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受欢迎的三角函数 >

sinh(a)= 1/2

解答

sinh (a) = \frac{1}{2}

解答

a = ln (\frac{1 + 5}{2})

+1

度数

a = 27.57140 \dots^{\circ}

求解步骤

sinh (a) = \frac{1}{2}

使用三角恒等式改写

sinh (a) = \frac{1}{2}

使用双曲函数恒等式:

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

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

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

\frac{e ^{a} - e ^{- a}}{2} = \frac{1}{2} : a = ln (\frac{1 + 5}{2})

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

在两边乘以

2

\frac{e ^{a} - e ^{- a}}{2} \cdot 2 = \frac{1}{2} \cdot 2

化简

e^{a} - e^{- a} = 1

使用指数运算法则

e^{a} - e^{- a} = 1

使用指数法则:

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

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

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

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

用

e^{a} = u

改写方程式

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

解

u - u^{- 1} = 1 : u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

u - u^{- 1} = 1

整理后得

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

在两边乘以

u

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

在两边乘以

u

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

化简

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

化简

1 \cdot u : u

1 \cdot u

乘以:

1 \cdot u = u

= u

u^{2} - 1 = u

u^{2} - 1 = u

u^{2} - 1 = u

解

u^{2} - 1 = u : u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

u^{2} - 1 = u

将

u

para o lado esquerdo

u^{2} - 1 = u

两边减去

u

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

化简

u^{2} - 1 - u = 0

u^{2} - 1 - u = 0

改写成标准形式

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

u^{2} - u - 1 = 0

使用求根公式求解

u^{2} - u - 1 = 0

二次方程求根公式:

若

a = 1, b = - 1, c = - 1

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

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

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

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

使用法则

- (- a) = a

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

(- 1)^{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

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 1}, u_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 1}

u = \frac{- ( - 1 ) + 5}{2 \cdot 1} : \frac{1 + 5}{2}

\frac{- ( - 1 ) + 5}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 + 5}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 + 5}{2}

u = \frac{- ( - 1 ) - 5}{2 \cdot 1} : \frac{1 - 5}{2}

\frac{- ( - 1 ) - 5}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 - 5}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 - 5}{2}

二次方程组的解是:

u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

验证解

找到无定义的点（奇点）:

u = 0

取

u - u^{- 1}

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 = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

代回

u = e^{a}

，求解

a

解

e^{a} = \frac{1 + 5}{2} : a = ln (\frac{1 + 5}{2})

e^{a} = \frac{1 + 5}{2}

使用指数运算法则

e^{a} = \frac{1 + 5}{2}

若

f (x) = g (x)

，则

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

ln (e^{a}) = ln (\frac{1 + 5}{2})

使用对数计算法则:

ln (e^{a}) = a

ln (e^{a}) = a

a = ln (\frac{1 + 5}{2})

a = ln (\frac{1 + 5}{2})

解

e^{a} = \frac{1 - 5}{2} : a \in R

无解

e^{a} = \frac{1 - 5}{2}

a^{f (a)}

对于

a

不能为零或负值

\in R

a \in R 无解

a = ln (\frac{1 + 5}{2})

a = ln (\frac{1 + 5}{2})

作图

流行的例子

sin(x)=(-1}{\frac{sqrt(5))/2}

sin (x) = \frac{- 1}{\frac{5}{2}}

2csc^2(θ)-3cot(θ)+3=0

2 csc^{2} (θ) - 3 cot (θ) + 3 = 0

(2sin(x)+1)/(sqrt(cos(x)))=0

\frac{2 sin ( x ) + 1}{cos ( x )} = 0

sin (x) = 0.03725

tan (x) = 2.747