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受欢迎的三角函数 >

cos(2arctan(-4/3))

解答

cos (2 arctan (- \frac{4}{3}))

解答

- \frac{7}{25}

+1

十进制

- 0.28

求解步骤

cos (2 arctan (- \frac{4}{3}))

使用三角恒等式改写:

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

cos (2 arctan (- \frac{4}{3}))

使用倍角公式:

cos (2 x) = 1 - 2 sin^{2} (x)

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

利用以下特性:

arctan (- x) = - arctan (x)

arctan (- \frac{4}{3}) = - arctan (\frac{4}{3})

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

利用以下特性:

sin (- x) = - sin (x)

sin (- arctan (\frac{4}{3})) = - sin (arctan (\frac{4}{3}))

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

化简

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

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

使用三角恒等式改写:

sin (arctan (\frac{4}{3})) = \frac{4}{5}

sin (arctan (\frac{4}{3}))

使用三角恒等式改写:

sin (arctan (\frac{4}{3})) = \frac{( \frac{4}{3} ) 1 + ( \frac{4}{3} ) ^{2}}{1 + ( \frac{4}{3} ) ^{2}}

利用以下特性:

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

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

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

化简

= \frac{4}{5}

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

化简

1 - 2 (\frac{4}{5})^{2} : - \frac{7}{25}

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

2 (\frac{4}{5})^{2} = \frac{32}{25}

2 (\frac{4}{5})^{2}

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

(\frac{4}{5})^{2}

使用指数法则:

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

= \frac{4 ^{2}}{5 ^{2}}

= 2 \cdot \frac{4 ^{2}}{5 ^{2}}

分式相乘:

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

= \frac{4 ^{2} \cdot 2}{5 ^{2}}

4^{2} \cdot 2 = 32

4^{2} \cdot 2

4^{2} = 16

= 16 \cdot 2

数字相乘:

16 \cdot 2 = 32

= 32

= \frac{32}{5 ^{2}}

5^{2} = 25

= \frac{32}{25}

= 1 - \frac{32}{25}

将项转换为分式:

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

= \frac{1 \cdot 25}{25} - \frac{32}{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{1 \cdot 25 - 32}{25}

1 \cdot 25 - 32 = - 7

1 \cdot 25 - 32

数字相乘:

1 \cdot 25 = 25

= 25 - 32

数字相减:

25 - 32 = - 7

= - 7

= \frac{- 7}{25}

使用分式法则:

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

= - \frac{7}{25}

= - \frac{7}{25}

流行的例子

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