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受欢迎的三角函数 >

solvefor y,x=sinh(y)

解答

求解

y, x = sinh (y)

解答

y = ln (x + x^{2} + 1), y = ln (x - x^{2} + 1)

求解步骤

x = sinh (y)

交换两边

sinh (y) = x

使用三角恒等式改写

sinh (y) = x

使用双曲函数恒等式:

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

\frac{e ^{y} - e ^{- y}}{2} = x

\frac{e ^{y} - e ^{- y}}{2} = x

\frac{e ^{y} - e ^{- y}}{2} = x : y = ln (x + x^{2} + 1), y = ln (x - x^{2} + 1)

\frac{e ^{y} - e ^{- y}}{2} = x

在两边乘以

2

\frac{e ^{y} - e ^{- y}}{2} \cdot 2 = x \cdot 2

化简

e^{y} - e^{- y} = x \cdot 2

使用指数运算法则

e^{y} - e^{- y} = x \cdot 2

使用指数法则:

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

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

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

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

用

e^{y} = u

改写方程式

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

解

u - u^{- 1} = x \cdot 2 : u = x + x^{2} + 1, u = x - x^{2} + 1

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

整理后得

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

在两边乘以

u

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

在两边乘以

u

uu - \frac{1}{u} u = x \cdot 2 u

化简

uu - \frac{1}{u} u = x \cdot 2 u

化简

uu : u^{2}

uu

使用指数法则:

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

uu = u^{1 + 1}

= u^{1 + 1}

数字相加:

1 + 1 = 2

= u^{2}

化简

- \frac{1}{u} u : - 1

- \frac{1}{u} u

分式相乘:

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

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

约分:

u

= - 1

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

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

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

解

u^{2} - 1 = x \cdot 2 u : u = x + x^{2} + 1, u = x - x^{2} + 1

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

将

x 2 u

para o lado esquerdo

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

两边减去

x 2 u

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

化简

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

化简

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

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

使用法则

- (- a) = a

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

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

(- x \cdot 2)^{2}

使用指数法则:

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

若

n

是偶数

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

= (2 x)^{2}

使用指数法则:

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

= 2^{2} x^{2}

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

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

分解

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

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

改写为

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

因式分解出通项

4

= 4 (x^{2} + 1)

= 4 (x^{2} + 1)

使用根式运算法则:

n ab = n a n b,

假定

a \geq 0, b \geq 0

= 4 x^{2} + 1

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= 2 x^{2} + 1

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

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

因式分解出通项

2

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

数字相除:

\frac{2}{2} = 1

= x + x^{2} + 1

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

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

因式分解出通项

2

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

数字相除:

\frac{2}{2} = 1

= x - x^{2} + 1

二次方程组的解是:

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

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

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

代回

u = e^{y}

，求解

y

解

e^{y} = x + x^{2} + 1 : y = ln (x + x^{2} + 1)

e^{y} = x + x^{2} + 1

使用指数运算法则

e^{y} = x + x^{2} + 1

若

f (x) = g (x)

，则

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

ln (e^{y}) = ln (x + x^{2} + 1)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{y}) = y

y = ln (x + x^{2} + 1)

y = ln (x + x^{2} + 1)

解

e^{y} = x - x^{2} + 1 : y = ln (x - x^{2} + 1)

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

使用指数运算法则

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

若

f (x) = g (x)

，则

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

ln (e^{y}) = ln (x - x^{2} + 1)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{y}) = y

y = ln (x - x^{2} + 1)

y = ln (x - x^{2} + 1)

y = ln (x + x^{2} + 1), y = ln (x - x^{2} + 1)

y = ln (x + x^{2} + 1), y = ln (x - x^{2} + 1)

作图

流行的例子

-3cos(2θ)+cos(θ)-10=-6cos(θ)-5

- 3 cos (2 θ) + cos (θ) - 10 = - 6 cos (θ) - 5

7cot(θ)=3csc(θ)

7 cot (θ) = 3 csc (θ)

5 sin (x) = 2

4(cot(θ)+1)=2(cot(θ)+2)

4 (cot (θ) + 1) = 2 (cot (θ) + 2)

sin (3 t) = 0