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受欢迎的三角函数 >

solvefor y,v=(1-4cos(5y))^{-1/2}

解答

求解

y, v = (1 - 4 cos (5 y))^{- \frac{1}{2}}

解答

y = \frac{arccos ( - \frac{1 - v ^{2}}{4 v ^{2}} )}{5} + \frac{2 πn}{5}, y = - \frac{arccos ( - \frac{1 - v ^{2}}{4 v ^{2}} )}{5} + \frac{2 πn}{5}

求解步骤

v = (1 - 4 cos (5 y))^{- \frac{1}{2}}

交换两边

(1 - 4 cos (5 y))^{- \frac{1}{2}} = v

对方程式两边

- 2

次方:

1 - 4 cos (5 y) = \frac{1}{v ^{2}}

(1 - 4 cos (5 y))^{- \frac{1}{2}} = v

((1 - 4 cos (5 y))^{- \frac{1}{2}})^{- 2} = v^{- 2}

展开

((1 - 4 cos (5 y))^{- \frac{1}{2}})^{- 2} : 1 - 4 cos (5 y)

((1 - 4 cos (5 y))^{- \frac{1}{2}})^{- 2}

使用指数法则:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{( ( 1 - 4 cos ( 5 y ) ) ^{- \frac{1}{2}} ) ^{2}}

((1 - 4 cos (5 y))^{- \frac{1}{2}})^{2} : (1 - 4 cos (5 y))^{- 1}

使用指数法则:

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

= (1 - 4 cos (5 y))^{(- \frac{1}{2}) \cdot 2}

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

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

去除括号:

(- a) = - a

= - \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 - 4 cos (5 y))^{- 1}

= \frac{1}{( 1 - 4 cos ( 5 y ) ) ^{- 1}}

使用指数法则:

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

(1 - 4 cos (5 y))^{- 1} = \frac{1}{( 1 - 4 cos ( 5 y ) )}

= \frac{1}{\frac{1}{1 - 4 c o s ( 5 y )}}

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{1 - 4 cos ( 5 y )}{1}

使用分式法则:

\frac{a}{1} = a

= 1 - 4 cos (5 y)

整理后得

= 1 - 4 cos (5 y)

展开

v^{- 2} : \frac{1}{v ^{2}}

v^{- 2}

使用指数法则:

a^{- b} = \frac{1}{a ^{b}}

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

1 - 4 cos (5 y) = \frac{1}{v ^{2}}

1 - 4 cos (5 y) = \frac{1}{v ^{2}}

解

1 - 4 cos (5 y) = \frac{1}{v ^{2}} : cos (5 y) = - \frac{1 - v ^{2}}{4 v ^{2}}

1 - 4 cos (5 y) = \frac{1}{v ^{2}}

将

1

到右边

1 - 4 cos (5 y) = \frac{1}{v ^{2}}

两边减去

1

1 - 4 cos (5 y) - 1 = \frac{1}{v ^{2}} - 1

化简

- 4 cos (5 y) = \frac{1}{v ^{2}} - 1

- 4 cos (5 y) = \frac{1}{v ^{2}} - 1

两边除以

- 4

- 4 cos (5 y) = \frac{1}{v ^{2}} - 1

两边除以

- 4

\frac{- 4 cos ( 5 y )}{- 4} = \frac{\frac{1}{v ^{2}}}{- 4} - \frac{1}{- 4}

化简

\frac{- 4 cos ( 5 y )}{- 4} = \frac{\frac{1}{v ^{2}}}{- 4} - \frac{1}{- 4}

化简

\frac{- 4 cos ( 5 y )}{- 4} : cos (5 y)

\frac{- 4 cos ( 5 y )}{- 4}

使用分式法则:

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

= \frac{4 cos ( 5 y )}{4}

数字相除:

\frac{4}{4} = 1

= cos (5 y)

化简

\frac{\frac{1}{v ^{2}}}{- 4} - \frac{1}{- 4} : - \frac{1 - v ^{2}}{4 v ^{2}}

\frac{\frac{1}{v ^{2}}}{- 4} - \frac{1}{- 4}

使用法则

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

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

使用分式法则:

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

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

化简

\frac{1}{v ^{2}} - 1 : \frac{1 - v ^{2}}{v ^{2}}

\frac{1}{v ^{2}} - 1

将项转换为分式:

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

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

乘以:

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

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

= - \frac{\frac{- v ^{2} + 1}{v ^{2}}}{4}

使用分式法则:

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

= - \frac{- v ^{2} + 1}{4 v ^{2}}

cos (5 y) = - \frac{1 - v ^{2}}{4 v ^{2}}

cos (5 y) = - \frac{1 - v ^{2}}{4 v ^{2}}

cos (5 y) = - \frac{1 - v ^{2}}{4 v ^{2}}

cos (5 y) = - \frac{1 - v ^{2}}{4 v ^{2}}

验证解:

cos (5 y) = - \frac{1 - v ^{2}}{4 v ^{2}} {v < 0 or v > 0}

将它们代入

(1 - 4 cos (5 y))^{- \frac{1}{2}} = v

检验解是否符合

去除与方程不符的解。

代入

cos (5 y) = - \frac{1 - v ^{2}}{4 v ^{2}} : (1 - 4 (- \frac{1 - v ^{2}}{4 v ^{2}}))^{- \frac{1}{2}} = v \Rightarrow v < 0 or v > 0

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

对方程式两边

- 2

次方:

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

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

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

展开

((1 - 4 (- \frac{1 - v ^{2}}{4 v ^{2}}))^{- \frac{1}{2}})^{- 2} : \frac{1}{v ^{2}}

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

使用指数法则:

a^{- b} = \frac{1}{a ^{b}}

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

((1 - 4 (- \frac{1 - v ^{2}}{4 v ^{2}}))^{- \frac{1}{2}})^{2} : (1 - 4 (- \frac{1 - v ^{2}}{4 v ^{2}}))^{- 1}

使用指数法则:

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

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

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

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

去除括号:

(- a) = - a

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

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

使用指数法则:

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

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

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

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

使用分式法则:

\frac{a}{1} = a

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

展开

1 - 4 (- \frac{1 - v ^{2}}{4 v ^{2}}) : \frac{1}{v ^{2}}

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

使用法则

- (- a) = a

= 1 + 4 \cdot \frac{1 - v ^{2}}{4 v ^{2}}

4 \cdot \frac{1 - v ^{2}}{4 v ^{2}} = \frac{1 - v ^{2}}{v ^{2}}

4 \cdot \frac{1 - v ^{2}}{4 v ^{2}}

分式相乘:

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

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

约分:

4

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

= 1 + \frac{- v ^{2} + 1}{v ^{2}}

使用分式法则:

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

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

= 1 + \frac{1}{v ^{2}} - \frac{v ^{2}}{v ^{2}}

使用法则

\frac{a}{a} = 1

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

= 1 + \frac{1}{v ^{2}} - 1

对同类项分组

= \frac{1}{v ^{2}} + 1 - 1

1 - 1 = 0

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

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

展开

v^{- 2} : \frac{1}{v ^{2}}

v^{- 2}

使用指数法则:

a^{- b} = \frac{1}{a ^{b}}

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

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

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

解

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

对所有

v

为真

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

两边减去

\frac{1}{v ^{2}}

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

化简

0 = 0

两侧相等

对所有 v 为真

验证解

找到无定义的点（奇点）:

v = 0

取

\frac{1}{v ^{2}}

的分母，令其等于零

解

v^{2} = 0 : v = 0

v^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

v = 0

取

\frac{1}{v ^{2}}

的分母，令其等于零

解

v^{2} = 0 : v = 0

v^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

v = 0

以下点无定义

v = 0

因为方程对以下值无定义:

0

对所有 v 为真

对所有 v 为真

验证解:

v < 0 or v > 0

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

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

的定义域

: v < 0 or v > 0

定义域定义

找到无定义的点（奇点）:

v = 0

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

取

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

的分母，令其等于零

v = 0

以下点无定义

v = 0

函数定义域

v < 0 or v > 0

将定义域区间与解区间合并:

对所有 v 为真 and (v < 0 or v > 0)

合并重叠的区间

v < 0 or v > 0

解是

v < 0 or v > 0

解是

cos (5 y) = - \frac{1 - v ^{2}}{4 v ^{2}} {v < 0 or v > 0}

使用反三角函数性质

cos (5 y) = - \frac{1 - v ^{2}}{4 v ^{2}}

cos (5 y) = - \frac{1 - v ^{2}}{4 v ^{2}}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

5 y = arccos (- \frac{1 - v ^{2}}{4 v ^{2}}) + 2 πn, 5 y = - arccos (- \frac{1 - v ^{2}}{4 v ^{2}}) + 2 πn

5 y = arccos (- \frac{1 - v ^{2}}{4 v ^{2}}) + 2 πn, 5 y = - arccos (- \frac{1 - v ^{2}}{4 v ^{2}}) + 2 πn

解

5 y = arccos (- \frac{1 - v ^{2}}{4 v ^{2}}) + 2 πn : y = \frac{arccos ( - \frac{1 - v ^{2}}{4 v ^{2}} )}{5} + \frac{2 πn}{5}

5 y = arccos (- \frac{1 - v ^{2}}{4 v ^{2}}) + 2 πn

两边除以

5

5 y = arccos (- \frac{1 - v ^{2}}{4 v ^{2}}) + 2 πn

两边除以

5

\frac{5 y}{5} = \frac{arccos ( - \frac{1 - v ^{2}}{4 v ^{2}} )}{5} + \frac{2 πn}{5}

化简

y = \frac{arccos ( - \frac{1 - v ^{2}}{4 v ^{2}} )}{5} + \frac{2 πn}{5}

y = \frac{arccos ( - \frac{1 - v ^{2}}{4 v ^{2}} )}{5} + \frac{2 πn}{5}

解

5 y = - arccos (- \frac{1 - v ^{2}}{4 v ^{2}}) + 2 πn : y = - \frac{arccos ( - \frac{1 - v ^{2}}{4 v ^{2}} )}{5} + \frac{2 πn}{5}

5 y = - arccos (- \frac{1 - v ^{2}}{4 v ^{2}}) + 2 πn

两边除以

5

5 y = - arccos (- \frac{1 - v ^{2}}{4 v ^{2}}) + 2 πn

两边除以

5

\frac{5 y}{5} = - \frac{arccos ( - \frac{1 - v ^{2}}{4 v ^{2}} )}{5} + \frac{2 πn}{5}

化简

y = - \frac{arccos ( - \frac{1 - v ^{2}}{4 v ^{2}} )}{5} + \frac{2 πn}{5}

y = - \frac{arccos ( - \frac{1 - v ^{2}}{4 v ^{2}} )}{5} + \frac{2 πn}{5}

y = \frac{arccos ( - \frac{1 - v ^{2}}{4 v ^{2}} )}{5} + \frac{2 πn}{5}, y = - \frac{arccos ( - \frac{1 - v ^{2}}{4 v ^{2}} )}{5} + \frac{2 πn}{5}

作图

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