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受欢迎的三角函数 >

8.5cos((2pi)/3 (2.5-1.3))-35.5

解答

8.5 cos (\frac{2 π}{3} (2.5 - 1.3)) - 35.5

解答

\frac{- 17 5 - 301}{8}

+1

十进制

- 42.37664 \dots

求解步骤

8.5 cos (\frac{2 π}{3} (2.5 - 1.3)) - 35.5

= \frac{17}{2} cos (\frac{2 π}{3} (\frac{5}{2} - \frac{13}{10})) - \frac{71}{2}

化简:

\frac{2 π}{3} (\frac{5}{2} - \frac{13}{10}) = \frac{4 π}{5}

\frac{2 π}{3} (\frac{5}{2} - \frac{13}{10})

分式相乘:

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

= \frac{2 π ( \frac{5}{2} - \frac{13}{10} )}{3}

化简

\frac{5}{2} - \frac{13}{10} : \frac{6}{5}

\frac{5}{2} - \frac{13}{10}

2, 10

的最小公倍数:

10

2, 10

最小公倍数 (LCM)

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

10

质因数分解:

2 \cdot 5

10

10

除以

210 = 5 \cdot 2

= 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 5

将每个因子乘以它在

2

或

10

中出现的最多次数

= 2 \cdot 5

数字相乘:

2 \cdot 5 = 10

= 10

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

10

对于

\frac{5}{2} :

将分母和分子乘以

5

\frac{5}{2} = \frac{5 \cdot 5}{2 \cdot 5} = \frac{25}{10}

= \frac{25}{10} - \frac{13}{10}

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{25 - 13}{10}

数字相减:

25 - 13 = 12

= \frac{12}{10}

约分:

2

= \frac{6}{5}

= \frac{2 π \frac{6}{5}}{3}

乘

2 π \frac{6}{5} : \frac{12 π}{5}

2 π \frac{6}{5}

分式相乘:

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

= \frac{6 \cdot 2 π}{5}

数字相乘:

6 \cdot 2 = 12

= \frac{12 π}{5}

= \frac{\frac{12 π}{5}}{3}

使用分式法则:

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

= \frac{12 π}{5 \cdot 3}

数字相乘:

5 \cdot 3 = 15

= \frac{12 π}{15}

约分:

3

= \frac{4 π}{5}

\frac{17}{2} cos (\frac{4 π}{5}) - \frac{71}{2} = \frac{17 cos ( \frac{4 π}{5} ) - 71}{2}

\frac{17}{2} cos (\frac{4 π}{5}) - \frac{71}{2}

乘

\frac{17}{2} cos (\frac{4 π}{5}) : \frac{17 cos ( \frac{4 π}{5} )}{2}

\frac{17}{2} cos (\frac{4 π}{5})

分式相乘:

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

= \frac{17 cos ( \frac{4 π}{5} )}{2}

= \frac{17 cos ( \frac{4 π}{5} )}{2} - \frac{71}{2}

使用法则

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

= \frac{17 cos ( \frac{4 π}{5} ) - 71}{2}

= \frac{17 cos ( \frac{4 π}{5} ) - 71}{2}

使用三角恒等式改写:

cos (\frac{4 π}{5}) = - \frac{5 + 1}{4}

cos (\frac{4 π}{5})

使用三角恒等式改写:

- cos (\frac{π}{5})

cos (\frac{4 π}{5})

使用基本三角恒等式:

cos (x) = - cos (π - x)

= - cos (π - \frac{4 π}{5})

化简:

π - \frac{4 π}{5} = \frac{π}{5}

π - \frac{4 π}{5}

将项转换为分式:

π = \frac{π 5}{5}

= \frac{π 5}{5} - \frac{4 π}{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{π 5 - 4 π}{5}

同类项相加:

5 π - 4 π = π

= \frac{π}{5}

= - cos (\frac{π}{5})

= - cos (\frac{π}{5})

使用三角恒等式改写:

cos (\frac{π}{5}) = \frac{5 + 1}{4}

cos (\frac{π}{5})

显示:

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

使用以下积化和差公式:

2 sin (x) cos (y) = sin (x + y) - sin (x - y)

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

显示:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

两边除以

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

代入

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

\frac{1}{2} = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

sin (\frac{3 π}{10}) = cos (\frac{π}{2} - \frac{3 π}{10})

\frac{1}{2} = cos (\frac{π}{2} - \frac{3 π}{10}) - sin (\frac{π}{10})

\frac{1}{2} = cos (\frac{π}{5}) - sin (\frac{π}{10})

显示:

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

使用因式分解法则:

a^{2} - b^{2} = (a + b) (a - b)

a = cos (\frac{π}{5}) + sin (\frac{π}{10})

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = ((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) ((cos (\frac{π}{5}) + sin (\frac{π}{10})) - (cos (\frac{π}{5}) - sin (\frac{π}{10})))

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 2 (2 cos (\frac{π}{5}) sin (\frac{π}{10}))

显示:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

两边除以

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

代入

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 1

代入

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (\frac{1}{2})^{2} = 1

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} = 1

两边加上

\frac{1}{4}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} = \frac{5}{4}

在两侧开平方

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \pm \frac{5}{4}

cos (\frac{π}{5})

不能为负

sin (\frac{π}{10})

不能为负

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

以下方程式相加

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{2}

((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) = (\frac{5}{2} + \frac{1}{2})

整理后得

cos (\frac{π}{5}) = \frac{5 + 1}{4}

= \frac{5 + 1}{4}

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

= \frac{17 ( - \frac{5 + 1}{4} ) - 71}{2}

化简

\frac{17 ( - \frac{5 + 1}{4} ) - 71}{2} : \frac{- 17 5 - 301}{8}

\frac{17 ( - \frac{5 + 1}{4} ) - 71}{2}

去除括号:

(- a) = - a

= \frac{- 17 \cdot \frac{5 + 1}{4} - 71}{2}

乘

17 \cdot \frac{5 + 1}{4} : \frac{17 ( 1 + 5 )}{4}

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

分式相乘:

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

= \frac{( 5 + 1 ) \cdot 17}{4}

= \frac{- \frac{17 ( 1 + 5 )}{4} - 71}{2}

化简

- \frac{( 5 + 1 ) \cdot 17}{4} - 71 : \frac{- 17 5 - 301}{4}

- \frac{( 5 + 1 ) \cdot 17}{4} - 71

将项转换为分式:

71 = \frac{71 \cdot 4}{4}

= - \frac{( 5 + 1 ) \cdot 17}{4} - \frac{71 \cdot 4}{4}

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{- ( 5 + 1 ) \cdot 17 - 71 \cdot 4}{4}

数字相乘:

71 \cdot 4 = 284

= \frac{- 17 ( 1 + 5 ) - 284}{4}

乘开

- (5 + 1) \cdot 17 - 284 : - 175 - 301

- (5 + 1) \cdot 17 - 284

= - 17 (5 + 1) - 284

乘开

- 17 (5 + 1) : - 175 - 17

- 17 (5 + 1)

使用分配律:

a (b + c) = ab + a c

a = - 17, b = 5, c = 1

= - 175 + (- 17) \cdot 1

使用加减运算法则

+ (- a) = - a

= - 175 - 17 \cdot 1

数字相乘:

17 \cdot 1 = 17

= - 175 - 17

= - 175 - 17 - 284

数字相减:

- 17 - 284 = - 301

= - 175 - 301

= \frac{- 17 5 - 301}{4}

= \frac{\frac{- 17 5 - 301}{4}}{2}

使用分式法则:

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

= \frac{- 17 5 - 301}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{- 17 5 - 301}{8}

= \frac{- 17 5 - 301}{8}

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