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受欢迎的三角函数 >

2-2cosh(-2)

解答

2 - 2 cosh (- 2)

解答

2 - \frac{e ^{4} + 1}{e ^{2}}

+1

十进制

- 5.52439 \dots

求解步骤

2 - 2 cosh (- 2)

使用三角恒等式改写:

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

cosh (- 2)

利用以下特性:

cosh (- x) = cosh (x)

cosh (- 2) = cosh (2)

= cosh (2)

使用三角恒等式改写:

\frac{e ^{4} + 1}{2 e ^{2}}

cosh (2)

使用双曲函数恒等式:

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

= \frac{e ^{2} + e ^{- 2}}{2}

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

\frac{e ^{2} + e ^{- 2}}{2}

使用指数法则:

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

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

化简

e^{2} + \frac{1}{e ^{2}} : \frac{e ^{4} + 1}{e ^{2}}

e^{2} + \frac{1}{e ^{2}}

将项转换为分式:

e^{2} = \frac{e ^{2} e ^{2}}{e ^{2}}

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

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

e^{2} e^{2} + 1

e^{2} e^{2} = e^{4}

e^{2} e^{2}

使用指数法则:

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

e^{2} e^{2} = e^{2 + 2}

= e^{2 + 2}

数字相加:

2 + 2 = 4

= e^{4}

= e^{4} + 1

= \frac{e ^{4} + 1}{e ^{2}}

= \frac{\frac{e ^{4} + 1}{e ^{2}}}{2}

使用分式法则:

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

= \frac{e ^{4} + 1}{e ^{2} \cdot 2}

= \frac{e ^{4} + 1}{2 e ^{2}}

= \frac{e ^{4} + 1}{2 e ^{2}}

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

化简

2 - 2 \cdot \frac{e ^{4} + 1}{2 e ^{2}} : 2 - \frac{e ^{4} + 1}{e ^{2}}

2 - 2 \cdot \frac{e ^{4} + 1}{2 e ^{2}}

2 \cdot \frac{e ^{4} + 1}{2 e ^{2}} = \frac{e ^{4} + 1}{e ^{2}}

2 \cdot \frac{e ^{4} + 1}{2 e ^{2}}

分式相乘:

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

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

约分:

2

= \frac{e ^{4} + 1}{e ^{2}}

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

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

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