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受欢迎的三角函数 >

tan(2arcsin(-12/13))

解答

tan (2 arcsin (- \frac{12}{13}))

解答

\frac{120}{119}

+1

十进制

1.00840 \dots

求解步骤

tan (2 arcsin (- \frac{12}{13}))

使用三角恒等式改写:

\frac{2 tan ( arcsin ( \frac{12}{13} ) )}{tan ^{2} ( arcsin ( \frac{12}{13} ) ) - 1}

tan (2 arcsin (- \frac{12}{13}))

使用倍角公式:

tan (2 x) = \frac{2 tan ( x )}{1 - tan ^{2} ( x )}

= \frac{2 tan ( arcsin ( - \frac{12}{13} ) )}{1 - tan ^{2} ( arcsin ( - \frac{12}{13} ) )}

\frac{2 tan ( arcsin ( - \frac{12}{13} ) )}{1 - tan ^{2} ( arcsin ( - \frac{12}{13} ) )} = - \frac{2 tan ( arcsin ( - \frac{12}{13} ) )}{tan ^{2} ( arcsin ( - \frac{12}{13} ) ) - 1}

\frac{2 tan ( arcsin ( - \frac{12}{13} ) )}{1 - tan ^{2} ( arcsin ( - \frac{12}{13} ) )}

使用分式法则:

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

1 - tan^{2} (arcsin (- \frac{12}{13})) = - (tan^{2} (arcsin (- \frac{12}{13})) - 1)

= \frac{- 2 tan ( arcsin ( - \frac{12}{13} ) )}{tan ^{2} ( arcsin ( - \frac{12}{13} ) ) - 1}

使用分式法则:

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

= - \frac{2 tan ( arcsin ( - \frac{12}{13} ) )}{tan ^{2} ( arcsin ( - \frac{12}{13} ) ) - 1}

= - \frac{2 tan ( arcsin ( - \frac{12}{13} ) )}{tan ^{2} ( arcsin ( - \frac{12}{13} ) ) - 1}

利用以下特性:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{12}{13}) = - arcsin (\frac{12}{13})

= - \frac{2 tan ( arcsin ( - \frac{12}{13} ) )}{tan ^{2} ( - arcsin ( \frac{12}{13} ) ) - 1}

利用以下特性:

tan (- x) = - tan (x)

tan (- arcsin (\frac{12}{13})) = - tan (arcsin (\frac{12}{13}))

= - \frac{2 tan ( arcsin ( - \frac{12}{13} ) )}{( - tan ( arcsin ( \frac{12}{13} ) ) ) ^{2} - 1}

利用以下特性:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{12}{13}) = - arcsin (\frac{12}{13})

= - \frac{2 tan ( - arcsin ( \frac{12}{13} ) )}{( - tan ( arcsin ( \frac{12}{13} ) ) ) ^{2} - 1}

利用以下特性:

tan (- x) = - tan (x)

tan (- arcsin (\frac{12}{13})) = - tan (arcsin (\frac{12}{13}))

= - \frac{2 ( - tan ( arcsin ( \frac{12}{13} ) ) )}{( - tan ( arcsin ( \frac{12}{13} ) ) ) ^{2} - 1}

化简

= \frac{2 tan ( arcsin ( \frac{12}{13} ) )}{tan ^{2} ( arcsin ( \frac{12}{13} ) ) - 1}

= \frac{2 tan ( arcsin ( \frac{12}{13} ) )}{tan ^{2} ( arcsin ( \frac{12}{13} ) ) - 1}

使用三角恒等式改写:

tan (arcsin (\frac{12}{13})) = \frac{12}{5}

tan (arcsin (\frac{12}{13}))

使用三角恒等式改写:

tan (arcsin (\frac{12}{13})) = \frac{( \frac{12}{13} ) 1 - ( \frac{12}{13} ) ^{2}}{1 - ( \frac{12}{13} ) ^{2}}

利用以下特性:

tan (arcsin (x)) = \frac{x 1 - x ^{2}}{1 - x ^{2}}

= \frac{( \frac{12}{13} ) 1 - ( \frac{12}{13} ) ^{2}}{1 - ( \frac{12}{13} ) ^{2}}

= \frac{\frac{12}{13} 1 - ( \frac{12}{13} ) ^{2}}{1 - ( \frac{12}{13} ) ^{2}}

化简

= \frac{12}{5}

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

化简

\frac{2 \cdot \frac{12}{5}}{( \frac{12}{5} ) ^{2} - 1} : \frac{120}{119}

\frac{2 \cdot \frac{12}{5}}{( \frac{12}{5} ) ^{2} - 1}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{2 \cdot \frac{12}{5}}{\frac{1 2 ^{2}}{5 ^{2}} - 1}

乘

2 \cdot \frac{12}{5} : \frac{24}{5}

2 \cdot \frac{12}{5}

分式相乘:

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

= \frac{12 \cdot 2}{5}

数字相乘:

12 \cdot 2 = 24

= \frac{24}{5}

= \frac{\frac{24}{5}}{\frac{1 2 ^{2}}{5 ^{2}} - 1}

\frac{1 2 ^{2}}{5 ^{2}} = \frac{144}{25}

\frac{1 2 ^{2}}{5 ^{2}}

1 2^{2} = 144

= \frac{144}{5 ^{2}}

5^{2} = 25

= \frac{144}{25}

= \frac{\frac{24}{5}}{\frac{144}{25} - 1}

使用分式法则:

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

= \frac{24}{5 ( \frac{144}{25} - 1 )}

化简

\frac{144}{25} - 1 : \frac{119}{25}

\frac{144}{25} - 1

将项转换为分式:

1 = \frac{1 \cdot 25}{25}

= \frac{144}{25} - \frac{1 \cdot 25}{25}

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{144 - 1 \cdot 25}{25}

144 - 1 \cdot 25 = 119

144 - 1 \cdot 25

数字相乘:

1 \cdot 25 = 25

= 144 - 25

数字相减:

144 - 25 = 119

= 119

= \frac{119}{25}

= \frac{24}{5 \cdot \frac{119}{25}}

乘

5 \cdot \frac{119}{25} : \frac{119}{5}

5 \cdot \frac{119}{25}

分式相乘:

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

= \frac{119 \cdot 5}{25}

数字相乘:

119 \cdot 5 = 595

= \frac{595}{25}

约分:

5

= \frac{119}{5}

= \frac{24}{\frac{119}{5}}

使用分式法则:

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

= \frac{24 \cdot 5}{119}

数字相乘:

24 \cdot 5 = 120

= \frac{120}{119}

= \frac{120}{119}

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