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受欢迎的三角函数 >

sin(2a+10)=cos(3a-20)

解答

sin (2 a + 10) = cos (3 a - 20)

解答

a = \frac{4 πn + 20 + π}{10}, a = - \frac{π + 4 πn - 60}{2}

+1

度数

a = 132.59155 \dots^{\circ} + 7 2^{\circ} n, a = 1628.87338 \dots^{\circ} - 36 0^{\circ} n

求解步骤

sin (2 a + 10) = cos (3 a - 20)

使用三角恒等式改写

sin (2 a + 10) = cos (3 a - 20)

利用以下特性:

cos (x) = sin (\frac{π}{2} - x)

sin (2 a + 10) = sin (\frac{π}{2} - (3 a - 20))

sin (2 a + 10) = sin (\frac{π}{2} - (3 a - 20))

使用反三角函数性质

sin (2 a + 10) = sin (\frac{π}{2} - (3 a - 20))

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

2 a + 10 = \frac{π}{2} - (3 a - 20) + 2 πn, 2 a + 10 = π - (\frac{π}{2} - (3 a - 20)) + 2 πn

2 a + 10 = \frac{π}{2} - (3 a - 20) + 2 πn, 2 a + 10 = π - (\frac{π}{2} - (3 a - 20)) + 2 πn

2 a + 10 = \frac{π}{2} - (3 a - 20) + 2 πn : a = \frac{4 πn + 20 + π}{10}

2 a + 10 = \frac{π}{2} - (3 a - 20) + 2 πn

展开

\frac{π}{2} - (3 a - 20) + 2 πn : \frac{π}{2} - 3 a + 20 + 2 πn

\frac{π}{2} - (3 a - 20) + 2 πn

- (3 a - 20) : - 3 a + 20

- (3 a - 20)

打开括号

= - (3 a) - (- 20)

使用加减运算法则

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

= - 3 a + 20

= \frac{π}{2} - 3 a + 20 + 2 πn

2 a + 10 = \frac{π}{2} - 3 a + 20 + 2 πn

将

10

到右边

2 a + 10 = \frac{π}{2} - 3 a + 20 + 2 πn

两边减去

10

2 a + 10 - 10 = \frac{π}{2} - 3 a + 20 + 2 πn - 10

化简

2 a + 10 - 10 = \frac{π}{2} - 3 a + 20 + 2 πn - 10

化简

2 a + 10 - 10 : 2 a

2 a + 10 - 10

同类项相加:

10 - 10 = 0

= 2 a

化简

\frac{π}{2} - 3 a + 20 + 2 πn - 10 : - 3 a + 2 πn + 10 + \frac{π}{2}

\frac{π}{2} - 3 a + 20 + 2 πn - 10

对同类项分组

= - 3 a + 2 πn + \frac{π}{2} + 20 - 10

数字相加/相减:

20 - 10 = 10

= - 3 a + 2 πn + 10 + \frac{π}{2}

2 a = - 3 a + 2 πn + 10 + \frac{π}{2}

2 a = - 3 a + 2 πn + 10 + \frac{π}{2}

2 a = - 3 a + 2 πn + 10 + \frac{π}{2}

将

3 a

para o lado esquerdo

2 a = - 3 a + 2 πn + 10 + \frac{π}{2}

两边加上

3 a

2 a + 3 a = - 3 a + 2 πn + 10 + \frac{π}{2} + 3 a

化简

5 a = 2 πn + 10 + \frac{π}{2}

5 a = 2 πn + 10 + \frac{π}{2}

两边除以

5

5 a = 2 πn + 10 + \frac{π}{2}

两边除以

5

\frac{5 a}{5} = \frac{2 πn}{5} + \frac{10}{5} + \frac{\frac{π}{2}}{5}

化简

\frac{5 a}{5} = \frac{2 πn}{5} + \frac{10}{5} + \frac{\frac{π}{2}}{5}

化简

\frac{5 a}{5} : a

\frac{5 a}{5}

数字相除:

\frac{5}{5} = 1

= a

化简

\frac{2 πn}{5} + \frac{10}{5} + \frac{\frac{π}{2}}{5} : \frac{4 πn + 20 + π}{10}

\frac{2 πn}{5} + \frac{10}{5} + \frac{\frac{π}{2}}{5}

使用法则

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

= \frac{2 πn + 10 + \frac{π}{2}}{5}

化简

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

2 πn + 10 + \frac{π}{2}

将项转换为分式:

2 πn = \frac{2 πn 2}{2}, 10 = \frac{10 \cdot 2}{2}

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

2 πn \cdot 2 + 10 \cdot 2 + π = 4 πn + 20 + π

2 πn \cdot 2 + 10 \cdot 2 + π

数字相乘:

2 \cdot 2 = 4

= 4 πn + 10 \cdot 2 + π

数字相乘:

10 \cdot 2 = 20

= 4 πn + 20 + π

= \frac{4 πn + 20 + π}{2}

= \frac{\frac{4 πn + 20 + π}{2}}{5}

使用分式法则:

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

= \frac{4 πn + 20 + π}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{4 πn + 20 + π}{10}

a = \frac{4 πn + 20 + π}{10}

a = \frac{4 πn + 20 + π}{10}

a = \frac{4 πn + 20 + π}{10}

2 a + 10 = π - (\frac{π}{2} - (3 a - 20)) + 2 πn : a = - \frac{π + 4 πn - 60}{2}

2 a + 10 = π - (\frac{π}{2} - (3 a - 20)) + 2 πn

展开

π - (\frac{π}{2} - (3 a - 20)) + 2 πn : π - \frac{π}{2} + 3 a - 20 + 2 πn

π - (\frac{π}{2} - (3 a - 20)) + 2 πn

- (3 a - 20) : - 3 a + 20

- (3 a - 20)

打开括号

= - (3 a) - (- 20)

使用加减运算法则

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

= - 3 a + 20

= π - (- 3 a + 20 + \frac{π}{2}) + 2 πn

- (\frac{π}{2} - 3 a + 20) : - \frac{π}{2} + 3 a - 20

- (\frac{π}{2} - 3 a + 20)

打开括号

= - (\frac{π}{2}) - (- 3 a) - (20)

使用加减运算法则

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

= - \frac{π}{2} + 3 a - 20

= π - \frac{π}{2} + 3 a - 20 + 2 πn

2 a + 10 = π - \frac{π}{2} + 3 a - 20 + 2 πn

将

10

到右边

2 a + 10 = π - \frac{π}{2} + 3 a - 20 + 2 πn

两边减去

10

2 a + 10 - 10 = π - \frac{π}{2} + 3 a - 20 + 2 πn - 10

化简

2 a + 10 - 10 = π - \frac{π}{2} + 3 a - 20 + 2 πn - 10

化简

2 a + 10 - 10 : 2 a

2 a + 10 - 10

同类项相加:

10 - 10 = 0

= 2 a

化简

π - \frac{π}{2} + 3 a - 20 + 2 πn - 10 : 3 a + 2 πn + π - 30 - \frac{π}{2}

π - \frac{π}{2} + 3 a - 20 + 2 πn - 10

对同类项分组

= 3 a + π + 2 πn - \frac{π}{2} - 20 - 10

数字相减:

- 20 - 10 = - 30

= 3 a + 2 πn + π - 30 - \frac{π}{2}

2 a = 3 a + 2 πn + π - 30 - \frac{π}{2}

2 a = 3 a + 2 πn + π - 30 - \frac{π}{2}

2 a = 3 a + 2 πn + π - 30 - \frac{π}{2}

将

3 a

para o lado esquerdo

2 a = 3 a + 2 πn + π - 30 - \frac{π}{2}

两边减去

3 a

2 a - 3 a = 3 a + 2 πn + π - 30 - \frac{π}{2} - 3 a

化简

- a = 2 πn + π - 30 - \frac{π}{2}

- a = 2 πn + π - 30 - \frac{π}{2}

两边除以

- 1

- a = 2 πn + π - 30 - \frac{π}{2}

两边除以

- 1

\frac{- a}{- 1} = \frac{2 πn}{- 1} + \frac{π}{- 1} - \frac{30}{- 1} - \frac{\frac{π}{2}}{- 1}

化简

\frac{- a}{- 1} = \frac{2 πn}{- 1} + \frac{π}{- 1} - \frac{30}{- 1} - \frac{\frac{π}{2}}{- 1}

化简

\frac{- a}{- 1} : a

\frac{- a}{- 1}

使用分式法则:

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

= \frac{a}{1}

使用法则

\frac{a}{1} = a

= a

化简

\frac{2 πn}{- 1} + \frac{π}{- 1} - \frac{30}{- 1} - \frac{\frac{π}{2}}{- 1} : - \frac{π + 4 πn - 60}{2}

\frac{2 πn}{- 1} + \frac{π}{- 1} - \frac{30}{- 1} - \frac{\frac{π}{2}}{- 1}

使用法则

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

= \frac{2 πn + π - 30 - \frac{π}{2}}{- 1}

使用分式法则:

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

= - \frac{2 πn + π - 30 - \frac{π}{2}}{1}

化简

2 πn + π - 30 - \frac{π}{2} : \frac{π + 4 πn - 60}{2}

2 πn + π - 30 - \frac{π}{2}

将项转换为分式:

2 πn = \frac{2 πn 2}{2}, π = \frac{π 2}{2}, 30 = \frac{30 \cdot 2}{2}

= \frac{2 πn \cdot 2}{2} + \frac{π 2}{2} - \frac{30 \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 - 30 \cdot 2 - π}{2}

2 πn \cdot 2 + π 2 - 30 \cdot 2 - π = π + 4 πn - 60

2 πn \cdot 2 + π 2 - 30 \cdot 2 - π

对同类项分组

= 2 π - π + 2 \cdot 2 πn - 30 \cdot 2

同类项相加:

2 π - π = π

= π + 2 \cdot 2 πn - 30 \cdot 2

数字相乘:

2 \cdot 2 = 4

= π + 4 πn - 30 \cdot 2

数字相乘:

30 \cdot 2 = 60

= π + 4 πn - 60

= \frac{π + 4 πn - 60}{2}

= - \frac{\frac{π + 4 πn - 60}{2}}{1}

使用分式法则:

\frac{a}{1} = a

= - \frac{π + 4 πn - 60}{2}

a = - \frac{π + 4 πn - 60}{2}

a = - \frac{π + 4 πn - 60}{2}

a = - \frac{π + 4 πn - 60}{2}

a = \frac{4 πn + 20 + π}{10}, a = - \frac{π + 4 πn - 60}{2}

a = \frac{4 πn + 20 + π}{10}, a = - \frac{π + 4 πn - 60}{2}

作图

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