zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

arctan(x)+arctan(1-x)=arctan(9/7)

解答

arctan (x) + arctan (1 - x) = arctan (\frac{9}{7})

解答

x = \frac{2}{3}, x = \frac{1}{3}

求解步骤

arctan (x) + arctan (1 - x) = arctan (\frac{9}{7})

使用三角恒等式改写

arctan (x) + arctan (1 - x)

使用和差化积恒等式:

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

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

arctan (\frac{x + 1 - x}{1 - x ( 1 - x )}) = arctan (\frac{9}{7})

使用反三角函数性质

arctan (\frac{x + 1 - x}{1 - x ( 1 - x )}) = arctan (\frac{9}{7})

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

\frac{x + 1 - x}{1 - x ( 1 - x )} = tan (arctan (\frac{9}{7}))

tan (arctan (\frac{9}{7})) = \frac{9}{7}

tan (arctan (\frac{9}{7}))

使用三角恒等式改写:

tan (arctan (\frac{9}{7})) = \frac{9}{7}

利用以下特性:

tan (arctan (x)) = x

= \frac{9}{7}

= \frac{9}{7}

\frac{x + 1 - x}{1 - x ( 1 - x )} = \frac{9}{7}

\frac{x + 1 - x}{1 - x ( 1 - x )} = \frac{9}{7}

解

\frac{x + 1 - x}{1 - x ( 1 - x )} = \frac{9}{7} : x = \frac{2}{3}, x = \frac{1}{3}

\frac{x + 1 - x}{1 - x ( 1 - x )} = \frac{9}{7}

使用分式交叉相乘

\frac{x + 1 - x}{1 - x ( 1 - x )} = \frac{9}{7}

化简

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

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

x + 1 - x = 1

x + 1 - x

对同类项分组

= x - x + 1

同类项相加:

x - x = 0

= 1

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

\frac{1}{1 - x ( 1 - x )} = \frac{9}{7}

使用分式交叉相乘: 若

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

则

a \cdot d = b \cdot c

1 \cdot 7 = (1 - x (1 - x)) \cdot 9

化简

1 \cdot 7 : 7

1 \cdot 7

数字相乘:

1 \cdot 7 = 7

= 7

7 = (1 - x (1 - x)) \cdot 9

7 = (1 - x (1 - x)) \cdot 9

解

7 = (1 - x (1 - x)) \cdot 9 : x = \frac{2}{3}, x = \frac{1}{3}

7 = (1 - x (1 - x)) \cdot 9

展开

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

(1 - x (1 - x)) \cdot 9

乘开

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

1 - x (1 - x)

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 - (- x) x

使用加减运算法则

- (- a) = a

= - 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}

= 1 - x + x^{2}

= 9 (x^{2} - x + 1)

= 9 (1 - x + x^{2})

打开括号

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

使用加减运算法则

+ (- a) = - a

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

数字相乘:

9 \cdot 1 = 9

= 9 - 9 x + 9 x^{2}

7 = 9 - 9 x + 9 x^{2}

交换两边

9 - 9 x + 9 x^{2} = 7

将

7

para o lado esquerdo

9 - 9 x + 9 x^{2} = 7

两边减去

7

9 - 9 x + 9 x^{2} - 7 = 7 - 7

化简

9 x^{2} - 9 x + 2 = 0

9 x^{2} - 9 x + 2 = 0

使用求根公式求解

9 x^{2} - 9 x + 2 = 0

二次方程求根公式:

若

a = 9, b = - 9, c = 2

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

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

(- 9)^{2} - 4 \cdot 9 \cdot 2 = 3

(- 9)^{2} - 4 \cdot 9 \cdot 2

使用指数法则:

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

若

n

是偶数

(- 9)^{2} = 9^{2}

= 9^{2} - 4 \cdot 9 \cdot 2

数字相乘:

4 \cdot 9 \cdot 2 = 72

= 9^{2} - 72

9^{2} = 81

= 81 - 72

数字相减:

81 - 72 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

3^{2} = 3

= 3

x_{1, 2} = \frac{- ( - 9 ) \pm 3}{2 \cdot 9}

将解分隔开

x_{1} = \frac{- ( - 9 ) + 3}{2 \cdot 9}, x_{2} = \frac{- ( - 9 ) - 3}{2 \cdot 9}

x = \frac{- ( - 9 ) + 3}{2 \cdot 9} : \frac{2}{3}

\frac{- ( - 9 ) + 3}{2 \cdot 9}

使用法则

- (- a) = a

= \frac{9 + 3}{2 \cdot 9}

数字相加:

9 + 3 = 12

= \frac{12}{2 \cdot 9}

数字相乘:

2 \cdot 9 = 18

= \frac{12}{18}

约分:

6

= \frac{2}{3}

x = \frac{- ( - 9 ) - 3}{2 \cdot 9} : \frac{1}{3}

\frac{- ( - 9 ) - 3}{2 \cdot 9}

使用法则

- (- a) = a

= \frac{9 - 3}{2 \cdot 9}

数字相减:

9 - 3 = 6

= \frac{6}{2 \cdot 9}

数字相乘:

2 \cdot 9 = 18

= \frac{6}{18}

约分:

6

= \frac{1}{3}

二次方程组的解是:

x = \frac{2}{3}, x = \frac{1}{3}

x = \frac{2}{3}, x = \frac{1}{3}

x = \frac{2}{3}, x = \frac{1}{3}

将解代入原方程进行验证

将它们代入

arctan (x) + arctan (1 - x) = arctan (\frac{9}{7})

检验解是否符合

去除与方程不符的解。

检验

\frac{2}{3}

的解:

真

\frac{2}{3}

代入

n = 1

\frac{2}{3}

对于

arctan (x) + arctan (1 - x) = arctan (\frac{9}{7})

代入

x = \frac{2}{3}

arctan (\frac{2}{3}) + arctan (1 - \frac{2}{3}) = arctan (\frac{9}{7})

整理后得

0.90975 \dots = 0.90975 \dots

\Rightarrow 真

检验

\frac{1}{3}

的解:

真

\frac{1}{3}

代入

n = 1

\frac{1}{3}

对于

arctan (x) + arctan (1 - x) = arctan (\frac{9}{7})

代入

x = \frac{1}{3}

arctan (\frac{1}{3}) + arctan (1 - \frac{1}{3}) = arctan (\frac{9}{7})

整理后得

0.90975 \dots = 0.90975 \dots

\Rightarrow 真

x = \frac{2}{3}, x = \frac{1}{3}

作图

流行的例子

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

3cos^2(θ)-sin^2(θ)=0

sin(x)=((sqrt(6))/4)

sqrt(3)*sin(x)+cos(x)=sqrt(3)

sin(θ)=(sqrt(5))/5