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受欢迎的三角函数 >

sin(4k-22)=cos(6k-13)

解答

sin (4 k - 22) = cos (6 k - 13)

解答

k = \frac{4 πn + 70 + π}{20}, k = - \frac{4 πn + 18 + π}{4}

+1

度数

k = 209.53522 \dots^{\circ} + 3 6^{\circ} n, k = - 302.83100 \dots^{\circ} - 18 0^{\circ} n

求解步骤

sin (4 k - 22) = cos (6 k - 13)

使用三角恒等式改写

sin (4 k - 22) = cos (6 k - 13)

利用以下特性:

cos (x) = sin (\frac{π}{2} - x)

sin (4 k - 22) = sin (\frac{π}{2} - (6 k - 13))

sin (4 k - 22) = sin (\frac{π}{2} - (6 k - 13))

使用反三角函数性质

sin (4 k - 22) = sin (\frac{π}{2} - (6 k - 13))

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

4 k - 22 = \frac{π}{2} - (6 k - 13) + 2 πn, 4 k - 22 = π - (\frac{π}{2} - (6 k - 13)) + 2 πn

4 k - 22 = \frac{π}{2} - (6 k - 13) + 2 πn, 4 k - 22 = π - (\frac{π}{2} - (6 k - 13)) + 2 πn

4 k - 22 = \frac{π}{2} - (6 k - 13) + 2 πn : k = \frac{4 πn + 70 + π}{20}

4 k - 22 = \frac{π}{2} - (6 k - 13) + 2 πn

展开

\frac{π}{2} - (6 k - 13) + 2 πn : \frac{π}{2} - 6 k + 13 + 2 πn

\frac{π}{2} - (6 k - 13) + 2 πn

- (6 k - 13) : - 6 k + 13

- (6 k - 13)

打开括号

= - (6 k) - (- 13)

使用加减运算法则

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

= - 6 k + 13

= \frac{π}{2} - 6 k + 13 + 2 πn

4 k - 22 = \frac{π}{2} - 6 k + 13 + 2 πn

将

22

到右边

4 k - 22 = \frac{π}{2} - 6 k + 13 + 2 πn

两边加上

22

4 k - 22 + 22 = \frac{π}{2} - 6 k + 13 + 2 πn + 22

化简

4 k - 22 + 22 = \frac{π}{2} - 6 k + 13 + 2 πn + 22

化简

4 k - 22 + 22 : 4 k

4 k - 22 + 22

同类项相加:

- 22 + 22 = 0

= 4 k

化简

\frac{π}{2} - 6 k + 13 + 2 πn + 22 : - 6 k + 2 πn + 35 + \frac{π}{2}

\frac{π}{2} - 6 k + 13 + 2 πn + 22

对同类项分组

= - 6 k + 2 πn + \frac{π}{2} + 13 + 22

数字相加:

13 + 22 = 35

= - 6 k + 2 πn + 35 + \frac{π}{2}

4 k = - 6 k + 2 πn + 35 + \frac{π}{2}

4 k = - 6 k + 2 πn + 35 + \frac{π}{2}

4 k = - 6 k + 2 πn + 35 + \frac{π}{2}

将

6 k

para o lado esquerdo

4 k = - 6 k + 2 πn + 35 + \frac{π}{2}

两边加上

6 k

4 k + 6 k = - 6 k + 2 πn + 35 + \frac{π}{2} + 6 k

化简

10 k = 2 πn + 35 + \frac{π}{2}

10 k = 2 πn + 35 + \frac{π}{2}

两边除以

10

10 k = 2 πn + 35 + \frac{π}{2}

两边除以

10

\frac{10 k}{10} = \frac{2 πn}{10} + \frac{35}{10} + \frac{\frac{π}{2}}{10}

化简

\frac{10 k}{10} = \frac{2 πn}{10} + \frac{35}{10} + \frac{\frac{π}{2}}{10}

化简

\frac{10 k}{10} : k

\frac{10 k}{10}

数字相除:

\frac{10}{10} = 1

= k

化简

\frac{2 πn}{10} + \frac{35}{10} + \frac{\frac{π}{2}}{10} : \frac{4 πn + 70 + π}{20}

\frac{2 πn}{10} + \frac{35}{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 + 35 + \frac{π}{2}}{10}

化简

2 πn + 35 + \frac{π}{2} : \frac{4 πn + 70 + π}{2}

2 πn + 35 + \frac{π}{2}

将项转换为分式:

2 πn = \frac{2 πn 2}{2}, 35 = \frac{35 \cdot 2}{2}

= \frac{2 πn \cdot 2}{2} + \frac{35 \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 + 35 \cdot 2 + π}{2}

2 πn \cdot 2 + 35 \cdot 2 + π = 4 πn + 70 + π

2 πn \cdot 2 + 35 \cdot 2 + π

数字相乘:

2 \cdot 2 = 4

= 4 πn + 35 \cdot 2 + π

数字相乘:

35 \cdot 2 = 70

= 4 πn + 70 + π

= \frac{4 πn + 70 + π}{2}

= \frac{\frac{4 πn + 70 + π}{2}}{10}

使用分式法则:

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

= \frac{4 πn + 70 + π}{2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{4 πn + 70 + π}{20}

k = \frac{4 πn + 70 + π}{20}

k = \frac{4 πn + 70 + π}{20}

k = \frac{4 πn + 70 + π}{20}

4 k - 22 = π - (\frac{π}{2} - (6 k - 13)) + 2 πn : k = - \frac{4 πn + 18 + π}{4}

4 k - 22 = π - (\frac{π}{2} - (6 k - 13)) + 2 πn

展开

π - (\frac{π}{2} - (6 k - 13)) + 2 πn : π - \frac{π}{2} + 6 k - 13 + 2 πn

π - (\frac{π}{2} - (6 k - 13)) + 2 πn

- (6 k - 13) : - 6 k + 13

- (6 k - 13)

打开括号

= - (6 k) - (- 13)

使用加减运算法则

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

= - 6 k + 13

= π - (- 6 k + 13 + \frac{π}{2}) + 2 πn

- (\frac{π}{2} - 6 k + 13) : - \frac{π}{2} + 6 k - 13

- (\frac{π}{2} - 6 k + 13)

打开括号

= - (\frac{π}{2}) - (- 6 k) - (13)

使用加减运算法则

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

= - \frac{π}{2} + 6 k - 13

= π - \frac{π}{2} + 6 k - 13 + 2 πn

4 k - 22 = π - \frac{π}{2} + 6 k - 13 + 2 πn

将

22

到右边

4 k - 22 = π - \frac{π}{2} + 6 k - 13 + 2 πn

两边加上

22

4 k - 22 + 22 = π - \frac{π}{2} + 6 k - 13 + 2 πn + 22

化简

4 k - 22 + 22 = π - \frac{π}{2} + 6 k - 13 + 2 πn + 22

化简

4 k - 22 + 22 : 4 k

4 k - 22 + 22

同类项相加:

- 22 + 22 = 0

= 4 k

化简

π - \frac{π}{2} + 6 k - 13 + 2 πn + 22 : 6 k + 2 πn + 9 + π - \frac{π}{2}

π - \frac{π}{2} + 6 k - 13 + 2 πn + 22

对同类项分组

= 6 k + π + 2 πn - \frac{π}{2} - 13 + 22

数字相加/相减:

- 13 + 22 = 9

= 6 k + 2 πn + 9 + π - \frac{π}{2}

4 k = 6 k + 2 πn + 9 + π - \frac{π}{2}

4 k = 6 k + 2 πn + 9 + π - \frac{π}{2}

4 k = 6 k + 2 πn + 9 + π - \frac{π}{2}

将

6 k

para o lado esquerdo

4 k = 6 k + 2 πn + 9 + π - \frac{π}{2}

两边减去

6 k

4 k - 6 k = 6 k + 2 πn + 9 + π - \frac{π}{2} - 6 k

化简

- 2 k = 2 πn + 9 + π - \frac{π}{2}

- 2 k = 2 πn + 9 + π - \frac{π}{2}

两边除以

- 2

- 2 k = 2 πn + 9 + π - \frac{π}{2}

两边除以

- 2

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

化简

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

化简

\frac{- 2 k}{- 2} : k

\frac{- 2 k}{- 2}

使用分式法则:

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

= \frac{2 k}{2}

数字相除:

\frac{2}{2} = 1

= k

化简

\frac{2 πn}{- 2} + \frac{9}{- 2} + \frac{π}{- 2} - \frac{\frac{π}{2}}{- 2} : - \frac{4 πn + 18 + π}{4}

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

使用法则

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

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

使用分式法则:

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

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

化简

2 πn + 9 + π - \frac{π}{2} : \frac{4 πn + 18 + π}{2}

2 πn + 9 + π - \frac{π}{2}

将项转换为分式:

2 πn = \frac{2 πn 2}{2}, 9 = \frac{9 \cdot 2}{2}, π = \frac{π 2}{2}

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

2 πn \cdot 2 + 9 \cdot 2 + π 2 - π = 4 πn + 18 + π

2 πn \cdot 2 + 9 \cdot 2 + π 2 - π

同类项相加:

2 π - π = π

= 2 \cdot 2 πn + 9 \cdot 2 + π

数字相乘:

2 \cdot 2 = 4

= 4 πn + 9 \cdot 2 + π

数字相乘:

9 \cdot 2 = 18

= 4 πn + 18 + π

= \frac{4 πn + 18 + π}{2}

= - \frac{\frac{4 πn + π + 18}{2}}{2}

化简

\frac{\frac{4 πn + 18 + π}{2}}{2} : \frac{4 πn + 18 + π}{4}

\frac{\frac{4 πn + 18 + π}{2}}{2}

使用分式法则:

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

= \frac{4 πn + 18 + π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{4 πn + 18 + π}{4}

= - \frac{4 πn + π + 18}{4}

= - \frac{4 πn + 18 + π}{4}

k = - \frac{4 πn + 18 + π}{4}

k = - \frac{4 πn + 18 + π}{4}

k = - \frac{4 πn + 18 + π}{4}

k = \frac{4 πn + 70 + π}{20}, k = - \frac{4 πn + 18 + π}{4}

k = \frac{4 πn + 70 + π}{20}, k = - \frac{4 πn + 18 + π}{4}

作图

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