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受欢迎的三角函数 >

tan(2arcsin(-4/5))

解答

tan (2 arcsin (- \frac{4}{5}))

解答

\frac{24}{7}

+1

十进制

3.42857 \dots

求解步骤

tan (2 arcsin (- \frac{4}{5}))

使用三角恒等式改写:

\frac{2 tan ( arcsin ( \frac{4}{5} ) )}{tan ^{2} ( arcsin ( \frac{4}{5} ) ) - 1}

tan (2 arcsin (- \frac{4}{5}))

使用倍角公式:

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

= \frac{2 tan ( arcsin ( - \frac{4}{5} ) )}{1 - tan ^{2} ( arcsin ( - \frac{4}{5} ) )}

\frac{2 tan ( arcsin ( - \frac{4}{5} ) )}{1 - tan ^{2} ( arcsin ( - \frac{4}{5} ) )} = - \frac{2 tan ( arcsin ( - \frac{4}{5} ) )}{tan ^{2} ( arcsin ( - \frac{4}{5} ) ) - 1}

\frac{2 tan ( arcsin ( - \frac{4}{5} ) )}{1 - tan ^{2} ( arcsin ( - \frac{4}{5} ) )}

使用分式法则:

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

1 - tan^{2} (arcsin (- \frac{4}{5})) = - (tan^{2} (arcsin (- \frac{4}{5})) - 1)

= \frac{- 2 tan ( arcsin ( - \frac{4}{5} ) )}{tan ^{2} ( arcsin ( - \frac{4}{5} ) ) - 1}

使用分式法则:

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

= - \frac{2 tan ( arcsin ( - \frac{4}{5} ) )}{tan ^{2} ( arcsin ( - \frac{4}{5} ) ) - 1}

= - \frac{2 tan ( arcsin ( - \frac{4}{5} ) )}{tan ^{2} ( arcsin ( - \frac{4}{5} ) ) - 1}

利用以下特性:

arcsin (- x) = - arcsin (x)

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

= - \frac{2 tan ( arcsin ( - \frac{4}{5} ) )}{tan ^{2} ( - arcsin ( \frac{4}{5} ) ) - 1}

利用以下特性:

tan (- x) = - tan (x)

tan (- arcsin (\frac{4}{5})) = - tan (arcsin (\frac{4}{5}))

= - \frac{2 tan ( arcsin ( - \frac{4}{5} ) )}{( - tan ( arcsin ( \frac{4}{5} ) ) ) ^{2} - 1}

利用以下特性:

arcsin (- x) = - arcsin (x)

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

= - \frac{2 tan ( - arcsin ( \frac{4}{5} ) )}{( - tan ( arcsin ( \frac{4}{5} ) ) ) ^{2} - 1}

利用以下特性:

tan (- x) = - tan (x)

tan (- arcsin (\frac{4}{5})) = - tan (arcsin (\frac{4}{5}))

= - \frac{2 ( - tan ( arcsin ( \frac{4}{5} ) ) )}{( - tan ( arcsin ( \frac{4}{5} ) ) ) ^{2} - 1}

化简

= \frac{2 tan ( arcsin ( \frac{4}{5} ) )}{tan ^{2} ( arcsin ( \frac{4}{5} ) ) - 1}

= \frac{2 tan ( arcsin ( \frac{4}{5} ) )}{tan ^{2} ( arcsin ( \frac{4}{5} ) ) - 1}

使用三角恒等式改写:

tan (arcsin (\frac{4}{5})) = \frac{4}{3}

tan (arcsin (\frac{4}{5}))

使用三角恒等式改写:

tan (arcsin (\frac{4}{5})) = \frac{( \frac{4}{5} ) 1 - ( \frac{4}{5} ) ^{2}}{1 - ( \frac{4}{5} ) ^{2}}

利用以下特性:

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

= \frac{( \frac{4}{5} ) 1 - ( \frac{4}{5} ) ^{2}}{1 - ( \frac{4}{5} ) ^{2}}

= \frac{\frac{4}{5} 1 - ( \frac{4}{5} ) ^{2}}{1 - ( \frac{4}{5} ) ^{2}}

化简

= \frac{4}{3}

= \frac{2 \cdot \frac{4}{3}}{( \frac{4}{3} ) ^{2} - 1}

化简

\frac{2 \cdot \frac{4}{3}}{( \frac{4}{3} ) ^{2} - 1} : \frac{24}{7}

\frac{2 \cdot \frac{4}{3}}{( \frac{4}{3} ) ^{2} - 1}

使用指数法则:

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

= \frac{2 \cdot \frac{4}{3}}{\frac{4 ^{2}}{3 ^{2}} - 1}

乘

2 \cdot \frac{4}{3} : \frac{8}{3}

2 \cdot \frac{4}{3}

分式相乘:

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

= \frac{4 \cdot 2}{3}

数字相乘:

4 \cdot 2 = 8

= \frac{8}{3}

= \frac{\frac{8}{3}}{\frac{4 ^{2}}{3 ^{2}} - 1}

\frac{4 ^{2}}{3 ^{2}} = \frac{16}{9}

\frac{4 ^{2}}{3 ^{2}}

4^{2} = 16

= \frac{16}{3 ^{2}}

3^{2} = 9

= \frac{16}{9}

= \frac{\frac{8}{3}}{\frac{16}{9} - 1}

使用分式法则:

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

= \frac{8}{3 ( \frac{16}{9} - 1 )}

化简

\frac{16}{9} - 1 : \frac{7}{9}

\frac{16}{9} - 1

将项转换为分式:

1 = \frac{1 \cdot 9}{9}

= \frac{16}{9} - \frac{1 \cdot 9}{9}

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{16 - 1 \cdot 9}{9}

16 - 1 \cdot 9 = 7

16 - 1 \cdot 9

数字相乘:

1 \cdot 9 = 9

= 16 - 9

数字相减:

16 - 9 = 7

= 7

= \frac{7}{9}

= \frac{8}{3 \cdot \frac{7}{9}}

乘

3 \cdot \frac{7}{9} : \frac{7}{3}

3 \cdot \frac{7}{9}

分式相乘:

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

= \frac{7 \cdot 3}{9}

数字相乘:

7 \cdot 3 = 21

= \frac{21}{9}

约分:

3

= \frac{7}{3}

= \frac{8}{\frac{7}{3}}

使用分式法则:

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

= \frac{8 \cdot 3}{7}

数字相乘:

8 \cdot 3 = 24

= \frac{24}{7}

= \frac{24}{7}

流行的例子

cos (84 0^{\circ})

tan (- \frac{4}{3})

sin^{2} (27 0^{\circ})

cot (- 2)

sec (\frac{9 π}{8})