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受欢迎的三角函数 >

cos(pi/2-x)= 4/5

解答

cos (\frac{π}{2} - x) = \frac{4}{5}

解答

x = - 2 πn + \frac{π}{2} - 0.64350 \dots, x = - 2 π - 2 πn + \frac{π}{2} + 0.64350 \dots

+1

度数

x = 53.13010 \dots^{\circ} - 36 0^{\circ} n, x = - 233.13010 \dots^{\circ} - 36 0^{\circ} n

求解步骤

cos (\frac{π}{2} - x) = \frac{4}{5}

使用反三角函数性质

cos (\frac{π}{2} - x) = \frac{4}{5}

cos (\frac{π}{2} - x) = \frac{4}{5}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

\frac{π}{2} - x = arccos (\frac{4}{5}) + 2 πn, \frac{π}{2} - x = 2 π - arccos (\frac{4}{5}) + 2 πn

\frac{π}{2} - x = arccos (\frac{4}{5}) + 2 πn, \frac{π}{2} - x = 2 π - arccos (\frac{4}{5}) + 2 πn

解

\frac{π}{2} - x = arccos (\frac{4}{5}) + 2 πn : x = - 2 πn + \frac{π}{2} - arccos (\frac{4}{5})

\frac{π}{2} - x = arccos (\frac{4}{5}) + 2 πn

将

\frac{π}{2}

到右边

\frac{π}{2} - x = arccos (\frac{4}{5}) + 2 πn

两边减去

\frac{π}{2}

\frac{π}{2} - x - \frac{π}{2} = arccos (\frac{4}{5}) + 2 πn - \frac{π}{2}

化简

- x = arccos (\frac{4}{5}) + 2 πn - \frac{π}{2}

- x = arccos (\frac{4}{5}) + 2 πn - \frac{π}{2}

两边除以

- 1

- x = arccos (\frac{4}{5}) + 2 πn - \frac{π}{2}

两边除以

- 1

\frac{- x}{- 1} = \frac{arccos ( \frac{4}{5} )}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1}

化简

\frac{- x}{- 1} = \frac{arccos ( \frac{4}{5} )}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1}

化简

\frac{- x}{- 1} : x

\frac{- x}{- 1}

使用分式法则:

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

= \frac{x}{1}

使用法则

\frac{a}{1} = a

= x

化简

\frac{arccos ( \frac{4}{5} )}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1} : - 2 πn + \frac{π}{2} - arccos (\frac{4}{5})

\frac{arccos ( \frac{4}{5} )}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1}

对同类项分组

= \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1} + \frac{arccos ( \frac{4}{5} )}{- 1}

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

使用分式法则:

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

= - \frac{2 πn}{1}

使用法则

\frac{a}{1} = a

= - 2 πn

= - 2 πn - \frac{\frac{π}{2}}{- 1} + \frac{arccos ( \frac{4}{5} )}{- 1}

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

\frac{\frac{π}{2}}{- 1}

使用分式法则:

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

= - \frac{\frac{π}{2}}{1}

使用分式法则:

\frac{a}{1} = a

\frac{\frac{π}{2}}{1} = \frac{π}{2}

= - \frac{π}{2}

\frac{arccos ( \frac{4}{5} )}{- 1} = - arccos (\frac{4}{5})

\frac{arccos ( \frac{4}{5} )}{- 1}

使用分式法则:

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

= - \frac{arccos ( \frac{4}{5} )}{1}

使用分式法则:

\frac{a}{1} = a

\frac{arccos ( \frac{4}{5} )}{1} = arccos (\frac{4}{5})

= - arccos (\frac{4}{5})

= - 2 πn - (- \frac{π}{2}) - arccos (\frac{4}{5})

使用法则

- (- a) = a

= - 2 πn + \frac{π}{2} - arccos (\frac{4}{5})

x = - 2 πn + \frac{π}{2} - arccos (\frac{4}{5})

x = - 2 πn + \frac{π}{2} - arccos (\frac{4}{5})

x = - 2 πn + \frac{π}{2} - arccos (\frac{4}{5})

解

\frac{π}{2} - x = 2 π - arccos (\frac{4}{5}) + 2 πn : x = - 2 π - 2 πn + \frac{π}{2} + arccos (\frac{4}{5})

\frac{π}{2} - x = 2 π - arccos (\frac{4}{5}) + 2 πn

将

\frac{π}{2}

到右边

\frac{π}{2} - x = 2 π - arccos (\frac{4}{5}) + 2 πn

两边减去

\frac{π}{2}

\frac{π}{2} - x - \frac{π}{2} = 2 π - arccos (\frac{4}{5}) + 2 πn - \frac{π}{2}

化简

- x = 2 π - arccos (\frac{4}{5}) + 2 πn - \frac{π}{2}

- x = 2 π - arccos (\frac{4}{5}) + 2 πn - \frac{π}{2}

两边除以

- 1

- x = 2 π - arccos (\frac{4}{5}) + 2 πn - \frac{π}{2}

两边除以

- 1

\frac{- x}{- 1} = \frac{2 π}{- 1} - \frac{arccos ( \frac{4}{5} )}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1}

化简

\frac{- x}{- 1} = \frac{2 π}{- 1} - \frac{arccos ( \frac{4}{5} )}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1}

化简

\frac{- x}{- 1} : x

\frac{- x}{- 1}

使用分式法则:

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

= \frac{x}{1}

使用法则

\frac{a}{1} = a

= x

化简

\frac{2 π}{- 1} - \frac{arccos ( \frac{4}{5} )}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1} : - 2 π - 2 πn + \frac{π}{2} + arccos (\frac{4}{5})

\frac{2 π}{- 1} - \frac{arccos ( \frac{4}{5} )}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1}

对同类项分组

= \frac{2 π}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1} - \frac{arccos ( \frac{4}{5} )}{- 1}

\frac{2 π}{- 1} = - 2 π

\frac{2 π}{- 1}

使用分式法则:

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

= - \frac{2 π}{1}

使用法则

\frac{a}{1} = a

= - 2 π

= - 2 π + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1} - \frac{arccos ( \frac{4}{5} )}{- 1}

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

使用分式法则:

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

= - \frac{2 πn}{1}

使用法则

\frac{a}{1} = a

= - 2 πn

= - 2 π - 2 πn - \frac{\frac{π}{2}}{- 1} - \frac{arccos ( \frac{4}{5} )}{- 1}

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

\frac{\frac{π}{2}}{- 1}

使用分式法则:

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

= - \frac{\frac{π}{2}}{1}

使用分式法则:

\frac{a}{1} = a

\frac{\frac{π}{2}}{1} = \frac{π}{2}

= - \frac{π}{2}

\frac{arccos ( \frac{4}{5} )}{- 1} = - arccos (\frac{4}{5})

\frac{arccos ( \frac{4}{5} )}{- 1}

使用分式法则:

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

= - \frac{arccos ( \frac{4}{5} )}{1}

使用分式法则:

\frac{a}{1} = a

\frac{arccos ( \frac{4}{5} )}{1} = arccos (\frac{4}{5})

= - arccos (\frac{4}{5})

= - 2 π - 2 πn - (- \frac{π}{2}) - (- arccos (\frac{4}{5}))

使用法则

- (- a) = a

= - 2 π - 2 πn + \frac{π}{2} + arccos (\frac{4}{5})

x = - 2 π - 2 πn + \frac{π}{2} + arccos (\frac{4}{5})

x = - 2 π - 2 πn + \frac{π}{2} + arccos (\frac{4}{5})

x = - 2 π - 2 πn + \frac{π}{2} + arccos (\frac{4}{5})

x = - 2 πn + \frac{π}{2} - arccos (\frac{4}{5}), x = - 2 π - 2 πn + \frac{π}{2} + arccos (\frac{4}{5})

以小数形式表示解

x = - 2 πn + \frac{π}{2} - 0.64350 \dots, x = - 2 π - 2 πn + \frac{π}{2} + 0.64350 \dots

作图

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