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受欢迎的三角函数 >

-arccos(x)+arctan(5/12)=0

解答

- arccos (x) + arctan (\frac{5}{12}) = 0

解答

x = \frac{12}{13}

求解步骤

- arccos (x) + arctan (\frac{5}{12}) = 0

将

arctan (\frac{5}{12})

到右边

- arccos (x) + arctan (\frac{5}{12}) = 0

两边减去

arctan (\frac{5}{12})

- arccos (x) + arctan (\frac{5}{12}) - arctan (\frac{5}{12}) = 0 - arctan (\frac{5}{12})

化简

- arccos (x) = - arctan (\frac{5}{12})

- arccos (x) = - arctan (\frac{5}{12})

两边除以

- 1

- arccos (x) = - arctan (\frac{5}{12})

两边除以

- 1

\frac{- arccos ( x )}{- 1} = \frac{- arctan ( \frac{5}{12} )}{- 1}

化简

\frac{- arccos ( x )}{- 1} = \frac{- arctan ( \frac{5}{12} )}{- 1}

化简

\frac{- arccos ( x )}{- 1} : arccos (x)

\frac{- arccos ( x )}{- 1}

使用分式法则:

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

= \frac{arccos ( x )}{1}

使用法则

\frac{a}{1} = a

= arccos (x)

化简

\frac{- arctan ( \frac{5}{12} )}{- 1} : arctan (\frac{5}{12})

\frac{- arctan ( \frac{5}{12} )}{- 1}

使用分式法则:

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

= \frac{arctan ( \frac{5}{12} )}{1}

使用分式法则:

\frac{a}{1} = a

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

arccos (x) = arctan (\frac{5}{12})

arccos (x) = arctan (\frac{5}{12})

arccos (x) = arctan (\frac{5}{12})

使用反三角函数性质

arccos (x) = arctan (\frac{5}{12})

arccos (x) = a \Rightarrow x = cos (a)

x = cos (arctan (\frac{5}{12}))

cos (arctan (\frac{5}{12})) = \frac{12}{13}

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

使用三角恒等式改写:

cos (arctan (\frac{5}{12})) = \frac{1 + ( \frac{5}{12} ) ^{2}}{1 + ( \frac{5}{12} ) ^{2}}

利用以下特性:

cos (arctan (x)) = \frac{1 + x ^{2}}{1 + x ^{2}}

= \frac{1 + ( \frac{5}{12} ) ^{2}}{1 + ( \frac{5}{12} ) ^{2}}

= \frac{1 + ( \frac{5}{12} ) ^{2}}{1 + ( \frac{5}{12} ) ^{2}}

化简

= \frac{12}{13}

x = \frac{12}{13}

x = \frac{12}{13}

作图

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