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受欢迎的三角函数 >

arccos(x)-arcsin(x)=arcsin(1-x)

解答

arccos (x) - arcsin (x) = arcsin (1 - x)

解答

x = 0, x = \frac{1}{2}

求解步骤

arccos (x) - arcsin (x) = arcsin (1 - x)

a = b \Rightarrow sin (a) = sin (b)

sin (arccos (x) - arcsin (x)) = sin (arcsin (1 - x))

利用以下特性:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

sin (arccos (x)) cos (arcsin (x)) - cos (arccos (x)) sin (arcsin (x)) = sin (arcsin (1 - x))

利用以下特性:

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

利用以下特性:

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

利用以下特性:

cos (arccos (x)) = x

利用以下特性:

sin (arcsin (x)) = x

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

解

1 - x^{2} 1 - x^{2} - xx = 1 - x : x = 0, x = \frac{1}{2}

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

展开

1 - x^{2} 1 - x^{2} - xx : 1 - 2 x^{2}

1 - x^{2} 1 - x^{2} - xx

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

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

使用根式运算法则:

a a = a

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

= 1 - x^{2}

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}

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

整理后得

= 1 - 2 x^{2}

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

解

1 - 2 x^{2} = 1 - x : x = 0, x = \frac{1}{2}

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

将

x

para o lado esquerdo

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

两边加上

x

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

化简

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

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

将

1

para o lado esquerdo

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

两边减去

1

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

化简

- 2 x^{2} + x = 0

- 2 x^{2} + x = 0

使用求根公式求解

- 2 x^{2} + x = 0

二次方程求根公式:

若

a = - 2, b = 1, c = 0

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

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 0

使用法则

- (- a) = a

= 1 + 4 \cdot 2 \cdot 0

使用法则

0 \cdot a = 0

= 1 + 0

数字相加:

1 + 0 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

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

去除括号:

(- a) = - a

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

数字相加/相减:

- 1 + 1 = 0

= \frac{0}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{0}{- 4}

使用分式法则:

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

= - \frac{0}{4}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

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

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

去除括号:

(- a) = - a

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

数字相减:

- 1 - 1 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

使用分式法则:

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

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

x = 0, x = \frac{1}{2}

x = 0, x = \frac{1}{2}

验证解:

x = 0

真

, x = \frac{1}{2}

真

将它们代入

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

检验解是否符合

去除与方程不符的解。

代入

x = 0 :

真

1 - 0^{2} 1 - 0^{2} - 0 \cdot 0 = 1 - 0

1 - 0^{2} 1 - 0^{2} - 0 \cdot 0 = 1 - 0

1 - 0^{2} 1 - 0^{2} - 0 \cdot 0

使用法则

0^{a} = 0

0^{2} = 0

= 1 - 0 1 - 0 - 0 \cdot 0

1 - 0 1 - 0 = 1

1 - 0 1 - 0

使用根式运算法则:

a a = a

1 - 0 1 - 0 = 1 - 0

= 1 - 0

数字相减:

1 - 0 = 1

= 1

0 \cdot 0 = 0

0 \cdot 0

数字相乘:

0 \cdot 0 = 0

= 0

= 1 - 0

1 - 0 = 1 - 0

真

代入

x = \frac{1}{2} :

真

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

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

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

去除括号:

(a) = a

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

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

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

使用根式运算法则:

a a = a

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

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

(\frac{1}{2})^{2} = \frac{1}{4}

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

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

2^{2} = 4

= \frac{1}{4}

= 1 - \frac{1}{4}

将项转换为分式:

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

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

1 \cdot 4 - 1 = 3

1 \cdot 4 - 1

数字相乘:

1 \cdot 4 = 4

= 4 - 1

数字相减:

4 - 1 = 3

= 3

= \frac{3}{4}

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

\frac{1}{2} \cdot \frac{1}{2}

分式相乘:

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

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

数字相乘:

1 \cdot 1 = 1

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

数字相乘:

2 \cdot 2 = 4

= \frac{1}{4}

= \frac{3}{4} - \frac{1}{4}

使用法则

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

= \frac{3 - 1}{4}

数字相减:

3 - 1 = 2

= \frac{2}{4}

约分:

2

= \frac{1}{2}

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

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

去除括号:

(a) = a

= 1 - \frac{1}{2}

将项转换为分式:

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

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

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 2 - 1}{2}

1 \cdot 2 - 1 = 1

1 \cdot 2 - 1

数字相乘:

1 \cdot 2 = 2

= 2 - 1

数字相减:

2 - 1 = 1

= 1

= \frac{1}{2}

\frac{1}{2} = \frac{1}{2}

真

解为

x = 0, x = \frac{1}{2}

x = 0, x = \frac{1}{2}

将解代入原方程进行验证

将它们代入

arccos (x) - arcsin (x) = arcsin (1 - x)

检验解是否符合

去除与方程不符的解。

检验

0

的解:

真

0

代入

n = 1

0

对于

arccos (x) - arcsin (x) = arcsin (1 - x)

代入

x = 0

arccos (0) - arcsin (0) = arcsin (1 - 0)

整理后得

1.57079 \dots = 1.57079 \dots

\Rightarrow 真

检验

\frac{1}{2}

的解:

真

\frac{1}{2}

代入

n = 1

\frac{1}{2}

对于

arccos (x) - arcsin (x) = arcsin (1 - x)

代入

x = \frac{1}{2}

arccos (\frac{1}{2}) - arcsin (\frac{1}{2}) = arcsin (1 - \frac{1}{2})

整理后得

0.52359 \dots = 0.52359 \dots

\Rightarrow 真

x = 0, x = \frac{1}{2}

作图

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