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受欢迎的三角函数 >

tanh(1)

解答

tanh (1)

解答

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

+1

十进制

0.76159 \dots

求解步骤

tanh (1)

使用双曲函数恒等式:

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

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

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

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

使用法则

a^{1} = a

e^{1} = e

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

使用指数法则:

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

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

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

化简

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

e + \frac{1}{e}

将项转换为分式:

e = \frac{ee}{e}

= \frac{ee}{e} + \frac{1}{e}

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{ee + 1}{e}

ee + 1 = e^{2} + 1

ee + 1

ee = e^{2}

ee

使用指数法则:

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

ee = e^{1 + 1}

= e^{1 + 1}

数字相加:

1 + 1 = 2

= e^{2}

= e^{2} + 1

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

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

化简

e - \frac{1}{e} : \frac{e ^{2} - 1}{e}

e - \frac{1}{e}

将项转换为分式:

e = \frac{ee}{e}

= \frac{ee}{e} - \frac{1}{e}

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{ee - 1}{e}

ee - 1 = e^{2} - 1

ee - 1

ee = e^{2}

ee

使用指数法则:

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

ee = e^{1 + 1}

= e^{1 + 1}

数字相加:

1 + 1 = 2

= e^{2}

= e^{2} - 1

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

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

分式相除:

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

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

约分:

e

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

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

流行的例子

tan (\frac{5 π}{4})

tan (22 5^{\circ})

tan (30 0^{\circ})

cos (- \frac{5 π}{6})

2 tan^{2} (x) - 2 = 0