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受欢迎的三角函数 >

arctan(1/(x-1))+arctan(2/(x+1))= pi/4

解答

arctan (\frac{1}{x - 1}) + arctan (\frac{2}{x + 1}) = \frac{π}{4}

解答

x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

求解步骤

arctan (\frac{1}{x - 1}) + arctan (\frac{2}{x + 1}) = \frac{π}{4}

使用三角恒等式改写

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

使用和差化积恒等式:

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

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

arctan (\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}}) = \frac{π}{4}

使用反三角函数性质

arctan (\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}}) = \frac{π}{4}

arctan (x) = a \Rightarrow x = tan (a)

\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}} = tan (\frac{π}{4})

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

使用以下普通恒等式:

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

= 1

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

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

解

\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}} = 1 : x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

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

化简

\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}} : \frac{3 x - 1}{x ^{2} - 3}

\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}}

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

\frac{1}{x - 1} \cdot \frac{2}{x + 1}

分式相乘:

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

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

数字相乘:

1 \cdot 2 = 2

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

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

化简

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

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

x - 1, x + 1

的最小公倍数:

(x - 1) (x + 1)

x - 1, x + 1

最小公倍数 (LCM)

计算出由出现在

x - 1

或

x + 1

中的因子组成的表达式

= (x - 1) (x + 1)

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

(x - 1) (x + 1)

对于

\frac{1}{x - 1} :

将分母和分子乘以

x + 1

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

对于

\frac{2}{x + 1} :

将分母和分子乘以

x - 1

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

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

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

乘开

x + 1 + 2 (x - 1) : 3 x - 1

x + 1 + 2 (x - 1)

乘开

2 (x - 1) : 2 x - 2

2 (x - 1)

使用分配律:

a (b - c) = ab - a c

a = 2, b = x, c = 1

= 2 x - 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 2 x - 2

= x + 1 + 2 x - 2

化简

x + 1 + 2 x - 2 : 3 x - 1

x + 1 + 2 x - 2

对同类项分组

= x + 2 x + 1 - 2

同类项相加:

x + 2 x = 3 x

= 3 x + 1 - 2

数字相加/相减:

1 - 2 = - 1

= 3 x - 1

= 3 x - 1

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

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

使用分式法则:

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

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

化简

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

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

将项转换为分式:

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

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

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

乘以:

1 \cdot (x - 1) = (x - 1)

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

乘开

(x - 1) (x + 1) - 2 : x^{2} - 3

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

乘开

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

(x - 1) (x + 1)

使用平方差公式:

(a - b) (a + b) = a^{2} - b^{2}

a = x, b = 1

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

使用法则

1^{a} = 1

1^{2} = 1

= x^{2} - 1

= x^{2} - 1 - 2

数字相减:

- 1 - 2 = - 3

= x^{2} - 3

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

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

乘

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

(x - 1) (x + 1) \frac{x ^{2} - 3}{( x - 1 ) ( x + 1 )}

分式相乘:

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

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

约分:

x - 1

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

约分:

x + 1

= x^{2} - 3

= \frac{3 x - 1}{x ^{2} - 3}

\frac{3 x - 1}{x ^{2} - 3} = 1

在两边乘以

x^{2} - 3

\frac{3 x - 1}{x ^{2} - 3} = 1

在两边乘以

x^{2} - 3

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

化简

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

化简

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

\frac{3 x - 1}{x ^{2} - 3} (x^{2} - 3)

分式相乘:

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

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

约分:

x^{2} - 3

= 3 x - 1

化简

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

1 \cdot (x^{2} - 3)

乘以:

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

= (x^{2} - 3)

去除括号:

(a) = a

= x^{2} - 3

3 x - 1 = x^{2} - 3

3 x - 1 = x^{2} - 3

3 x - 1 = x^{2} - 3

解

3 x - 1 = x^{2} - 3 : x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

3 x - 1 = x^{2} - 3

交换两边

x^{2} - 3 = 3 x - 1

将

1

para o lado esquerdo

x^{2} - 3 = 3 x - 1

两边加上

1

x^{2} - 3 + 1 = 3 x - 1 + 1

化简

x^{2} - 2 = 3 x

x^{2} - 2 = 3 x

将

3 x

para o lado esquerdo

x^{2} - 2 = 3 x

两边减去

3 x

x^{2} - 2 - 3 x = 3 x - 3 x

化简

x^{2} - 2 - 3 x = 0

x^{2} - 2 - 3 x = 0

改写成标准形式

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

x^{2} - 3 x - 2 = 0

使用求根公式求解

x^{2} - 3 x - 2 = 0

二次方程求根公式:

若

a = 1, b = - 3, c = - 2

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 2) = 17

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 3^{2} + 8

3^{2} = 9

= 9 + 8

数字相加:

9 + 8 = 17

= 17

x_{1, 2} = \frac{- ( - 3 ) \pm 17}{2 \cdot 1}

将解分隔开

x_{1} = \frac{- ( - 3 ) + 17}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 17}{2 \cdot 1}

x = \frac{- ( - 3 ) + 17}{2 \cdot 1} : \frac{3 + 17}{2}

\frac{- ( - 3 ) + 17}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{3 + 17}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{3 + 17}{2}

x = \frac{- ( - 3 ) - 17}{2 \cdot 1} : \frac{3 - 17}{2}

\frac{- ( - 3 ) - 17}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{3 - 17}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{3 - 17}{2}

二次方程组的解是:

x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

验证解

找到无定义的点（奇点）:

x = 3, x = - 3, x = 1, x = - 1

取

\frac{\frac{1}{x - 1} + \frac{2}{x + 1}}{1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}}

的分母，令其等于零

解

1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1} = 0 : x = 3, x = - 3

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

化简

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

- \frac{1}{x - 1} \cdot \frac{2}{x + 1}

分式相乘:

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

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

数字相乘:

1 \cdot 2 = 2

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

1 - \frac{2}{( x - 1 ) ( x + 1 )} = 0

在两边乘以

(x - 1) (x + 1)

1 - \frac{2}{( x - 1 ) ( x + 1 )} = 0

在两边乘以

(x - 1) (x + 1)

1 \cdot (x - 1) (x + 1) - \frac{2}{( x - 1 ) ( x + 1 )} (x - 1) (x + 1) = 0 \cdot (x - 1) (x + 1)

化简

1 \cdot (x - 1) (x + 1) - \frac{2}{( x - 1 ) ( x + 1 )} (x - 1) (x + 1) = 0 \cdot (x - 1) (x + 1)

化简

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

1 \cdot (x - 1) (x + 1)

乘以:

1 \cdot (x - 1) = (x - 1)

= (x - 1) (x + 1)

化简

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

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

分式相乘:

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

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

约分:

x - 1

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

约分:

x + 1

= - 2

化简

0 \cdot (x - 1) (x + 1) : 0

0 \cdot (x - 1) (x + 1)

使用法则

0 \cdot a = 0

= 0

(x - 1) (x + 1) - 2 = 0

(x - 1) (x + 1) - 2 = 0

(x - 1) (x + 1) - 2 = 0

解

(x - 1) (x + 1) - 2 = 0 : x = 3, x = - 3

(x - 1) (x + 1) - 2 = 0

展开

(x - 1) (x + 1) - 2 : x^{2} - 3

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

乘开

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

(x - 1) (x + 1)

使用平方差公式:

(a - b) (a + b) = a^{2} - b^{2}

a = x, b = 1

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

使用法则

1^{a} = 1

1^{2} = 1

= x^{2} - 1

= x^{2} - 1 - 2

数字相减:

- 1 - 2 = - 3

= x^{2} - 3

x^{2} - 3 = 0

使用求根公式求解

x^{2} - 3 = 0

二次方程求根公式:

若

a = 1, b = 0, c = - 3

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

0^{2} - 4 \cdot 1 \cdot (- 3) = 23

0^{2} - 4 \cdot 1 \cdot (- 3)

使用法则

0^{a} = 0

0^{2} = 0

= 0 - 4 \cdot 1 \cdot (- 3)

使用法则

- (- a) = a

= 0 + 4 \cdot 1 \cdot 3

数字相乘:

4 \cdot 1 \cdot 3 = 12

= 0 + 12

数字相加:

0 + 12 = 12

= 12

12

质因数分解:

2^{2} \cdot 3

12

12

除以

212 = 6 \cdot 2

= 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

使用根式运算法则:

n ab = n a n b

= 3 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 23

x_{1, 2} = \frac{- 0 \pm 2 3}{2 \cdot 1}

将解分隔开

x_{1} = \frac{- 0 + 2 3}{2 \cdot 1}, x_{2} = \frac{- 0 - 2 3}{2 \cdot 1}

x = \frac{- 0 + 2 3}{2 \cdot 1} : 3

\frac{- 0 + 2 3}{2 \cdot 1}

- 0 + 23 = 23

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

数字相乘:

2 \cdot 1 = 2

= \frac{2 3}{2}

数字相除:

\frac{2}{2} = 1

= 3

x = \frac{- 0 - 2 3}{2 \cdot 1} : - 3

\frac{- 0 - 2 3}{2 \cdot 1}

- 0 - 23 = - 23

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2 3}{2}

使用分式法则:

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

= - \frac{2 3}{2}

数字相除:

\frac{2}{2} = 1

= - 3

二次方程组的解是:

x = 3, x = - 3

x = 3, x = - 3

验证解

找到无定义的点（奇点）:

x = 1, x = - 1

取

1 - \frac{1}{x - 1} \cdot \frac{2}{x + 1}

的分母，令其等于零

解

x - 1 = 0 : x = 1

x - 1 = 0

将

1

到右边

x - 1 = 0

两边加上

1

x - 1 + 1 = 0 + 1

化简

x = 1

x = 1

解

x + 1 = 0 : x = - 1

x + 1 = 0

将

1

到右边

x + 1 = 0

两边减去

1

x + 1 - 1 = 0 - 1

化简

x = - 1

x = - 1

以下点无定义

x = 1, x = - 1

将不在定义域的点与解相综合:

x = 3, x = - 3

解

x - 1 = 0 : x = 1

x - 1 = 0

将

1

到右边

x - 1 = 0

两边加上

1

x - 1 + 1 = 0 + 1

化简

x = 1

x = 1

解

x + 1 = 0 : x = - 1

x + 1 = 0

将

1

到右边

x + 1 = 0

两边减去

1

x + 1 - 1 = 0 - 1

化简

x = - 1

x = - 1

以下点无定义

x = 3, x = - 3, x = 1, x = - 1

将不在定义域的点与解相综合:

x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

将解代入原方程进行验证

将它们代入

arctan (\frac{1}{x - 1}) + arctan (\frac{2}{x + 1}) = \frac{π}{4}

检验解是否符合

去除与方程不符的解。

检验

\frac{3 + 17}{2}

的解:

真

\frac{3 + 17}{2}

代入

n = 1

\frac{3 + 17}{2}

对于

arctan (\frac{1}{x - 1}) + arctan (\frac{2}{x + 1}) = \frac{π}{4}

代入

x = \frac{3 + 17}{2}

arctan (\frac{1}{\frac{3 + 17}{2} - 1}) + arctan (\frac{2}{\frac{3 + 17}{2} + 1}) = \frac{π}{4}

整理后得

0.78539 \dots = 0.78539 \dots

\Rightarrow 真

检验

\frac{3 - 17}{2}

的解:

真

\frac{3 - 17}{2}

代入

n = 1

\frac{3 - 17}{2}

对于

arctan (\frac{1}{x - 1}) + arctan (\frac{2}{x + 1}) = \frac{π}{4}

代入

x = \frac{3 - 17}{2}

arctan (\frac{1}{\frac{3 - 17}{2} - 1}) + arctan (\frac{2}{\frac{3 - 17}{2} + 1}) = \frac{π}{4}

整理后得

0.78539 \dots = 0.78539 \dots

\Rightarrow 真

x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

作图

流行的例子

cot(3x)=-tan(-(2pi)/5)

cot (3 x) = - tan (- \frac{2 π}{5})

cos (x) = \frac{60}{61}

cot (x) = 2

cos(θ)cot(θ)=-cos(θ)

cos (θ) cot (θ) = - cos (θ)

tan(x/2)=-1/(sqrt(3))

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