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受欢迎的三角函数 >

arccot(x)+arccot(1+x)= pi/4

解答

arccot (x) + arccot (1 + x) = \frac{π}{4}

解答

x = 2

求解步骤

arccot (x) + arccot (1 + x) = \frac{π}{4}

使用三角恒等式改写

arccot (x) + arccot (1 + x)

使用和差化积恒等式:

arccot (s) + arccot (t) = arccot (\frac{s t - 1}{t + s})

= arccot (\frac{x ( 1 + x ) - 1}{1 + x + x})

arccot (\frac{x ( 1 + x ) - 1}{1 + x + x}) = \frac{π}{4}

使用反三角函数性质

arccot (\frac{x ( 1 + x ) - 1}{1 + x + x}) = \frac{π}{4}

arccot (x) = a \Rightarrow x = cot (a)

\frac{x ( 1 + x ) - 1}{1 + x + x} = cot (\frac{π}{4})

cot (\frac{π}{4}) = 1

cot (\frac{π}{4})

使用以下普通恒等式:

cot (\frac{π}{4}) = 1

cot (\frac{π}{4})

cot (x)

周期表（周期为

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

= 1

= 1

\frac{x ( 1 + x ) - 1}{1 + x + x} = 1

\frac{x ( 1 + x ) - 1}{1 + x + x} = 1

解

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

\frac{x ( 1 + x ) - 1}{1 + x + x} = 1

化简

\frac{x ( 1 + x ) - 1}{1 + x + x} : \frac{x ( 1 + x ) - 1}{1 + 2 x}

\frac{x ( 1 + x ) - 1}{1 + x + x}

同类项相加:

x + x = 2 x

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

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

在两边乘以

1 + 2 x

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

在两边乘以

1 + 2 x

\frac{x ( 1 + x ) - 1}{1 + 2 x} (1 + 2 x) = 1 \cdot (1 + 2 x)

化简

\frac{x ( 1 + x ) - 1}{1 + 2 x} (1 + 2 x) = 1 \cdot (1 + 2 x)

化简

\frac{x ( 1 + x ) - 1}{1 + 2 x} (1 + 2 x) : x (1 + x) - 1

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

分式相乘:

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

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

约分:

1 + 2 x

= x (1 + x) - 1

化简

1 \cdot (1 + 2 x) : 1 + 2 x

1 \cdot (1 + 2 x)

乘以:

1 \cdot (1 + 2 x) = (1 + 2 x)

= (1 + 2 x)

去除括号:

(a) = a

= 1 + 2 x

x (1 + x) - 1 = 1 + 2 x

x (1 + x) - 1 = 1 + 2 x

x (1 + x) - 1 = 1 + 2 x

解

x (1 + x) - 1 = 1 + 2 x : x = 2, x = - 1

x (1 + x) - 1 = 1 + 2 x

展开

x (1 + x) - 1 : x + x^{2} - 1

x (1 + x) - 1

乘开

x (1 + x) : x + x^{2}

x (1 + x)

使用分配律:

a (b + c) = ab + a c

a = x, b = 1, c = x

= x \cdot 1 + xx

= 1 \cdot x + xx

化简

1 \cdot x + xx : x + x^{2}

1 \cdot x + xx

1 \cdot x = x

1 \cdot x

乘以:

1 \cdot x = x

= x

xx = x^{2}

xx

使用指数法则:

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

xx = x^{1 + 1}

= x^{1 + 1}

数字相加:

1 + 1 = 2

= x^{2}

= x + x^{2}

= x + x^{2}

= x + x^{2} - 1

x + x^{2} - 1 = 1 + 2 x

将

2 x

para o lado esquerdo

x + x^{2} - 1 = 1 + 2 x

两边减去

2 x

x + x^{2} - 1 - 2 x = 1 + 2 x - 2 x

化简

x^{2} - x - 1 = 1

x^{2} - x - 1 = 1

将

1

para o lado esquerdo

x^{2} - x - 1 = 1

两边减去

1

x^{2} - x - 1 - 1 = 1 - 1

化简

x^{2} - x - 2 = 0

x^{2} - x - 2 = 0

使用求根公式求解

x^{2} - x - 2 = 0

二次方程求根公式:

若

a = 1, b = - 1, c = - 2

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

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 2) = 3

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

使用法则

- (- a) = a

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

(- 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 2 = 8

4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

x_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 \cdot 1}

将解分隔开

x_{1} = \frac{- ( - 1 ) + 3}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 3}{2 \cdot 1}

x = \frac{- ( - 1 ) + 3}{2 \cdot 1} : 2

\frac{- ( - 1 ) + 3}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 + 3}{2 \cdot 1}

数字相加:

1 + 3 = 4

= \frac{4}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

x = \frac{- ( - 1 ) - 3}{2 \cdot 1} : - 1

\frac{- ( - 1 ) - 3}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 - 3}{2 \cdot 1}

数字相减:

1 - 3 = - 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{2}

使用分式法则:

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

= - \frac{2}{2}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

x = 2, x = - 1

x = 2, x = - 1

验证解

找到无定义的点（奇点）:

x = - \frac{1}{2}

取

\frac{x ( 1 + x ) - 1}{1 + x + x}

的分母，令其等于零

解

1 + x + x = 0 : x = - \frac{1}{2}

1 + x + x = 0

同类项相加:

x + x = 2 x

1 + 2 x = 0

将

1

到右边

1 + 2 x = 0

两边减去

1

1 + 2 x - 1 = 0 - 1

化简

2 x = - 1

2 x = - 1

两边除以

2

2 x = - 1

两边除以

2

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

化简

x = - \frac{1}{2}

x = - \frac{1}{2}

以下点无定义

x = - \frac{1}{2}

将不在定义域的点与解相综合:

x = 2, x = - 1

x = 2, x = - 1

将解代入原方程进行验证

将它们代入

arccot (x) + arccot (1 + x) = \frac{π}{4}

检验解是否符合

去除与方程不符的解。

检验

2

的解:

真

2

代入

n = 1

2

对于

arccot (x) + arccot (1 + x) = \frac{π}{4}

代入

x = 2

arccot (2) + arccot (1 + 2) = \frac{π}{4}

整理后得

0.78539 \dots = 0.78539 \dots

\Rightarrow 真

检验

- 1

的解:

假

- 1

代入

n = 1

- 1

对于

arccot (x) + arccot (1 + x) = \frac{π}{4}

代入

x = - 1

arccot (- 1) + arccot (1 - 1) = \frac{π}{4}

整理后得

3.92699 \dots = 0.78539 \dots

\Rightarrow 假

x = 2

作图

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