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受欢迎的三角函数 >

sin^2(1/2 arccos(-4/5))

解答

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

解答

\frac{9}{10}

+1

十进制

0.9

求解步骤

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

化简:

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

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

分式相乘:

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

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

乘以:

1 \cdot arccos (- \frac{4}{5}) = arccos (- \frac{4}{5})

= \frac{arccos ( - \frac{4}{5} )}{2}

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

使用三角恒等式改写:

\frac{1 - cos ( arccos ( - \frac{4}{5} ) )}{2}

sin^{2} (\frac{arccos ( - \frac{4}{5} )}{2})

利用以下特性:

sin^{2} (θ) = \frac{1 - cos ( 2 θ )}{2}

使用倍角公式

cos (2 θ) = 1 - 2 sin^{2} (θ)

交换两边

2 sin^{2} (θ) - 1 = - cos (2 θ)

两边加上

1

2 sin^{2} (θ) = 1 - cos (2 θ)

两边除以

2

sin^{2} (θ) = \frac{1 - cos ( 2 θ )}{2}

= \frac{1 - cos ( 2 \cdot \frac{a r c c o s ( - \frac{4}{5} )}{2} )}{2}

化简:

2 \cdot \frac{arccos ( - \frac{4}{5} )}{2} = arccos (- \frac{4}{5})

2 \cdot \frac{arccos ( - \frac{4}{5} )}{2}

分式相乘:

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

= \frac{arccos ( - \frac{4}{5} ) \cdot 2}{2}

约分:

2

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

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

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

使用三角恒等式改写:

cos (arccos (- \frac{4}{5})) = - \frac{4}{5}

利用以下特性:

cos (arccos (x)) = x

= - \frac{4}{5}

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

化简

\frac{1 - ( - \frac{4}{5} )}{2} : \frac{9}{10}

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

使用法则

- (- a) = a

= \frac{1 + \frac{4}{5}}{2}

化简

1 + \frac{4}{5} : \frac{9}{5}

1 + \frac{4}{5}

将项转换为分式:

1 = \frac{1 \cdot 5}{5}

= \frac{1 \cdot 5}{5} + \frac{4}{5}

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 5 + 4}{5}

1 \cdot 5 + 4 = 9

1 \cdot 5 + 4

数字相乘:

1 \cdot 5 = 5

= 5 + 4

数字相加:

5 + 4 = 9

= 9

= \frac{9}{5}

= \frac{\frac{9}{5}}{2}

使用分式法则:

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

= \frac{9}{5 \cdot 2}

数字相乘:

5 \cdot 2 = 10

= \frac{9}{10}

= \frac{9}{10}

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