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受欢迎的三角函数 >

sin(x)<cos(x)<tan(x)

解答

sin (x) < cos (x) < tan (x)

解答

0.66623 \dots + 2 πn < x < \frac{π}{4} + 2 πn or \frac{5 π}{4} + 2 πn < x < \frac{3 π}{2} + 2 πn

+2

间隔符号

(0.66623 \dots + 2 πn, \frac{π}{4} + 2 πn) \cup (\frac{5 π}{4} + 2 πn, \frac{3 π}{2} + 2 πn)

十进制

0.66623 \dots + 2 πn < x < 0.78539 \dots + 2 πn or 3.92699 \dots + 2 πn < x < 4.71238 \dots + 2 πn

求解步骤

sin (x) < cos (x) < tan (x)

若

a < u < b

，则

a < u and u < b

sin (x) < cos (x) and cos (x) < tan (x)

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

sin (x) < cos (x)

将

cos (x)

para o lado esquerdo

sin (x) < cos (x)

两边减去

cos (x)

sin (x) - cos (x) < cos (x) - cos (x)

sin (x) - cos (x) < 0

sin (x) - cos (x) < 0

利用以下特性:

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

- 2 cos (\frac{π}{4} + x) < 0

在两边乘以

- 1

- 2 cos (\frac{π}{4} + x) < 0

两边乘以 -1（不等式变号）

(- 2 cos (\frac{π}{4} + x)) (- 1) > 0 \cdot (- 1)

化简

2 cos (\frac{π}{4} + x) > 0

2 cos (\frac{π}{4} + x) > 0

两边除以

2

2 cos (\frac{π}{4} + x) > 0

两边除以

2

\frac{2 cos ( \frac{π}{4} + x )}{2} > \frac{0}{2}

化简

cos (\frac{π}{4} + x) > 0

cos (\frac{π}{4} + x) > 0

对于

cos (x) > a

，若

- 1 \leq a < 1

，则

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (0) + 2 πn < (\frac{π}{4} + x) < arccos (0) + 2 πn

若

a < u < b

，则

a < u and u < b

- arccos (0) + 2 πn < \frac{π}{4} + x and \frac{π}{4} + x < arccos (0) + 2 πn

- arccos (0) + 2 πn < \frac{π}{4} + x : x > 2 πn - \frac{3 π}{4}

- arccos (0) + 2 πn < \frac{π}{4} + x

交换两边

\frac{π}{4} + x > - arccos (0) + 2 πn

化简

- arccos (0) + 2 πn : - \frac{π}{2} + 2 πn

- arccos (0) + 2 πn

使用以下普通恒等式:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{π}{2} + 2 πn

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

将

\frac{π}{4}

到右边

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

两边减去

\frac{π}{4}

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

化简

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

化简

\frac{π}{4} + x - \frac{π}{4} : x

\frac{π}{4} + x - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} > 0

= x

化简

- \frac{π}{2} + 2 πn - \frac{π}{4} : 2 πn - \frac{3 π}{4}

- \frac{π}{2} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn - \frac{π}{2} - \frac{π}{4}

2, 4

的最小公倍数:

4

2, 4

最小公倍数 (LCM)

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

将每个因子乘以它在

2

或

4

中出现的最多次数

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{π}{2} :

将分母和分子乘以

2

\frac{π}{2} = \frac{π 2}{2 \cdot 2} = \frac{π 2}{4}

= - \frac{π 2}{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 2 - π}{4}

同类项相加:

- 2 π - π = - 3 π

= \frac{- 3 π}{4}

使用分式法则:

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

= 2 πn - \frac{3 π}{4}

x > 2 πn - \frac{3 π}{4}

x > 2 πn - \frac{3 π}{4}

x > 2 πn - \frac{3 π}{4}

\frac{π}{4} + x < arccos (0) + 2 πn : x < 2 πn + \frac{π}{4}

\frac{π}{4} + x < arccos (0) + 2 πn

化简

arccos (0) + 2 πn : \frac{π}{2} + 2 πn

arccos (0) + 2 πn

使用以下普通恒等式:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{2} + 2 πn

\frac{π}{4} + x < \frac{π}{2} + 2 πn

将

\frac{π}{4}

到右边

\frac{π}{4} + x < \frac{π}{2} + 2 πn

两边减去

\frac{π}{4}

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

化简

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

化简

\frac{π}{4} + x - \frac{π}{4} : x

\frac{π}{4} + x - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} < 0

= x

化简

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

\frac{π}{2} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn + \frac{π}{2} - \frac{π}{4}

2, 4

的最小公倍数:

4

2, 4

最小公倍数 (LCM)

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

将每个因子乘以它在

2

或

4

中出现的最多次数

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{π}{2} :

将分母和分子乘以

2

\frac{π}{2} = \frac{π 2}{2 \cdot 2} = \frac{π 2}{4}

= \frac{π 2}{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 2 - π}{4}

同类项相加:

2 π - π = π

= 2 πn + \frac{π}{4}

x < 2 πn + \frac{π}{4}

x < 2 πn + \frac{π}{4}

x < 2 πn + \frac{π}{4}

合并区间

x > 2 πn - \frac{3 π}{4} and x < 2 πn + \frac{π}{4}

合并重叠的区间

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

cos (x) < tan (x) : 0.66623 \dots + 2 πn < x < \frac{π}{2} + 2 πn or π - 0.66623 \dots + 2 πn < x < \frac{3 π}{2} + 2 πn

cos (x) < tan (x)

将

tan (x)

para o lado esquerdo

cos (x) < tan (x)

两边减去

tan (x)

cos (x) - tan (x) < tan (x) - tan (x)

cos (x) - tan (x) < 0

cos (x) - tan (x) < 0

cos (x) - tan (x)

的周期:

2 π

周期函数和的复合周期是这些周期的最小公倍数

cos (x), tan (x)

cos (x)

的周期:

2 π

cos (x)

的周期是

2 π

= 2 π

tan (x)

的周期:

π

tan (x)

的周期是

π

= π

合并周期:

2 π, π

= 2 π

用 sin, cos 表示

cos (x) - tan (x) < 0

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

cos (x) - \frac{sin ( x )}{cos ( x )} < 0

cos (x) - \frac{sin ( x )}{cos ( x )} < 0

化简

cos (x) - \frac{sin ( x )}{cos ( x )} : \frac{cos ^{2} ( x ) - sin ( x )}{cos ( x )}

cos (x) - \frac{sin ( x )}{cos ( x )}

将项转换为分式:

cos (x) = \frac{cos ( x ) cos ( x )}{cos ( x )}

= \frac{cos ( x ) cos ( x )}{cos ( x )} - \frac{sin ( x )}{cos ( x )}

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{cos ( x ) cos ( x ) - sin ( x )}{cos ( x )}

cos (x) cos (x) - sin (x) = cos^{2} (x) - sin (x)

cos (x) cos (x) - sin (x)

cos (x) cos (x) = cos^{2} (x)

cos (x) cos (x)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

= cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= cos^{2} (x)

= cos^{2} (x) - sin (x)

= \frac{cos ^{2} ( x ) - sin ( x )}{cos ( x )}

\frac{cos ^{2} ( x ) - sin ( x )}{cos ( x )} < 0

确定

0 \leq x < 2 π

时

\frac{cos ^{2} ( x ) - sin ( x )}{cos ( x )}

的零点和无定义点

要找到零点，将不等式设置为零

\frac{cos ^{2} ( x ) - sin ( x )}{cos ( x )} = 0

\frac{cos ^{2} ( x ) - sin ( x )}{cos ( x )} = 0, 0 \leq x < 2 π : x = 0.66623 \dots, x = π - 0.66623 \dots

\frac{cos ^{2} ( x ) - sin ( x )}{cos ( x )} = 0, 0 \leq x < 2 π

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cos^{2} (x) - sin (x) = 0

使用三角恒等式改写

cos^{2} (x) - sin (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= 1 - sin^{2} (x) - sin (x)

1 - sin (x) - sin^{2} (x) = 0

用替代法求解

1 - sin (x) - sin^{2} (x) = 0

令

: sin (x) = u

1 - u - u^{2} = 0

1 - u - u^{2} = 0 : u = - \frac{1 + 5}{2}, u = \frac{5 - 1}{2}

1 - u - u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

- u^{2} - u + 1 = 0

使用求根公式求解

- u^{2} - u + 1 = 0

二次方程求根公式:

若

a = - 1, b = - 1, c = 1

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 1 ) \cdot 1}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 1 ) \cdot 1}{2 ( - 1 )}

(- 1)^{2} - 4 (- 1) \cdot 1 = 5

(- 1)^{2} - 4 (- 1) \cdot 1

使用法则

- (- a) = a

= (- 1)^{2} + 4 \cdot 1 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 5}{2 ( - 1 )}, u_{2} = \frac{- ( - 1 ) - 5}{2 ( - 1 )}

u = \frac{- ( - 1 ) + 5}{2 ( - 1 )} : - \frac{1 + 5}{2}

\frac{- ( - 1 ) + 5}{2 ( - 1 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{1 + 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 + 5}{- 2}

使用分式法则:

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

= - \frac{1 + 5}{2}

u = \frac{- ( - 1 ) - 5}{2 ( - 1 )} : \frac{5 - 1}{2}

\frac{- ( - 1 ) - 5}{2 ( - 1 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{1 - 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 - 5}{- 2}

使用分式法则:

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

1 - 5 = - (5 - 1)

= \frac{5 - 1}{2}

二次方程组的解是:

u = - \frac{1 + 5}{2}, u = \frac{5 - 1}{2}

u = sin (x)

代回

sin (x) = - \frac{1 + 5}{2}, sin (x) = \frac{5 - 1}{2}

sin (x) = - \frac{1 + 5}{2}, sin (x) = \frac{5 - 1}{2}

sin (x) = - \frac{1 + 5}{2}, 0 \leq x < 2 π :

无解

sin (x) = - \frac{1 + 5}{2}, 0 \leq x < 2 π

- 1 \leq sin (x) \leq 1

无解

sin (x) = \frac{5 - 1}{2}, 0 \leq x < 2 π : x = arcsin (\frac{5 - 1}{2}), x = π - arcsin (\frac{5 - 1}{2})

sin (x) = \frac{5 - 1}{2}, 0 \leq x < 2 π

使用反三角函数性质

sin (x) = \frac{5 - 1}{2}

sin (x) = \frac{5 - 1}{2}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

在

0 \leq x < 2 π

范围内的解

x = arcsin (\frac{5 - 1}{2}), x = π - arcsin (\frac{5 - 1}{2})

合并所有解

x = arcsin (\frac{5 - 1}{2}), x = π - arcsin (\frac{5 - 1}{2})

以小数形式表示解

x = 0.66623 \dots, x = π - 0.66623 \dots

确定无定义点:

x = \frac{π}{2}, x = \frac{3 π}{2}

找到分母的零解

cos (x) = 0

cos (x) = 0

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

在

0 \leq x < 2 π

范围内的解

x = \frac{π}{2}, x = \frac{3 π}{2}

0.66623 \dots, \frac{π}{2}, π - 0.66623 \dots, \frac{3 π}{2}

确定区间

0 < x < 0.66623 \dots, 0.66623 \dots < x < \frac{π}{2}, \frac{π}{2} < x < π - 0.66623 \dots, π - 0.66623 \dots < x < \frac{3 π}{2}, \frac{3 π}{2} < x < 2 π

总结如下表：

cos^{2} (x) - sin (x) cos (x) \frac{c o s ^{2} ( x ) - s i n ( x )}{c o s ( x )} x = 0 + + + 0 < x < 0.66623 \dots + + + x = 0.66623 \dots 0 + 0 0.66623 \dots < x < \frac{π}{2} - + - x = \frac{π}{2} - 0 未定义 \frac{π}{2} < x < π - 0.66623 \dots - - + x = π - 0.66623 \dots 0 - 0 π - 0.66623 \dots < x < \frac{3 π}{2} + - - x = \frac{3 π}{2} + 0 未定义 \frac{3 π}{2} < x < 2 π + + + x = 2 π + + +

确定满足所需条件的区间：

< 0

0.66623 \dots < x < \frac{π}{2} or π - 0.66623 \dots < x < \frac{3 π}{2}

使用周期

cos (x) - tan (x)

0.66623 \dots + 2 πn < x < \frac{π}{2} + 2 πn or π - 0.66623 \dots + 2 πn < x < \frac{3 π}{2} + 2 πn

合并区间

- \frac{3 π}{4} + 2 πn < x < \frac{π}{4} + 2 πn and (0.66623 \dots + 2 πn < x < \frac{π}{2} + 2 πn or π - 0.66623 \dots + 2 πn < x < \frac{3 π}{2} + 2 πn)

合并重叠的区间

0.66623 \dots + 2 πn < x < \frac{π}{4} + 2 πn or \frac{5 π}{4} + 2 πn < x < \frac{3 π}{2} + 2 πn

流行的例子

sin(x)0<= x<= 2pi

sin (x) 0 \leq x \leq 2 π

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

- \frac{π}{2} < sin (x) < \frac{π}{2}

sin(2x)>= 0\land cos(x)>0

sin (2 x) \geq 0 and cos (x) > 0

cos(x)= 5/13 \land sin(x)<0,tan(2x)

cos (x) = \frac{5}{13} and sin (x) < 0, tan (2 x)

sin(θ)sec(θ)>0\land sin(θ)<4

sin (θ) sec (θ) > 0 and sin (θ) < 4