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受欢迎的三角函数 >

coth(ln(2))

解答

coth (ln (2))

解答

\frac{5}{3}

+1

十进制

1.66666 \dots

求解步骤

coth (ln (2))

使用双曲函数恒等式:

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

= \frac{e ^{l n (2)} + e ^{- l n (2)}}{e ^{l n (2)} - e ^{- l n (2)}}

\frac{e ^{l n (2)} + e ^{- l n (2)}}{e ^{l n (2)} - e ^{- l n (2)}} = \frac{5}{3}

\frac{e ^{l n (2)} + e ^{- l n (2)}}{e ^{l n (2)} - e ^{- l n (2)}}

e^{l n (2)} = 2

e^{l n (2)}

使用对数计算法则:

a^{l o g_{a} (b)} = b

= 2

e^{- l n (2)} = 2^{- 1}

e^{- l n (2)}

使用指数法则:

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

= (e^{l n (2)})^{- 1}

使用对数计算法则:

a^{l o g_{a} (b)} = b

e^{l n (2)} = 2

= 2^{- 1}

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

e^{l n (2)} = 2

e^{l n (2)}

使用对数计算法则:

a^{l o g_{a} (b)} = b

= 2

e^{- l n (2)} = 2^{- 1}

e^{- l n (2)}

使用指数法则:

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

= (e^{l n (2)})^{- 1}

使用对数计算法则:

a^{l o g_{a} (b)} = b

e^{l n (2)} = 2

= 2^{- 1}

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

化简

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

使用指数法则:

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

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

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

使用指数法则:

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

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

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

化简

2 - \frac{1}{2} : \frac{3}{2}

2 - \frac{1}{2}

将项转换为分式:

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

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

2 \cdot 2 - 1 = 3

2 \cdot 2 - 1

数字相乘:

2 \cdot 2 = 4

= 4 - 1

数字相减:

4 - 1 = 3

= 3

= \frac{3}{2}

= \frac{2 + \frac{1}{2}}{\frac{3}{2}}

化简

2 + \frac{1}{2} : \frac{5}{2}

2 + \frac{1}{2}

将项转换为分式:

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

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

2 \cdot 2 + 1 = 5

2 \cdot 2 + 1

数字相乘:

2 \cdot 2 = 4

= 4 + 1

数字相加:

4 + 1 = 5

= 5

= \frac{5}{2}

= \frac{\frac{5}{2}}{\frac{3}{2}}

分式相除:

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

= \frac{5 \cdot 2}{2 \cdot 3}

约分:

2

= \frac{5}{3}

= \frac{5}{3}

= \frac{5}{3}

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