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受欢迎的三角函数 >

sin(a)=cos(pi/2-θ)

解答

sin (a) = cos (\frac{π}{2} - θ)

解答

θ = - arccos (sin (a)) - 2 πn + \frac{π}{2}, θ = arccos (sin (a)) - 2 πn + \frac{π}{2}

求解步骤

sin (a) = cos (\frac{π}{2} - θ)

交换两边

cos (\frac{π}{2} - θ) = sin (a)

使用反三角函数性质

cos (\frac{π}{2} - θ) = sin (a)

cos (\frac{π}{2} - θ) = sin (a)

的通解

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

\frac{π}{2} - θ = arccos (sin (a)) + 2 πn, \frac{π}{2} - θ = - arccos (sin (a)) + 2 πn

\frac{π}{2} - θ = arccos (sin (a)) + 2 πn, \frac{π}{2} - θ = - arccos (sin (a)) + 2 πn

解

\frac{π}{2} - θ = arccos (sin (a)) + 2 πn : θ = - arccos (sin (a)) - 2 πn + \frac{π}{2}

\frac{π}{2} - θ = arccos (sin (a)) + 2 πn

将

\frac{π}{2}

到右边

\frac{π}{2} - θ = arccos (sin (a)) + 2 πn

两边减去

\frac{π}{2}

\frac{π}{2} - θ - \frac{π}{2} = arccos (sin (a)) + 2 πn - \frac{π}{2}

化简

- θ = arccos (sin (a)) + 2 πn - \frac{π}{2}

- θ = arccos (sin (a)) + 2 πn - \frac{π}{2}

两边除以

- 1

- θ = arccos (sin (a)) + 2 πn - \frac{π}{2}

两边除以

- 1

\frac{- θ}{- 1} = \frac{arccos ( sin ( a ) )}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1}

化简

\frac{- θ}{- 1} = \frac{arccos ( sin ( a ) )}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1}

化简

\frac{- θ}{- 1} : θ

\frac{- θ}{- 1}

使用分式法则:

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

= \frac{θ}{1}

使用法则

\frac{a}{1} = a

= θ

化简

\frac{arccos ( sin ( a ) )}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1} : - arccos (sin (a)) - 2 πn + \frac{π}{2}

\frac{arccos ( sin ( a ) )}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1}

\frac{arccos ( sin ( a ) )}{- 1} = - arccos (sin (a))

\frac{arccos ( sin ( a ) )}{- 1}

使用分式法则:

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

= - \frac{arccos ( sin ( a ) )}{1}

使用法则

\frac{a}{1} = a

= - arccos (sin (a))

= - arccos (sin (a)) + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 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

= - arccos (sin (a)) - 2 πn - \frac{\frac{π}{2}}{- 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}

= - arccos (sin (a)) - 2 πn - (- \frac{π}{2})

使用法则

- (- a) = a

= - arccos (sin (a)) - 2 πn + \frac{π}{2}

θ = - arccos (sin (a)) - 2 πn + \frac{π}{2}

θ = - arccos (sin (a)) - 2 πn + \frac{π}{2}

θ = - arccos (sin (a)) - 2 πn + \frac{π}{2}

解

\frac{π}{2} - θ = - arccos (sin (a)) + 2 πn : θ = arccos (sin (a)) - 2 πn + \frac{π}{2}

\frac{π}{2} - θ = - arccos (sin (a)) + 2 πn

将

\frac{π}{2}

到右边

\frac{π}{2} - θ = - arccos (sin (a)) + 2 πn

两边减去

\frac{π}{2}

\frac{π}{2} - θ - \frac{π}{2} = - arccos (sin (a)) + 2 πn - \frac{π}{2}

化简

- θ = - arccos (sin (a)) + 2 πn - \frac{π}{2}

- θ = - arccos (sin (a)) + 2 πn - \frac{π}{2}

两边除以

- 1

- θ = - arccos (sin (a)) + 2 πn - \frac{π}{2}

两边除以

- 1

\frac{- θ}{- 1} = - \frac{arccos ( sin ( a ) )}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1}

化简

\frac{- θ}{- 1} = - \frac{arccos ( sin ( a ) )}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1}

化简

\frac{- θ}{- 1} : θ

\frac{- θ}{- 1}

使用分式法则:

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

= \frac{θ}{1}

使用法则

\frac{a}{1} = a

= θ

化简

- \frac{arccos ( sin ( a ) )}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1} : arccos (sin (a)) - 2 πn + \frac{π}{2}

- \frac{arccos ( sin ( a ) )}{- 1} + \frac{2 πn}{- 1} - \frac{\frac{π}{2}}{- 1}

\frac{arccos ( sin ( a ) )}{- 1} = - arccos (sin (a))

\frac{arccos ( sin ( a ) )}{- 1}

使用分式法则:

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

= - \frac{arccos ( sin ( a ) )}{1}

使用法则

\frac{a}{1} = a

= - arccos (sin (a))

\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

= - (- arccos (sin (a))) - 2 πn - \frac{\frac{π}{2}}{- 1}

使用法则

- (- a) = a

= arccos (sin (a)) - 2 πn - \frac{\frac{π}{2}}{- 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}

= arccos (sin (a)) - 2 πn - (- \frac{π}{2})

使用法则

- (- a) = a

= arccos (sin (a)) - 2 πn + \frac{π}{2}

θ = arccos (sin (a)) - 2 πn + \frac{π}{2}

θ = arccos (sin (a)) - 2 πn + \frac{π}{2}

θ = arccos (sin (a)) - 2 πn + \frac{π}{2}

θ = - arccos (sin (a)) - 2 πn + \frac{π}{2}, θ = arccos (sin (a)) - 2 πn + \frac{π}{2}

作图

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