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受欢迎的三角函数 >

solvefor u,x=4sinh(u)

解答

求解

u, x = 4 sinh (u)

解答

u = ln (\frac{x + x ^{2} + 16}{4}), u = ln (\frac{x - x ^{2} + 16}{4})

求解步骤

x = 4 sinh (u)

交换两边

4 sinh (u) = x

使用三角恒等式改写

4 sinh (u) = x

使用双曲函数恒等式:

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

4 \cdot \frac{e ^{u} - e ^{- u}}{2} = x

4 \cdot \frac{e ^{u} - e ^{- u}}{2} = x

4 \cdot \frac{e ^{u} - e ^{- u}}{2} = x : u = ln (\frac{x + x ^{2} + 16}{4}), u = ln (\frac{x - x ^{2} + 16}{4})

4 \cdot \frac{e ^{u} - e ^{- u}}{2} = x

使用指数运算法则

4 \cdot \frac{e ^{u} - e ^{- u}}{2} = x

使用指数法则:

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

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

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

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

用

e^{u} = v

改写方程式

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

解

4 \cdot \frac{v - v ^{- 1}}{2} = x : v = \frac{x + x ^{2} + 16}{4}, v = \frac{x - x ^{2} + 16}{4}

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

整理后得

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

在两边乘以

v

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

在两边乘以

v

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

化简

2 (v^{2} - 1) = xv

2 (v^{2} - 1) = xv

解

2 (v^{2} - 1) = xv : v = \frac{x + x ^{2} + 16}{4}, v = \frac{x - x ^{2} + 16}{4}

2 (v^{2} - 1) = xv

展开

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

2 (v^{2} - 1)

使用分配律:

a (b - c) = ab - a c

a = 2, b = v^{2}, c = 1

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

数字相乘:

2 \cdot 1 = 2

= 2 v^{2} - 2

2 v^{2} - 2 = xv

将

xv

para o lado esquerdo

2 v^{2} - 2 = xv

两边减去

xv

2 v^{2} - 2 - xv = xv - xv

化简

2 v^{2} - 2 - xv = 0

2 v^{2} - 2 - xv = 0

改写成标准形式

a x^{2} + b x + c = 0

2 v^{2} - xv - 2 = 0

使用求根公式求解

2 v^{2} - xv - 2 = 0

二次方程求根公式:

若

a = 2, b = - x, c = - 2

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

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

化简

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

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

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

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

数字相乘:

4 \cdot 2 \cdot 2 = 16

= x^{2} + 16

v_{1, 2} = \frac{- ( - x ) \pm x ^{2} + 16}{2 \cdot 2}

将解分隔开

v_{1} = \frac{- ( - x ) + x ^{2} + 16}{2 \cdot 2}, v_{2} = \frac{- ( - x ) - x ^{2} + 16}{2 \cdot 2}

v = \frac{- ( - x ) + x ^{2} + 16}{2 \cdot 2} : \frac{x + x ^{2} + 16}{4}

\frac{- ( - x ) + x ^{2} + 16}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{x + x ^{2} + 16}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{x + x ^{2} + 16}{4}

v = \frac{- ( - x ) - x ^{2} + 16}{2 \cdot 2} : \frac{x - x ^{2} + 16}{4}

\frac{- ( - x ) - x ^{2} + 16}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{x - x ^{2} + 16}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{x - x ^{2} + 16}{4}

二次方程组的解是:

v = \frac{x + x ^{2} + 16}{4}, v = \frac{x - x ^{2} + 16}{4}

v = \frac{x + x ^{2} + 16}{4}, v = \frac{x - x ^{2} + 16}{4}

v = \frac{x + x ^{2} + 16}{4}, v = \frac{x - x ^{2} + 16}{4}

代回

v = e^{u}

，求解

u

解

e^{u} = \frac{x + x ^{2} + 16}{4} : u = ln (\frac{x + x ^{2} + 16}{4})

e^{u} = \frac{x + x ^{2} + 16}{4}

使用指数运算法则

e^{u} = \frac{x + x ^{2} + 16}{4}

若

f (x) = g (x)

，则

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

ln (e^{u}) = ln (\frac{x + x ^{2} + 16}{4})

使用对数计算法则:

ln (e^{a}) = a

ln (e^{u}) = u

u = ln (\frac{x + x ^{2} + 16}{4})

u = ln (\frac{x + x ^{2} + 16}{4})

解

e^{u} = \frac{x - x ^{2} + 16}{4} : u = ln (\frac{x - x ^{2} + 16}{4})

e^{u} = \frac{x - x ^{2} + 16}{4}

使用指数运算法则

e^{u} = \frac{x - x ^{2} + 16}{4}

若

f (x) = g (x)

，则

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

ln (e^{u}) = ln (\frac{x - x ^{2} + 16}{4})

使用对数计算法则:

ln (e^{a}) = a

ln (e^{u}) = u

u = ln (\frac{x - x ^{2} + 16}{4})

u = ln (\frac{x - x ^{2} + 16}{4})

u = ln (\frac{x + x ^{2} + 16}{4}), u = ln (\frac{x - x ^{2} + 16}{4})

u = ln (\frac{x + x ^{2} + 16}{4}), u = ln (\frac{x - x ^{2} + 16}{4})

作图

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