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受欢迎的三角函数 >

arctan(x+1/3)+arctan(x-1/3)=arctan(2)

解答

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

解答

x = \frac{2}{3}

求解步骤

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

使用三角恒等式改写

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

使用和差化积恒等式:

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

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

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

使用反三角函数性质

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

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

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

tan (arctan (2)) = 2

tan (arctan (2))

使用三角恒等式改写:

tan (arctan (2)) = 2

利用以下特性:

tan (arctan (x)) = x

= 2

= 2

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

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

解

\frac{x + \frac{1}{3} + x - \frac{1}{3}}{1 - ( x + \frac{1}{3} ) ( x - \frac{1}{3} )} = 2 : x = - \frac{5}{3}, x = \frac{2}{3}

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

化简

\frac{x + \frac{1}{3} + x - \frac{1}{3}}{1 - ( x + \frac{1}{3} ) ( x - \frac{1}{3} )} : \frac{18 x}{- 9 x ^{2} + 10}

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

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

x + \frac{1}{3} + x - \frac{1}{3}

对同类项分组

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

同类项相加:

x + x = 2 x

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

同类项相加:

\frac{1}{3} - \frac{1}{3} = 0

= 2 x

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

乘开

1 - (x + \frac{1}{3}) (x - \frac{1}{3}) : - x^{2} + \frac{10}{9}

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

乘开

- (x + \frac{1}{3}) (x - \frac{1}{3}) : - x^{2} + \frac{1}{9}

乘开

(x + \frac{1}{3}) (x - \frac{1}{3}) : x^{2} - \frac{1}{9}

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

使用平方差公式:

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

a = x, b = \frac{1}{3}

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

(\frac{1}{3})^{2} = \frac{1}{9}

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

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

使用法则

1^{a} = 1

1^{2} = 1

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

3^{2} = 9

= \frac{1}{9}

= x^{2} - \frac{1}{9}

= - (x^{2} - \frac{1}{9})

打开括号

= - (x^{2}) - (- \frac{1}{9})

使用加减运算法则

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

= - x^{2} + \frac{1}{9}

= 1 - x^{2} + \frac{1}{9}

合并分式

1 + \frac{1}{9} : \frac{10}{9}

1 + \frac{1}{9}

将项转换为分式:

1 = \frac{1 \cdot 9}{9}

= \frac{1 \cdot 9}{9} + \frac{1}{9}

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 9 + 1}{9}

1 \cdot 9 + 1 = 10

1 \cdot 9 + 1

数字相乘:

1 \cdot 9 = 9

= 9 + 1

数字相加:

9 + 1 = 10

= 10

= \frac{10}{9}

= - x^{2} + \frac{10}{9}

= \frac{2 x}{- x ^{2} + \frac{10}{9}}

化简

- x^{2} + \frac{10}{9} : \frac{- 9 x ^{2} + 10}{9}

- x^{2} + \frac{10}{9}

将项转换为分式:

x^{2} = \frac{x ^{2} 9}{9}

= - \frac{x ^{2} \cdot 9}{9} + \frac{10}{9}

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 ^{2} \cdot 9 + 10}{9}

= \frac{2 x}{\frac{- 9 x ^{2} + 10}{9}}

使用分式法则:

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

= \frac{2 x \cdot 9}{- x ^{2} \cdot 9 + 10}

数字相乘:

2 \cdot 9 = 18

= \frac{18 x}{- 9 x ^{2} + 10}

\frac{18 x}{- 9 x ^{2} + 10} = 2

在两边乘以

- 9 x^{2} + 10

\frac{18 x}{- 9 x ^{2} + 10} = 2

在两边乘以

- 9 x^{2} + 10

\frac{18 x}{- 9 x ^{2} + 10} (- 9 x^{2} + 10) = 2 (- 9 x^{2} + 10)

化简

18 x = 2 (- 9 x^{2} + 10)

18 x = 2 (- 9 x^{2} + 10)

解

18 x = 2 (- 9 x^{2} + 10) : x = - \frac{5}{3}, x = \frac{2}{3}

18 x = 2 (- 9 x^{2} + 10)

展开

2 (- 9 x^{2} + 10) : - 18 x^{2} + 20

2 (- 9 x^{2} + 10)

使用分配律:

a (b + c) = ab + a c

a = 2, b = - 9 x^{2}, c = 10

= 2 (- 9 x^{2}) + 2 \cdot 10

使用加减运算法则

+ (- a) = - a

= - 2 \cdot 9 x^{2} + 2 \cdot 10

化简

- 2 \cdot 9 x^{2} + 2 \cdot 10 : - 18 x^{2} + 20

- 2 \cdot 9 x^{2} + 2 \cdot 10

数字相乘:

2 \cdot 9 = 18

= - 18 x^{2} + 2 \cdot 10

数字相乘:

2 \cdot 10 = 20

= - 18 x^{2} + 20

= - 18 x^{2} + 20

18 x = - 18 x^{2} + 20

交换两边

- 18 x^{2} + 20 = 18 x

将

18 x

para o lado esquerdo

- 18 x^{2} + 20 = 18 x

两边减去

18 x

- 18 x^{2} + 20 - 18 x = 18 x - 18 x

化简

- 18 x^{2} + 20 - 18 x = 0

- 18 x^{2} + 20 - 18 x = 0

改写成标准形式

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

- 18 x^{2} - 18 x + 20 = 0

使用求根公式求解

- 18 x^{2} - 18 x + 20 = 0

二次方程求根公式:

若

a = - 18, b = - 18, c = 20

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 ( - 18 ) \cdot 20}{2 ( - 18 )}

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 ( - 18 ) \cdot 20}{2 ( - 18 )}

(- 18)^{2} - 4 (- 18) \cdot 20 = 42

(- 18)^{2} - 4 (- 18) \cdot 20

使用法则

- (- a) = a

= (- 18)^{2} + 4 \cdot 18 \cdot 20

使用指数法则:

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

若

n

是偶数

(- 18)^{2} = 1 8^{2}

= 1 8^{2} + 4 \cdot 18 \cdot 20

数字相乘:

4 \cdot 18 \cdot 20 = 1440

= 1 8^{2} + 1440

1 8^{2} = 324

= 324 + 1440

数字相加:

324 + 1440 = 1764

= 1764

因式分解数字:

1764 = 4 2^{2}

= 4 2^{2}

使用根式运算法则:

n a^{n} = a

4 2^{2} = 42

= 42

x_{1, 2} = \frac{- ( - 18 ) \pm 42}{2 ( - 18 )}

将解分隔开

x_{1} = \frac{- ( - 18 ) + 42}{2 ( - 18 )}, x_{2} = \frac{- ( - 18 ) - 42}{2 ( - 18 )}

x = \frac{- ( - 18 ) + 42}{2 ( - 18 )} : - \frac{5}{3}

\frac{- ( - 18 ) + 42}{2 ( - 18 )}

去除括号:

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

= \frac{18 + 42}{- 2 \cdot 18}

数字相加:

18 + 42 = 60

= \frac{60}{- 2 \cdot 18}

数字相乘:

2 \cdot 18 = 36

= \frac{60}{- 36}

使用分式法则:

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

= - \frac{60}{36}

约分:

12

= - \frac{5}{3}

x = \frac{- ( - 18 ) - 42}{2 ( - 18 )} : \frac{2}{3}

\frac{- ( - 18 ) - 42}{2 ( - 18 )}

去除括号:

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

= \frac{18 - 42}{- 2 \cdot 18}

数字相减:

18 - 42 = - 24

= \frac{- 24}{- 2 \cdot 18}

数字相乘:

2 \cdot 18 = 36

= \frac{- 24}{- 36}

使用分式法则:

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

= \frac{24}{36}

约分:

12

= \frac{2}{3}

二次方程组的解是:

x = - \frac{5}{3}, x = \frac{2}{3}

x = - \frac{5}{3}, x = \frac{2}{3}

验证解

找到无定义的点（奇点）:

x = - \frac{10}{3}, x = \frac{10}{3}

取

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

的分母，令其等于零

解

1 - (x + \frac{1}{3}) (x - \frac{1}{3}) = 0 : x = - \frac{10}{3}, x = \frac{10}{3}

1 - (x + \frac{1}{3}) (x - \frac{1}{3}) = 0

展开

1 - (x + \frac{1}{3}) (x - \frac{1}{3}) : - x^{2} + \frac{10}{9}

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

乘开

- (x + \frac{1}{3}) (x - \frac{1}{3}) : - x^{2} + \frac{1}{9}

乘开

(x + \frac{1}{3}) (x - \frac{1}{3}) : x^{2} - \frac{1}{9}

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

使用平方差公式:

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

a = x, b = \frac{1}{3}

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

(\frac{1}{3})^{2} = \frac{1}{9}

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

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

使用法则

1^{a} = 1

1^{2} = 1

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

3^{2} = 9

= \frac{1}{9}

= x^{2} - \frac{1}{9}

= - (x^{2} - \frac{1}{9})

打开括号

= - (x^{2}) - (- \frac{1}{9})

使用加减运算法则

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

= - x^{2} + \frac{1}{9}

= 1 - x^{2} + \frac{1}{9}

合并分式

1 + \frac{1}{9} : \frac{10}{9}

1 + \frac{1}{9}

将项转换为分式:

1 = \frac{1 \cdot 9}{9}

= \frac{1 \cdot 9}{9} + \frac{1}{9}

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 9 + 1}{9}

1 \cdot 9 + 1 = 10

1 \cdot 9 + 1

数字相乘:

1 \cdot 9 = 9

= 9 + 1

数字相加:

9 + 1 = 10

= 10

= \frac{10}{9}

= - x^{2} + \frac{10}{9}

- x^{2} + \frac{10}{9} = 0

使用求根公式求解

- x^{2} + \frac{10}{9} = 0

二次方程求根公式:

若

a = - 1, b = 0, c = \frac{10}{9}

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 ( - 1 ) \frac{10}{9}}{2 ( - 1 )}

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 ( - 1 ) \frac{10}{9}}{2 ( - 1 )}

0^{2} - 4 (- 1) \frac{10}{9} = \frac{2 10}{3}

0^{2} - 4 (- 1) \frac{10}{9}

使用法则

0^{a} = 0

0^{2} = 0

= 0 - 4 (- 1) \frac{10}{9}

使用法则

- (- a) = a

= 0 + 4 \cdot 1 \cdot \frac{10}{9}

4 \cdot 1 \cdot \frac{10}{9} = \frac{40}{9}

4 \cdot 1 \cdot \frac{10}{9}

分式相乘:

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

= 1 \cdot \frac{10 \cdot 4}{9}

数字相乘:

10 \cdot 4 = 40

= 1 \cdot \frac{40}{9}

乘以:

1 \cdot \frac{40}{9} = \frac{40}{9}

= \frac{40}{9}

= 0 + \frac{40}{9}

0 + \frac{40}{9} = \frac{40}{9}

= \frac{40}{9}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{40}{9}

9 = 3

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

= \frac{40}{3}

40 = 210

40

40

质因数分解:

2^{3} \cdot 5

40

40

除以

240 = 20 \cdot 2

= 2 \cdot 20

20

除以

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 10

10

除以

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 5

= 2^{3} \cdot 5

= 2^{3} \cdot 5

使用指数法则:

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

= 2^{2} \cdot 2 \cdot 5

使用根式运算法则:

n ab = n a n b

= 2^{2} 2 \cdot 5

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 2 \cdot 5

整理后得

= 210

= \frac{2 10}{3}

x_{1, 2} = \frac{- 0 \pm \frac{2 10}{3}}{2 ( - 1 )}

将解分隔开

x_{1} = \frac{- 0 + \frac{2 10}{3}}{2 ( - 1 )}, x_{2} = \frac{- 0 - \frac{2 10}{3}}{2 ( - 1 )}

x = \frac{- 0 + \frac{2 10}{3}}{2 ( - 1 )} : - \frac{10}{3}

\frac{- 0 + \frac{2 10}{3}}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 0 + \frac{2 10}{3}}{- 2 \cdot 1}

- 0 + \frac{2 10}{3} = \frac{2 10}{3}

= \frac{\frac{2 10}{3}}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{\frac{2 10}{3}}{- 2}

使用分式法则:

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

= - \frac{\frac{2 10}{3}}{2}

使用分式法则:

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

\frac{\frac{2 10}{3}}{2} = \frac{2 10}{3 \cdot 2}

= - \frac{2 10}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= - \frac{2 10}{6}

约分:

2

= - \frac{10}{3}

x = \frac{- 0 - \frac{2 10}{3}}{2 ( - 1 )} : \frac{10}{3}

\frac{- 0 - \frac{2 10}{3}}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 0 - \frac{2 10}{3}}{- 2 \cdot 1}

- 0 - \frac{2 10}{3} = - \frac{2 10}{3}

= \frac{- \frac{2 10}{3}}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- \frac{2 10}{3}}{- 2}

使用分式法则:

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

= \frac{\frac{2 10}{3}}{2}

使用分式法则:

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

= \frac{2 10}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{2 10}{6}

约分:

2

= \frac{10}{3}

二次方程组的解是:

x = - \frac{10}{3}, x = \frac{10}{3}

以下点无定义

x = - \frac{10}{3}, x = \frac{10}{3}

将不在定义域的点与解相综合:

x = - \frac{5}{3}, x = \frac{2}{3}

x = - \frac{5}{3}, x = \frac{2}{3}

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

- \frac{5}{3}

的解:

假

- \frac{5}{3}

代入

n = 1

- \frac{5}{3}

对于

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

代入

x = - \frac{5}{3}

arctan (- \frac{5}{3} + \frac{1}{3}) + arctan (- \frac{5}{3} - \frac{1}{3}) = arctan (2)

整理后得

- 2.03444 \dots = 1.10714 \dots

\Rightarrow 假

检验

\frac{2}{3}

的解:

真

\frac{2}{3}

代入

n = 1

\frac{2}{3}

对于

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

代入

x = \frac{2}{3}

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

整理后得

1.10714 \dots = 1.10714 \dots

\Rightarrow 真

x = \frac{2}{3}

作图

流行的例子

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