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受欢迎的三角函数 >

sin(x-4)=cos(9x+4)

解答

sin (x - 4) = cos (9 x + 4)

解答

x = \frac{4 πn + π}{20}, x = - \frac{4 πn + 16 + π}{16}

+1

度数

x = 9^{\circ} + 3 6^{\circ} n, x = - 68.54577 \dots^{\circ} - 4 5^{\circ} n

求解步骤

sin (x - 4) = cos (9 x + 4)

使用三角恒等式改写

sin (x - 4) = cos (9 x + 4)

利用以下特性:

cos (x) = sin (\frac{π}{2} - x)

sin (x - 4) = sin (\frac{π}{2} - (9 x + 4))

sin (x - 4) = sin (\frac{π}{2} - (9 x + 4))

使用反三角函数性质

sin (x - 4) = sin (\frac{π}{2} - (9 x + 4))

sin (x) = sin (y) \Rightarrow x = y + 2 π n, x = π - y + 2 π n

x - 4 = \frac{π}{2} - (9 x + 4) + 2 πn, x - 4 = π - (\frac{π}{2} - (9 x + 4)) + 2 πn

x - 4 = \frac{π}{2} - (9 x + 4) + 2 πn, x - 4 = π - (\frac{π}{2} - (9 x + 4)) + 2 πn

x - 4 = \frac{π}{2} - (9 x + 4) + 2 πn : x = \frac{4 πn + π}{20}

x - 4 = \frac{π}{2} - (9 x + 4) + 2 πn

展开

\frac{π}{2} - (9 x + 4) + 2 πn : \frac{π}{2} - 9 x - 4 + 2 πn

\frac{π}{2} - (9 x + 4) + 2 πn

- (9 x + 4) : - 9 x - 4

- (9 x + 4)

打开括号

= - (9 x) - (4)

使用加减运算法则

+ (- a) = - a

= - 9 x - 4

= \frac{π}{2} - 9 x - 4 + 2 πn

x - 4 = \frac{π}{2} - 9 x - 4 + 2 πn

将

4

到右边

x - 4 = \frac{π}{2} - 9 x - 4 + 2 πn

两边加上

4

x - 4 + 4 = \frac{π}{2} - 9 x - 4 + 2 πn + 4

化简

x - 4 + 4 = \frac{π}{2} - 9 x - 4 + 2 πn + 4

化简

x - 4 + 4 : x

x - 4 + 4

同类项相加:

- 4 + 4 = 0

= x

化简

\frac{π}{2} - 9 x - 4 + 2 πn + 4 : - 9 x + 2 πn + \frac{π}{2}

\frac{π}{2} - 9 x - 4 + 2 πn + 4

对同类项分组

= - 9 x + 2 πn + \frac{π}{2} - 4 + 4

- 4 + 4 = 0

= - 9 x + 2 πn + \frac{π}{2}

x = - 9 x + 2 πn + \frac{π}{2}

x = - 9 x + 2 πn + \frac{π}{2}

x = - 9 x + 2 πn + \frac{π}{2}

将

9 x

para o lado esquerdo

x = - 9 x + 2 πn + \frac{π}{2}

两边加上

9 x

x + 9 x = - 9 x + 2 πn + \frac{π}{2} + 9 x

化简

10 x = 2 πn + \frac{π}{2}

10 x = 2 πn + \frac{π}{2}

两边除以

10

10 x = 2 πn + \frac{π}{2}

两边除以

10

\frac{10 x}{10} = \frac{2 πn}{10} + \frac{\frac{π}{2}}{10}

化简

\frac{10 x}{10} = \frac{2 πn}{10} + \frac{\frac{π}{2}}{10}

化简

\frac{10 x}{10} : x

\frac{10 x}{10}

数字相除:

\frac{10}{10} = 1

= x

化简

\frac{2 πn}{10} + \frac{\frac{π}{2}}{10} : \frac{4 πn + π}{20}

\frac{2 πn}{10} + \frac{\frac{π}{2}}{10}

使用法则

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

= \frac{2 πn + \frac{π}{2}}{10}

化简

2 πn + \frac{π}{2} : \frac{4 πn + π}{2}

2 πn + \frac{π}{2}

将项转换为分式:

2 πn = \frac{2 πn 2}{2}

= \frac{2 πn \cdot 2}{2} + \frac{π}{2}

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{2 πn \cdot 2 + π}{2}

数字相乘:

2 \cdot 2 = 4

= \frac{4 πn + π}{2}

= \frac{\frac{4 πn + π}{2}}{10}

使用分式法则:

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

= \frac{4 πn + π}{2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{4 πn + π}{20}

x = \frac{4 πn + π}{20}

x = \frac{4 πn + π}{20}

x = \frac{4 πn + π}{20}

x - 4 = π - (\frac{π}{2} - (9 x + 4)) + 2 πn : x = - \frac{4 πn + 16 + π}{16}

x - 4 = π - (\frac{π}{2} - (9 x + 4)) + 2 πn

展开

π - (\frac{π}{2} - (9 x + 4)) + 2 πn : π - \frac{π}{2} + 9 x + 4 + 2 πn

π - (\frac{π}{2} - (9 x + 4)) + 2 πn

- (9 x + 4) : - 9 x - 4

- (9 x + 4)

打开括号

= - (9 x) - (4)

使用加减运算法则

+ (- a) = - a

= - 9 x - 4

= π - (- 9 x + \frac{π}{2} - 4) + 2 πn

- (\frac{π}{2} - 9 x - 4) : - \frac{π}{2} + 9 x + 4

- (\frac{π}{2} - 9 x - 4)

打开括号

= - (\frac{π}{2}) - (- 9 x) - (- 4)

使用加减运算法则

- (- a) = a, - (a) = - a

= - \frac{π}{2} + 9 x + 4

= π - \frac{π}{2} + 9 x + 4 + 2 πn

x - 4 = π - \frac{π}{2} + 9 x + 4 + 2 πn

将

4

到右边

x - 4 = π - \frac{π}{2} + 9 x + 4 + 2 πn

两边加上

4

x - 4 + 4 = π - \frac{π}{2} + 9 x + 4 + 2 πn + 4

化简

x - 4 + 4 = π - \frac{π}{2} + 9 x + 4 + 2 πn + 4

化简

x - 4 + 4 : x

x - 4 + 4

同类项相加:

- 4 + 4 = 0

= x

化简

π - \frac{π}{2} + 9 x + 4 + 2 πn + 4 : 9 x + 2 πn + 8 + π - \frac{π}{2}

π - \frac{π}{2} + 9 x + 4 + 2 πn + 4

对同类项分组

= 9 x + π + 2 πn - \frac{π}{2} + 4 + 4

数字相加:

4 + 4 = 8

= 9 x + 2 πn + 8 + π - \frac{π}{2}

x = 9 x + 2 πn + 8 + π - \frac{π}{2}

x = 9 x + 2 πn + 8 + π - \frac{π}{2}

x = 9 x + 2 πn + 8 + π - \frac{π}{2}

将

9 x

para o lado esquerdo

x = 9 x + 2 πn + 8 + π - \frac{π}{2}

两边减去

9 x

x - 9 x = 9 x + 2 πn + 8 + π - \frac{π}{2} - 9 x

化简

- 8 x = 2 πn + 8 + π - \frac{π}{2}

- 8 x = 2 πn + 8 + π - \frac{π}{2}

两边除以

- 8

- 8 x = 2 πn + 8 + π - \frac{π}{2}

两边除以

- 8

\frac{- 8 x}{- 8} = \frac{2 πn}{- 8} + \frac{8}{- 8} + \frac{π}{- 8} - \frac{\frac{π}{2}}{- 8}

化简

\frac{- 8 x}{- 8} = \frac{2 πn}{- 8} + \frac{8}{- 8} + \frac{π}{- 8} - \frac{\frac{π}{2}}{- 8}

化简

\frac{- 8 x}{- 8} : x

\frac{- 8 x}{- 8}

使用分式法则:

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

= \frac{8 x}{8}

数字相除:

\frac{8}{8} = 1

= x

化简

\frac{2 πn}{- 8} + \frac{8}{- 8} + \frac{π}{- 8} - \frac{\frac{π}{2}}{- 8} : - \frac{4 πn + 16 + π}{16}

\frac{2 πn}{- 8} + \frac{8}{- 8} + \frac{π}{- 8} - \frac{\frac{π}{2}}{- 8}

使用法则

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

= \frac{2 πn + 8 + π - \frac{π}{2}}{- 8}

使用分式法则:

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

= - \frac{2 πn + 8 + π - \frac{π}{2}}{8}

化简

2 πn + 8 + π - \frac{π}{2} : \frac{4 πn + 16 + π}{2}

2 πn + 8 + π - \frac{π}{2}

将项转换为分式:

2 πn = \frac{2 πn 2}{2}, 8 = \frac{8 \cdot 2}{2}, π = \frac{π 2}{2}

= \frac{2 πn \cdot 2}{2} + \frac{8 \cdot 2}{2} + \frac{π 2}{2} - \frac{π}{2}

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{2 πn \cdot 2 + 8 \cdot 2 + π 2 - π}{2}

2 πn \cdot 2 + 8 \cdot 2 + π 2 - π = 4 πn + 16 + π

2 πn \cdot 2 + 8 \cdot 2 + π 2 - π

同类项相加:

2 π - π = π

= 2 \cdot 2 πn + 8 \cdot 2 + π

数字相乘:

2 \cdot 2 = 4

= 4 πn + 8 \cdot 2 + π

数字相乘:

8 \cdot 2 = 16

= 4 πn + 16 + π

= \frac{4 πn + 16 + π}{2}

= - \frac{\frac{4 πn + π + 16}{2}}{8}

化简

\frac{\frac{4 πn + 16 + π}{2}}{8} : \frac{4 πn + 16 + π}{16}

\frac{\frac{4 πn + 16 + π}{2}}{8}

使用分式法则:

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

= \frac{4 πn + 16 + π}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{4 πn + 16 + π}{16}

= - \frac{4 πn + π + 16}{16}

= - \frac{4 πn + 16 + π}{16}

x = - \frac{4 πn + 16 + π}{16}

x = - \frac{4 πn + 16 + π}{16}

x = - \frac{4 πn + 16 + π}{16}

x = \frac{4 πn + π}{20}, x = - \frac{4 πn + 16 + π}{16}

x = \frac{4 πn + π}{20}, x = - \frac{4 πn + 16 + π}{16}

作图

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