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受欢迎的三角函数 >

sqrt((9.8(10))/(2pi)tanh(2pi 8/10))

解答

\frac{9.8 ( 10 )}{2 π} tanh (2 π \frac{8}{10})

解答

\frac{7 π e ^{\frac{16 π}{5}} - 1 e ^{\frac{16 π}{5}} + 1}{π e ^{\frac{16 π}{5}} + π}

+1

十进制

3.94915 \dots

求解步骤

\frac{9.8 ( 10 )}{2 π} tanh (2 π \frac{8}{10})

= \frac{\frac{49}{5} \cdot 10}{2 π} tanh (2 π \frac{8}{10})

化简:

2 π \frac{8}{10} = \frac{8 π}{5}

2 π \frac{8}{10}

分式相乘:

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

= \frac{8 \cdot 2 π}{10}

数字相乘:

8 \cdot 2 = 16

= \frac{16 π}{10}

约分:

2

= \frac{8 π}{5}

\frac{\frac{49}{5} \cdot 10}{2 π} tanh (\frac{8 π}{5}) = \frac{7 π tanh ( \frac{8 π}{5} )}{π}

\frac{\frac{49}{5} \cdot 10}{2 π} tanh (\frac{8 π}{5})

乘

\frac{\frac{49}{5} \cdot 10}{2 π} tanh (\frac{8 π}{5}) : \frac{49 tanh ( \frac{8 π}{5} )}{π}

\frac{\frac{49}{5} \cdot 10}{2 π} tanh (\frac{8 π}{5})

分式相乘:

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

= \frac{\frac{49}{5} \cdot 10 tanh ( \frac{8 π}{5} )}{2 π}

乘

\frac{49}{5} \cdot 10 tanh (\frac{8 π}{5}) : 98 tanh (\frac{8 π}{5})

\frac{49}{5} \cdot 10 tanh (\frac{8 π}{5})

分式相乘:

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

= \frac{49 \cdot 10 tanh ( \frac{8 π}{5} )}{5}

数字相乘:

49 \cdot 10 = 490

= \frac{490 tanh ( \frac{8 π}{5} )}{5}

数字相除:

\frac{490}{5} = 98

= 98 tanh (\frac{8 π}{5})

= \frac{98 tanh ( \frac{8 π}{5} )}{2 π}

数字相除:

\frac{98}{2} = 49

= \frac{49 tanh ( \frac{8 π}{5} )}{π}

= \frac{49 tanh ( \frac{8 π}{5} )}{π}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{49 tanh ( \frac{8 π}{5} )}{π}

使用根式运算法则:

n ab = n a n b,

假定

a \geq 0, b \geq 0

49 tanh (\frac{8 π}{5}) = 49 tanh (\frac{8 π}{5})

= \frac{49 tanh ( \frac{8 π}{5} )}{π}

49 = 7

49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

= \frac{7 tanh ( \frac{8 π}{5} )}{π}

\frac{7 tanh ( \frac{8 π}{5} )}{π}

有理化:

\frac{7 π tanh ( \frac{8 π}{5} )}{π}

\frac{7 tanh ( \frac{8 π}{5} )}{π}

乘以共轭根式

\frac{π}{π}

= \frac{7 tanh ( \frac{8 π}{5} ) π}{π π}

π π = π

π π

使用根式运算法则:

a a = a

π π = π

= π

= \frac{7 π tanh ( \frac{8 π}{5} )}{π}

= \frac{7 π tanh ( \frac{8 π}{5} )}{π}

= \frac{7 π tanh ( \frac{8 π}{5} )}{π}

使用三角恒等式改写:

tanh (\frac{8 π}{5}) = \frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{16 π}{5}} + 1}

tanh (\frac{8 π}{5})

使用双曲函数恒等式:

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

= \frac{e ^{\frac{8 π}{5}} - e ^{- \frac{8 π}{5}}}{e ^{\frac{8 π}{5}} + e ^{- \frac{8 π}{5}}}

\frac{e ^{\frac{8 π}{5}} - e ^{- \frac{8 π}{5}}}{e ^{\frac{8 π}{5}} + e ^{- \frac{8 π}{5}}} = \frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{16 π}{5}} + 1}

\frac{e ^{\frac{8 π}{5}} - e ^{- \frac{8 π}{5}}}{e ^{\frac{8 π}{5}} + e ^{- \frac{8 π}{5}}}

使用指数法则:

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

e^{- \frac{8 π}{5}} = \frac{1}{e ^{\frac{8 π}{5}}}

= \frac{e ^{\frac{8 π}{5}} - \frac{1}{e ^{\frac{8 π}{5}}}}{e ^{\frac{8 π}{5}} + \frac{1}{e ^{\frac{8 π}{5}}}}

化简

e^{\frac{8 π}{5}} + \frac{1}{e ^{\frac{8 π}{5}}} : \frac{e ^{\frac{16 π}{5}} + 1}{e ^{\frac{8 π}{5}}}

e^{\frac{8 π}{5}} + \frac{1}{e ^{\frac{8 π}{5}}}

将项转换为分式:

e^{\frac{8 π}{5}} = \frac{e ^{\frac{8 π}{5}} e ^{\frac{8 π}{5}}}{e ^{\frac{8 π}{5}}}

= \frac{e ^{\frac{8 π}{5}} e ^{\frac{8 π}{5}}}{e ^{\frac{8 π}{5}}} + \frac{1}{e ^{\frac{8 π}{5}}}

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 ^{\frac{8 π}{5}} e ^{\frac{8 π}{5}} + 1}{e ^{\frac{8 π}{5}}}

e^{\frac{8 π}{5}} e^{\frac{8 π}{5}} + 1 = e^{2 \cdot \frac{8 π}{5}} + 1

e^{\frac{8 π}{5}} e^{\frac{8 π}{5}} + 1

e^{\frac{8 π}{5}} e^{\frac{8 π}{5}} = e^{2 \cdot \frac{8 π}{5}}

e^{\frac{8 π}{5}} e^{\frac{8 π}{5}}

使用指数法则:

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

e^{\frac{8 π}{5}} e^{\frac{8 π}{5}} = e^{\frac{8 π}{5} + \frac{8 π}{5}}

= e^{\frac{8 π}{5} + \frac{8 π}{5}}

同类项相加:

\frac{8 π}{5} + \frac{8 π}{5} = 2 \cdot \frac{8 π}{5}

= e^{2 \cdot \frac{8 π}{5}}

= e^{2 \cdot \frac{8 π}{5}} + 1

= \frac{e ^{2 \cdot \frac{8 π}{5}} + 1}{e ^{\frac{8 π}{5}}}

乘

2 \cdot \frac{8 π}{5} : \frac{16 π}{5}

2 \cdot \frac{8 π}{5}

分式相乘:

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

= \frac{8 π 2}{5}

数字相乘:

8 \cdot 2 = 16

= \frac{16 π}{5}

= \frac{e ^{\frac{16 π}{5}} + 1}{e ^{\frac{8 π}{5}}}

= \frac{e ^{\frac{8 π}{5}} - \frac{1}{e ^{\frac{8 π}{5}}}}{\frac{e ^{\frac{16 π}{5}} + 1}{e ^{\frac{8 π}{5}}}}

化简

e^{\frac{8 π}{5}} - \frac{1}{e ^{\frac{8 π}{5}}} : \frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{8 π}{5}}}

e^{\frac{8 π}{5}} - \frac{1}{e ^{\frac{8 π}{5}}}

将项转换为分式:

e^{\frac{8 π}{5}} = \frac{e ^{\frac{8 π}{5}} e ^{\frac{8 π}{5}}}{e ^{\frac{8 π}{5}}}

= \frac{e ^{\frac{8 π}{5}} e ^{\frac{8 π}{5}}}{e ^{\frac{8 π}{5}}} - \frac{1}{e ^{\frac{8 π}{5}}}

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 ^{\frac{8 π}{5}} e ^{\frac{8 π}{5}} - 1}{e ^{\frac{8 π}{5}}}

e^{\frac{8 π}{5}} e^{\frac{8 π}{5}} - 1 = e^{2 \cdot \frac{8 π}{5}} - 1

e^{\frac{8 π}{5}} e^{\frac{8 π}{5}} - 1

e^{\frac{8 π}{5}} e^{\frac{8 π}{5}} = e^{2 \cdot \frac{8 π}{5}}

e^{\frac{8 π}{5}} e^{\frac{8 π}{5}}

使用指数法则:

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

e^{\frac{8 π}{5}} e^{\frac{8 π}{5}} = e^{\frac{8 π}{5} + \frac{8 π}{5}}

= e^{\frac{8 π}{5} + \frac{8 π}{5}}

同类项相加:

\frac{8 π}{5} + \frac{8 π}{5} = 2 \cdot \frac{8 π}{5}

= e^{2 \cdot \frac{8 π}{5}}

= e^{2 \cdot \frac{8 π}{5}} - 1

= \frac{e ^{2 \cdot \frac{8 π}{5}} - 1}{e ^{\frac{8 π}{5}}}

乘

2 \cdot \frac{8 π}{5} : \frac{16 π}{5}

2 \cdot \frac{8 π}{5}

分式相乘:

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

= \frac{8 π 2}{5}

数字相乘:

8 \cdot 2 = 16

= \frac{16 π}{5}

= \frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{8 π}{5}}}

= \frac{\frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{8 π}{5}}}}{\frac{e ^{\frac{16 π}{5}} + 1}{e ^{\frac{8 π}{5}}}}

分式相除:

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

= \frac{( e ^{\frac{16 π}{5}} - 1 ) e ^{\frac{8 π}{5}}}{e ^{\frac{8 π}{5}} ( e ^{\frac{16 π}{5}} + 1 )}

约分:

e^{\frac{8 π}{5}}

= \frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{16 π}{5}} + 1}

= \frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{16 π}{5}} + 1}

= \frac{7 π \frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{16 π}{5}} + 1}}{π}

\frac{7 π \frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{16 π}{5}} + 1}}{π} = \frac{7 π e ^{\frac{16 π}{5}} - 1 e ^{\frac{16 π}{5}} + 1}{π e ^{\frac{16 π}{5}} + π}

\frac{7 π \frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{16 π}{5}} + 1}}{π}

\frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{16 π}{5}} + 1} = \frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{16 π}{5}} + 1}

\frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{16 π}{5}} + 1}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{16 π}{5}} + 1}

= \frac{7 π \frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{16 π}{5}} + 1}}{π}

乘

7 π \frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{16 π}{5}} + 1} : \frac{7 π e ^{\frac{16 π}{5}} - 1}{e ^{\frac{16 π}{5}} + 1}

7 π \frac{e ^{\frac{16 π}{5}} - 1}{e ^{\frac{16 π}{5}} + 1}

分式相乘:

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

= \frac{e ^{\frac{16 π}{5}} - 1 \cdot 7 π}{e ^{\frac{16 π}{5}} + 1}

= \frac{\frac{7 π e ^{\frac{16 π}{5}} - 1}{e ^{\frac{16 π}{5}} + 1}}{π}

使用分式法则:

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

= \frac{e ^{\frac{16 π}{5}} - 1 \cdot 7 π}{e ^{\frac{16 π}{5}} + 1 π}

\frac{7 π e ^{\frac{16 π}{5}} - 1}{π e ^{\frac{16 π}{5}} + 1}

有理化:

\frac{7 π e ^{\frac{16 π}{5}} - 1 e ^{\frac{16 π}{5}} + 1}{π e ^{\frac{16 π}{5}} + π}

\frac{7 π e ^{\frac{16 π}{5}} - 1}{π e ^{\frac{16 π}{5}} + 1}

乘以共轭根式

\frac{e ^{\frac{16 π}{5}} + 1}{e ^{\frac{16 π}{5}} + 1}

= \frac{e ^{\frac{16 π}{5}} - 1 \cdot 7 π e ^{\frac{16 π}{5}} + 1}{e ^{\frac{16 π}{5}} + 1 π e ^{\frac{16 π}{5}} + 1}

e^{\frac{16 π}{5}} + 1 π e^{\frac{16 π}{5}} + 1 = π e^{\frac{16 π}{5}} + π

e^{\frac{16 π}{5}} + 1 π e^{\frac{16 π}{5}} + 1

使用根式运算法则:

a a = a

e^{\frac{16 π}{5}} + 1 e^{\frac{16 π}{5}} + 1 = e^{\frac{16 π}{5}} + 1

= π (e^{\frac{16 π}{5}} + 1)

使用分配律:

a (b + c) = ab + a c

a = π, b = e^{\frac{16 π}{5}}, c = 1

= π e^{\frac{16 π}{5}} + π 1

= π e^{\frac{16 π}{5}} + 1 π

乘以:

1 π = π

= π e^{\frac{16 π}{5}} + π

= \frac{7 π e ^{\frac{16 π}{5}} - 1 e ^{\frac{16 π}{5}} + 1}{π e ^{\frac{16 π}{5}} + π}

= \frac{7 π e ^{\frac{16 π}{5}} - 1 e ^{\frac{16 π}{5}} + 1}{π e ^{\frac{16 π}{5}} + π}

= \frac{7 π e ^{\frac{16 π}{5}} - 1 e ^{\frac{16 π}{5}} + 1}{π e ^{\frac{16 π}{5}} + π}

流行的例子

sec(arctan(5/12))

sec (arctan (\frac{5}{12}))

sin (210 \circ)

\frac{tan ( π )}{12}

2(cos(150)+isin(150))

2 (cos (15 0^{\circ}) + i sin (15 0^{\circ}))

(1-cos(315))/(sin(315))

\frac{1 - cos ( 31 5 ^{\circ} )}{sin ( 31 5 ^{\circ} )}