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受欢迎的三角函数 >

1+tanh^2(x)=sech^2(x)

解答

1 + tanh^{2} (x) = sech^{2} (x)

解答

x = 0

+1

度数

x = 0^{\circ}

求解步骤

1 + tanh^{2} (x) = sech^{2} (x)

使用三角恒等式改写

1 + tanh^{2} (x) = sech^{2} (x)

使用双曲函数恒等式:

tanh (x) = \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}

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

使用双曲函数恒等式:

sech (x) = \frac{2}{e ^{x} + e ^{- x}}

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

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

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

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

使用指数运算法则

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

使用指数法则:

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

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

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

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

用

e^{x} = u

改写方程式

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

解

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

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

整理后得

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

在两边乘以

(u^{2} + 1)^{2}

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

在两边乘以

(u^{2} + 1)^{2}

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

化简

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

化简

1 \cdot (u^{2} + 1)^{2} : (u^{2} + 1)^{2}

1 \cdot (u^{2} + 1)^{2}

乘以:

1 \cdot (u^{2} + 1)^{2} = (u^{2} + 1)^{2}

= (u^{2} + 1)^{2}

化简

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

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

分式相乘:

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

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

约分:

(u^{2} + 1)^{2}

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

化简

\frac{4 u ^{2}}{( u ^{2} + 1 ) ^{2}} (u^{2} + 1)^{2} : 4 u^{2}

\frac{4 u ^{2}}{( u ^{2} + 1 ) ^{2}} (u^{2} + 1)^{2}

分式相乘:

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

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

约分:

(u^{2} + 1)^{2}

= 4 u^{2}

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

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

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

解

(u^{2} + 1)^{2} + (u^{2} - 1)^{2} = 4 u^{2} : u = 1, u = - 1

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

展开

(u^{2} + 1)^{2} + (u^{2} - 1)^{2} : 2 u^{4} + 2

(u^{2} + 1)^{2} + (u^{2} - 1)^{2}

(u^{2} + 1)^{2} : u^{4} + 2 u^{2} + 1

使用完全平方公式:

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

a = u^{2}, b = 1

= (u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2}

化简

(u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2} : u^{4} + 2 u^{2} + 1

(u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

= (u^{2})^{2} + 2 \cdot 1 \cdot u^{2} + 1

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

(u^{2})^{2}

使用指数法则:

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

= u^{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= u^{4}

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

2 u^{2} \cdot 1

数字相乘:

2 \cdot 1 = 2

= 2 u^{2}

= u^{4} + 2 u^{2} + 1

= u^{4} + 2 u^{2} + 1

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

(u^{2} - 1)^{2} : u^{4} - 2 u^{2} + 1

使用完全平方公式:

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

a = u^{2}, b = 1

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

化简

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

(u^{2})^{2} - 2 u^{2} \cdot 1 + 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

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

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

(u^{2})^{2}

使用指数法则:

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

= u^{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= u^{4}

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

2 u^{2} \cdot 1

数字相乘:

2 \cdot 1 = 2

= 2 u^{2}

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

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

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

化简

u^{4} + 2 u^{2} + 1 + u^{4} - 2 u^{2} + 1 : 2 u^{4} + 2

u^{4} + 2 u^{2} + 1 + u^{4} - 2 u^{2} + 1

对同类项分组

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

同类项相加:

2 u^{2} - 2 u^{2} = 0

= u^{4} + u^{4} + 1 + 1

同类项相加:

u^{4} + u^{4} = 2 u^{4}

= 2 u^{4} + 1 + 1

数字相加:

1 + 1 = 2

= 2 u^{4} + 2

= 2 u^{4} + 2

2 u^{4} + 2 = 4 u^{2}

将

4 u^{2}

para o lado esquerdo

2 u^{4} + 2 = 4 u^{2}

两边减去

4 u^{2}

2 u^{4} + 2 - 4 u^{2} = 4 u^{2} - 4 u^{2}

化简

2 u^{4} + 2 - 4 u^{2} = 0

2 u^{4} + 2 - 4 u^{2} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

2 u^{4} - 4 u^{2} + 2 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

2 v^{2} - 4 v + 2 = 0

解

2 v^{2} - 4 v + 2 = 0 : v = 1

2 v^{2} - 4 v + 2 = 0

使用求根公式求解

2 v^{2} - 4 v + 2 = 0

二次方程求根公式:

若

a = 2, b = - 4, c = 2

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

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

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

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

使用指数法则:

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

若

n

是偶数

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

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

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

数字相减:

16 - 16 = 0

= 0

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

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

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

\frac{- ( - 4 )}{2 \cdot 2}

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 2 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

v = 1

二次方程组的解是:

v = 1

v = 1

代回

v = u^{2}

，求解

u

解

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

对于

x^{2} = f (a)

解为

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

u = 1, u = - 1

1 = 1

1

使用根式运算法则:

1 = 1

= 1

- 1 = - 1

- 1

使用根式运算法则:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

解为

u = 1, u = - 1

u = 1, u = - 1

验证解

找到无定义的点（奇点）:

u = 0

取

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

的分母，令其等于零

u = 0

取

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

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = 1, u = - 1

u = 1, u = - 1

代回

u = e^{x}

，求解

x

解

e^{x} = 1 : x = 0

e^{x} = 1

使用指数运算法则

e^{x} = 1

若

f (x) = g (x)

，则

ln (f (x)) = ln (g (x))

ln (e^{x}) = ln (1)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (1)

化简

ln (1) : 0

ln (1)

使用对数计算法则:

lo g_{a} (1) = 0

= 0

x = 0

x = 0

解

e^{x} = - 1 : x \in R

无解

e^{x} = - 1

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

x = 0

x = 0

作图

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