arcsin (- \frac{2}{2}) + 2 πn < \frac{π}{4} + θ and \frac{π}{4} + θ < π - arcsin (- \frac{2}{2}) + 2 πn

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

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

交换两边

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

将

\frac{π}{4}

到右边

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

两边减去

\frac{π}{4}

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

化简

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

化简

\frac{π}{4} + θ - \frac{π}{4} : θ

\frac{π}{4} + θ - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} > 0

= θ

化简

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

- \frac{π}{4} + 2 πn - \frac{π}{4}

对同类项分组

= - \frac{π}{4} - \frac{π}{4} + 2 πn

合并分式

- \frac{π}{4} - \frac{π}{4} : - \frac{π}{2}

使用法则

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

= \frac{- π - π}{4}

同类项相加:

- π - π = - 2 π

= \frac{- 2 π}{4}

使用分式法则:

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

= - \frac{2 π}{4}

约分:

2

= - \frac{π}{2}

= - \frac{π}{2} + 2 πn

θ > - \frac{π}{2} + 2 πn

θ > - \frac{π}{2} + 2 πn

θ > - \frac{π}{2} + 2 πn

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

\frac{π}{4} + θ < 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{π}{4} + θ < π + \frac{π}{4} + 2 πn

将

\frac{π}{4}

到右边

\frac{π}{4} + θ < π + \frac{π}{4} + 2 πn

两边减去

\frac{π}{4}

\frac{π}{4} + θ - \frac{π}{4} < π + \frac{π}{4} + 2 πn - \frac{π}{4}

化简

\frac{π}{4} + θ - \frac{π}{4} < π + \frac{π}{4} + 2 πn - \frac{π}{4}

化简

\frac{π}{4} + θ - \frac{π}{4} : θ

\frac{π}{4} + θ - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} < 0

= θ

化简

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

π + \frac{π}{4} + 2 πn - \frac{π}{4}

对同类项分组

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

同类项相加:

\frac{π}{4} - \frac{π}{4} = 0

= π + 2 πn

θ < π + 2 πn

θ < π + 2 πn

θ < π + 2 πn

合并区间

θ > - \frac{π}{2} + 2 πn and θ < π + 2 πn

合并重叠的区间

- \frac{π}{2} + 2 πn < θ < π + 2 πn

流行的例子

-cos(x)-4sin(2x)<0

- cos (x) - 4 sin (2 x) < 0

50sin(-pi/2 x-pi/2)>=-15

50 sin (- \frac{π}{2} x - \frac{π}{2}) \geq - 15

sin(x)cos(y)-cos(x)sin(y)>= 0

sin (x) cos (y) - cos (x) sin (y) \geq 0

cos(θ)>0,tan(θ)>0

cos (θ) > 0, tan (θ) > 0

sqrt(3)cot(x)<-1

3 cot (x) < - 1