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受欢迎的三角函数 >

arccos(2x)-arccos(x)= pi/3

解答

arccos (2 x) - arccos (x) = \frac{π}{3}

解答

x = - \frac{1}{2}

求解步骤

arccos (2 x) - arccos (x) = \frac{π}{3}

使用三角恒等式改写

arccos (2 x) - arccos (x)

使用和差化积恒等式:

arccos (s) - arccos (t) = arccos (s t + (1 - s^{2}) (1 - t^{2}))

= arccos (2 xx + (1 - (2 x)^{2}) (1 - x^{2}))

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

使用反三角函数性质

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

arccos (x) = a \Rightarrow x = cos (a)

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

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

使用以下普通恒等式:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

cos (x)

周期表（周期为

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{1}{2}

= \frac{1}{2}

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

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

解

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

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

在两边乘以

2

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

化简

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

去除平方根

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

两边减去

4 x^{2}

4 x^{2} + 2 (1 - (2 x)^{2}) (1 - x^{2}) - 4 x^{2} = 1 - 4 x^{2}

化简

2 (1 - (2 x)^{2}) (1 - x^{2}) = 1 - 4 x^{2}

两边进行平方:

4 - 20 x^{2} + 16 x^{4} = 1 - 8 x^{2} + 16 x^{4}

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

(2 (1 - (2 x)^{2}) (1 - x^{2}))^{2} = (1 - 4 x^{2})^{2}

展开

(2 (1 - (2 x)^{2}) (1 - x^{2}))^{2} : 4 - 20 x^{2} + 16 x^{4}

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

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

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

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

使用根式运算法则:

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

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

使用指数法则:

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

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

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

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

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

2^{2} = 4

= 4 (1 - (2 x)^{2}) (1 - x^{2})

展开

4 (1 - (2 x)^{2}) (1 - x^{2}) : 4 - 20 x^{2} + 16 x^{4}

4 (1 - (2 x)^{2}) (1 - x^{2})

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 4 (- 2^{2} x^{2} + 1) (- x^{2} + 1)

2^{2} = 4

= 4 (- 4 x^{2} + 1) (- x^{2} + 1)

乘开

(1 - 4 x^{2}) (1 - x^{2}) : 1 - 5 x^{2} + 4 x^{4}

(1 - 4 x^{2}) (1 - x^{2})

使用 FOIL 方法:

(a + b) (c + d) = a c + a d + b c + b d

a = 1, b = - 4 x^{2}, c = 1, d = - x^{2}

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

使用加减运算法则

+ (- a) = - a, (- a) (- b) = ab

= 1 \cdot 1 - 1 \cdot x^{2} - 4 \cdot 1 \cdot x^{2} + 4 x^{2} x^{2}

化简

1 \cdot 1 - 1 \cdot x^{2} - 4 \cdot 1 \cdot x^{2} + 4 x^{2} x^{2} : 1 - 5 x^{2} + 4 x^{4}

1 \cdot 1 - 1 \cdot x^{2} - 4 \cdot 1 \cdot x^{2} + 4 x^{2} x^{2}

1 \cdot 1 = 1

1 \cdot 1

数字相乘:

1 \cdot 1 = 1

= 1

1 \cdot x^{2} = x^{2}

1 \cdot x^{2}

乘以:

1 \cdot x^{2} = x^{2}

= x^{2}

4 \cdot 1 \cdot x^{2} = 4 x^{2}

4 \cdot 1 \cdot x^{2}

数字相乘:

4 \cdot 1 = 4

= 4 x^{2}

4 x^{2} x^{2} = 4 x^{4}

4 x^{2} x^{2}

使用指数法则:

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

x^{2} x^{2} = x^{2 + 2}

= 4 x^{2 + 2}

数字相加:

2 + 2 = 4

= 4 x^{4}

= 1 - x^{2} - 4 x^{2} + 4 x^{4}

同类项相加:

- x^{2} - 4 x^{2} = - 5 x^{2}

= 1 - 5 x^{2} + 4 x^{4}

= 1 - 5 x^{2} + 4 x^{4}

= 4 (1 - 5 x^{2} + 4 x^{4})

乘开

4 (1 - 5 x^{2} + 4 x^{4}) : 4 - 20 x^{2} + 16 x^{4}

4 (1 - 5 x^{2} + 4 x^{4})

打开括号

= 4 \cdot 1 + 4 (- 5 x^{2}) + 4 \cdot 4 x^{4}

使用加减运算法则

+ (- a) = - a

= 4 \cdot 1 - 4 \cdot 5 x^{2} + 4 \cdot 4 x^{4}

化简

4 \cdot 1 - 4 \cdot 5 x^{2} + 4 \cdot 4 x^{4} : 4 - 20 x^{2} + 16 x^{4}

4 \cdot 1 - 4 \cdot 5 x^{2} + 4 \cdot 4 x^{4}

数字相乘:

4 \cdot 1 = 4

= 4 - 4 \cdot 5 x^{2} + 4 \cdot 4 x^{4}

数字相乘:

4 \cdot 5 = 20

= 4 - 20 x^{2} + 4 \cdot 4 x^{4}

数字相乘:

4 \cdot 4 = 16

= 4 - 20 x^{2} + 16 x^{4}

= 4 - 20 x^{2} + 16 x^{4}

= 4 - 20 x^{2} + 16 x^{4}

= 4 - 20 x^{2} + 16 x^{4}

展开

(1 - 4 x^{2})^{2} : 1 - 8 x^{2} + 16 x^{4}

(1 - 4 x^{2})^{2}

使用完全平方公式:

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

a = 1, b = 4 x^{2}

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

化简

1^{2} - 2 \cdot 1 \cdot 4 x^{2} + (4 x^{2})^{2} : 1 - 8 x^{2} + 16 x^{4}

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

使用法则

1^{a} = 1

1^{2} = 1

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

2 \cdot 1 \cdot 4 x^{2} = 8 x^{2}

2 \cdot 1 \cdot 4 x^{2}

数字相乘:

2 \cdot 1 \cdot 4 = 8

= 8 x^{2}

(4 x^{2})^{2} = 16 x^{4}

(4 x^{2})^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 4^{2} (x^{2})^{2}

(x^{2})^{2} : x^{4}

使用指数法则:

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

= x^{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= x^{4}

= 4^{2} x^{4}

4^{2} = 16

= 16 x^{4}

= 1 - 8 x^{2} + 16 x^{4}

= 1 - 8 x^{2} + 16 x^{4}

4 - 20 x^{2} + 16 x^{4} = 1 - 8 x^{2} + 16 x^{4}

4 - 20 x^{2} + 16 x^{4} = 1 - 8 x^{2} + 16 x^{4}

4 - 20 x^{2} + 16 x^{4} = 1 - 8 x^{2} + 16 x^{4}

解

4 - 20 x^{2} + 16 x^{4} = 1 - 8 x^{2} + 16 x^{4} : x = \frac{1}{2}, x = - \frac{1}{2}

4 - 20 x^{2} + 16 x^{4} = 1 - 8 x^{2} + 16 x^{4}

将

4

到右边

4 - 20 x^{2} + 16 x^{4} = 1 - 8 x^{2} + 16 x^{4}

两边减去

4

4 - 20 x^{2} + 16 x^{4} - 4 = 1 - 8 x^{2} + 16 x^{4} - 4

化简

- 20 x^{2} + 16 x^{4} = 16 x^{4} - 8 x^{2} - 3

- 20 x^{2} + 16 x^{4} = 16 x^{4} - 8 x^{2} - 3

将

8 x^{2}

para o lado esquerdo

- 20 x^{2} + 16 x^{4} = 16 x^{4} - 8 x^{2} - 3

两边加上

8 x^{2}

- 20 x^{2} + 16 x^{4} + 8 x^{2} = 16 x^{4} - 8 x^{2} - 3 + 8 x^{2}

化简

16 x^{4} - 12 x^{2} = 16 x^{4} - 3

16 x^{4} - 12 x^{2} = 16 x^{4} - 3

将

16 x^{4}

para o lado esquerdo

16 x^{4} - 12 x^{2} = 16 x^{4} - 3

两边减去

16 x^{4}

16 x^{4} - 12 x^{2} - 16 x^{4} = 16 x^{4} - 3 - 16 x^{4}

化简

- 12 x^{2} = - 3

- 12 x^{2} = - 3

两边除以

- 12

- 12 x^{2} = - 3

两边除以

- 12

\frac{- 12 x ^{2}}{- 12} = \frac{- 3}{- 12}

化简

\frac{- 12 x ^{2}}{- 12} = \frac{- 3}{- 12}

化简

\frac{- 12 x ^{2}}{- 12} : x^{2}

\frac{- 12 x ^{2}}{- 12}

使用分式法则:

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

= \frac{12 x ^{2}}{12}

数字相除:

\frac{12}{12} = 1

= x^{2}

化简

\frac{- 3}{- 12} : \frac{1}{4}

\frac{- 3}{- 12}

使用分式法则:

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

= \frac{3}{12}

约分:

3

= \frac{1}{4}

x^{2} = \frac{1}{4}

x^{2} = \frac{1}{4}

x^{2} = \frac{1}{4}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

x = \frac{1}{4}, x = - \frac{1}{4}

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

\frac{1}{4}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{4}

使用根式运算法则:

1 = 1

1 = 1

= \frac{1}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

= \frac{1}{2}

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

- \frac{1}{4}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{1}{4}

使用根式运算法则:

1 = 1

1 = 1

= - \frac{1}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

= - \frac{1}{2}

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

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

验证解:

x = \frac{1}{2}

真

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

真

将它们代入

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

检验解是否符合

去除与方程不符的解。

代入

x = \frac{1}{2} :

真

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

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

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

去除括号:

(a) = a

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

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

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

分式相乘:

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

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

约分:

2

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

数字相乘:

1 \cdot 1 = 1

= \frac{1}{2}

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

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

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

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

乘

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

2 \cdot \frac{1}{2}

分式相乘:

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

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

约分:

2

= 1

= 1^{2}

使用法则

1^{a} = 1

= 1

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

(\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 - 1) (- \frac{1}{4} + 1)

数字相减:

1 - 1 = 0

= 0 \cdot (- \frac{1}{4} + 1)

化简

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

= 0 \cdot \frac{3}{4}

使用法则

0 \cdot a = 0

= 0

使用法则

0 = 0

= 0

= \frac{1}{2} + 0

\frac{1}{2} + 0 = \frac{1}{2}

= \frac{1}{2}

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

真

代入

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

真

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

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

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

去除括号:

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

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

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

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

分式相乘:

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

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

约分:

2

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

数字相乘:

1 \cdot 1 = 1

= \frac{1}{2}

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

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

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

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

乘

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

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

分式相乘:

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

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

约分:

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

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

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

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

使用指数法则:

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

若

n

是偶数

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

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

数字相减:

1 - 1 = 0

= 0 \cdot (- \frac{1}{4} + 1)

化简

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

= 0 \cdot \frac{3}{4}

使用法则

0 \cdot a = 0

= 0

使用法则

0 = 0

= 0

= \frac{1}{2} + 0

\frac{1}{2} + 0 = \frac{1}{2}

= \frac{1}{2}

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

真

解为

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

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

将解代入原方程进行验证

将它们代入

arccos (2 x) - arccos (x) = \frac{π}{3}

检验解是否符合

去除与方程不符的解。

检验

\frac{1}{2}

的解:

假

\frac{1}{2}

代入

n = 1

\frac{1}{2}

对于

arccos (2 x) - arccos (x) = \frac{π}{3}

代入

x = \frac{1}{2}

arccos (2 \cdot \frac{1}{2}) - arccos (\frac{1}{2}) = \frac{π}{3}

整理后得

- 1.04719 \dots = 1.04719 \dots

\Rightarrow 假

检验

- \frac{1}{2}

的解:

真

- \frac{1}{2}

代入

n = 1

- \frac{1}{2}

对于

arccos (2 x) - arccos (x) = \frac{π}{3}

代入

x = - \frac{1}{2}

arccos (2 (- \frac{1}{2})) - arccos (- \frac{1}{2}) = \frac{π}{3}

整理后得

1.04719 \dots = 1.04719 \dots

\Rightarrow 真

x = - \frac{1}{2}

作图

流行的例子

cos(2θ)= 1/2 ,0<= θ<= 2pi

cos (2 θ) = \frac{1}{2}, 0 \leq θ \leq 2 π

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

tan (x - \frac{π}{2}) = \frac{3}{3}

2 = 2 cos (3 x)

cos(2x)=sqrt(3)

cos (2 x) = 3

tan(θ)=((2.5)/(1.5))^{1/3}

tan (θ) = (\frac{2.5}{1.5})^{\frac{1}{3}}