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受欢迎的三角函数 >

sin(θ+1)=cos(θ)

解答

sin (θ + 1) = cos (θ)

解答

θ = - 2 πn - \frac{1}{2} + \frac{π}{4}, θ = - \frac{1}{2} - \frac{3 π}{4} - 2 πn

+1

度数

θ = 16.35211 \dots^{\circ} - 36 0^{\circ} n, θ = - 163.64788 \dots^{\circ} - 36 0^{\circ} n

求解步骤

sin (θ + 1) = cos (θ)

两边减去

cos (θ)

sin (θ + 1) - cos (θ) = 0

使用三角恒等式改写

- cos (θ) + sin (1 + θ)

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

= - cos (θ) + cos (\frac{π}{2} - (1 + θ))

- (1 + θ) : - 1 - θ

- (1 + θ)

打开括号

= - (1) - (θ)

使用加减运算法则

+ (- a) = - a

= - 1 - θ

= - cos (θ) + cos (\frac{π}{2} - 1 - θ)

使用和差化积恒等式:

cos (s) - cos (t) = - 2 sin (\frac{s + t}{2}) sin (\frac{s - t}{2})

= - 2 sin (\frac{- 1 + \frac{π}{2} - θ + θ}{2}) sin (\frac{- 1 + \frac{π}{2} - θ - θ}{2})

化简

- 2 sin (\frac{- 1 + \frac{π}{2} - θ + θ}{2}) sin (\frac{- 1 + \frac{π}{2} - θ - θ}{2}) : - 2 sin (\frac{- 2 + π}{4}) sin (\frac{- 2 + π - 4 θ}{4})

- 2 sin (\frac{- 1 + \frac{π}{2} - θ + θ}{2}) sin (\frac{- 1 + \frac{π}{2} - θ - θ}{2})

\frac{- 1 + \frac{π}{2} - θ + θ}{2} = \frac{- 2 + π}{4}

\frac{- 1 + \frac{π}{2} - θ + θ}{2}

同类项相加:

- θ + θ = 0

= \frac{- 1 + \frac{π}{2}}{2}

化简

- 1 + \frac{π}{2} : \frac{- 2 + π}{2}

- 1 + \frac{π}{2}

将项转换为分式:

1 = \frac{1 \cdot 2}{2}

= - \frac{1 \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{- 1 \cdot 2 + π}{2}

数字相乘:

1 \cdot 2 = 2

= \frac{- 2 + π}{2}

= \frac{\frac{- 2 + π}{2}}{2}

使用分式法则:

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

= \frac{- 2 + π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 + π}{4}

= - 2 sin (\frac{π - 2}{4}) sin (\frac{- θ - θ + \frac{π}{2} - 1}{2})

\frac{- 1 + \frac{π}{2} - θ - θ}{2} = \frac{- 2 + π - 4 θ}{4}

\frac{- 1 + \frac{π}{2} - θ - θ}{2}

同类项相加:

- θ - θ = - 2 θ

= \frac{- 1 + \frac{π}{2} - 2 θ}{2}

化简

- 1 + \frac{π}{2} - 2 θ : \frac{- 2 + π - 4 θ}{2}

- 1 + \frac{π}{2} - 2 θ

将项转换为分式:

1 = \frac{1 \cdot 2}{2}, 2 θ = \frac{2 θ 2}{2}

= - \frac{1 \cdot 2}{2} + \frac{π}{2} - \frac{2 θ \cdot 2}{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{- 1 \cdot 2 + π - 2 θ \cdot 2}{2}

- 1 \cdot 2 + π - 2 θ \cdot 2 = - 2 + π - 4 θ

- 1 \cdot 2 + π - 2 θ \cdot 2

数字相乘:

1 \cdot 2 = 2

= - 2 + π - 2 \cdot 2 θ

数字相乘:

2 \cdot 2 = 4

= - 2 + π - 4 θ

= \frac{- 2 + π - 4 θ}{2}

= \frac{\frac{- 2 + π - 4 θ}{2}}{2}

使用分式法则:

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

= \frac{- 2 + π - 4 θ}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 + π - 4 θ}{4}

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

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

- 2 sin (\frac{- 2 + π}{4}) sin (\frac{- 2 + π - 4 θ}{4}) = 0

两边除以

- 2 sin (\frac{- 2 + π}{4})

- 2 sin (\frac{- 2 + π}{4}) sin (\frac{- 2 + π - 4 θ}{4}) = 0

两边除以

- 2 sin (\frac{- 2 + π}{4})

\frac{- 2 sin ( \frac{- 2 + π}{4} ) sin ( \frac{- 2 + π - 4 θ}{4} )}{- 2 sin ( \frac{- 2 + π}{4} )} = \frac{0}{- 2 sin ( \frac{- 2 + π}{4} )}

化简

sin (\frac{- 2 + π - 4 θ}{4}) = 0

sin (\frac{- 2 + π - 4 θ}{4}) = 0

sin (\frac{- 2 + π - 4 θ}{4}) = 0

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

\frac{- 2 + π - 4 θ}{4} = 0 + 2 πn, \frac{- 2 + π - 4 θ}{4} = π + 2 πn

\frac{- 2 + π - 4 θ}{4} = 0 + 2 πn, \frac{- 2 + π - 4 θ}{4} = π + 2 πn

解

\frac{- 2 + π - 4 θ}{4} = 0 + 2 πn : θ = - 2 πn - \frac{1}{2} + \frac{π}{4}

\frac{- 2 + π - 4 θ}{4} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{- 2 + π - 4 θ}{4} = 2 πn

在两边乘以

4

\frac{- 2 + π - 4 θ}{4} = 2 πn

在两边乘以

4

\frac{4 ( - 2 + π - 4 θ )}{4} = 4 \cdot 2 πn

化简

- 2 + π - 4 θ = 8 πn

- 2 + π - 4 θ = 8 πn

将

2

到右边

- 2 + π - 4 θ = 8 πn

两边加上

2

- 2 + π - 4 θ + 2 = 8 πn + 2

化简

π - 4 θ = 8 πn + 2

π - 4 θ = 8 πn + 2

将

π

到右边

π - 4 θ = 8 πn + 2

两边减去

π

π - 4 θ - π = 8 πn + 2 - π

化简

- 4 θ = 8 πn + 2 - π

- 4 θ = 8 πn + 2 - π

两边除以

- 4

- 4 θ = 8 πn + 2 - π

两边除以

- 4

\frac{- 4 θ}{- 4} = \frac{8 πn}{- 4} + \frac{2}{- 4} - \frac{π}{- 4}

化简

\frac{- 4 θ}{- 4} = \frac{8 πn}{- 4} + \frac{2}{- 4} - \frac{π}{- 4}

化简

\frac{- 4 θ}{- 4} : θ

\frac{- 4 θ}{- 4}

使用分式法则:

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

= \frac{4 θ}{4}

数字相除:

\frac{4}{4} = 1

= θ

化简

\frac{8 πn}{- 4} + \frac{2}{- 4} - \frac{π}{- 4} : - 2 πn - \frac{1}{2} + \frac{π}{4}

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

\frac{8 πn}{- 4} = - 2 πn

\frac{8 πn}{- 4}

使用分式法则:

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

= - \frac{8 πn}{4}

数字相除:

\frac{8}{4} = 2

= - 2 πn

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

\frac{2}{- 4} = - \frac{1}{2}

\frac{2}{- 4}

使用分式法则:

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

= - \frac{2}{4}

约分:

2

= - \frac{1}{2}

= - 2 πn - \frac{1}{2} - \frac{π}{- 4}

整理后得

= - 2 πn - \frac{1}{2} + \frac{π}{4}

θ = - 2 πn - \frac{1}{2} + \frac{π}{4}

θ = - 2 πn - \frac{1}{2} + \frac{π}{4}

θ = - 2 πn - \frac{1}{2} + \frac{π}{4}

解

\frac{- 2 + π - 4 θ}{4} = π + 2 πn : θ = - \frac{1}{2} - \frac{3 π}{4} - 2 πn

\frac{- 2 + π - 4 θ}{4} = π + 2 πn

在两边乘以

4

\frac{- 2 + π - 4 θ}{4} = π + 2 πn

在两边乘以

4

\frac{4 ( - 2 + π - 4 θ )}{4} = 4 π + 4 \cdot 2 πn

化简

- 2 + π - 4 θ = 4 π + 8 πn

- 2 + π - 4 θ = 4 π + 8 πn

将

2

到右边

- 2 + π - 4 θ = 4 π + 8 πn

两边加上

2

- 2 + π - 4 θ + 2 = 4 π + 8 πn + 2

化简

π - 4 θ = 4 π + 8 πn + 2

π - 4 θ = 4 π + 8 πn + 2

将

π

到右边

π - 4 θ = 4 π + 8 πn + 2

两边减去

π

π - 4 θ - π = 4 π + 8 πn + 2 - π

化简

- 4 θ = 3 π + 8 πn + 2

- 4 θ = 3 π + 8 πn + 2

两边除以

- 4

- 4 θ = 3 π + 8 πn + 2

两边除以

- 4

\frac{- 4 θ}{- 4} = \frac{3 π}{- 4} + \frac{8 πn}{- 4} + \frac{2}{- 4}

化简

\frac{- 4 θ}{- 4} = \frac{3 π}{- 4} + \frac{8 πn}{- 4} + \frac{2}{- 4}

化简

\frac{- 4 θ}{- 4} : θ

\frac{- 4 θ}{- 4}

使用分式法则:

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

= \frac{4 θ}{4}

数字相除:

\frac{4}{4} = 1

= θ

化简

\frac{3 π}{- 4} + \frac{8 πn}{- 4} + \frac{2}{- 4} : - \frac{1}{2} - \frac{3 π}{4} - 2 πn

\frac{3 π}{- 4} + \frac{8 πn}{- 4} + \frac{2}{- 4}

对同类项分组

= \frac{2}{- 4} + \frac{3 π}{- 4} + \frac{8 πn}{- 4}

\frac{2}{- 4} = - \frac{1}{2}

\frac{2}{- 4}

使用分式法则:

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

= - \frac{2}{4}

约分:

2

= - \frac{1}{2}

= - \frac{1}{2} + \frac{3 π}{- 4} + \frac{8 πn}{- 4}

使用分式法则:

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

= - \frac{1}{2} - \frac{3 π}{4} + \frac{8 πn}{- 4}

\frac{8 πn}{- 4} = - 2 πn

\frac{8 πn}{- 4}

使用分式法则:

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

= - \frac{8 πn}{4}

数字相除:

\frac{8}{4} = 2

= - 2 πn

= - \frac{1}{2} - \frac{3 π}{4} - 2 πn

θ = - \frac{1}{2} - \frac{3 π}{4} - 2 πn

θ = - \frac{1}{2} - \frac{3 π}{4} - 2 πn

θ = - \frac{1}{2} - \frac{3 π}{4} - 2 πn

θ = - 2 πn - \frac{1}{2} + \frac{π}{4}, θ = - \frac{1}{2} - \frac{3 π}{4} - 2 πn

作图

流行的例子

0 = sin (x + c)

7sin(2θ)-2sin(θ)=0

7 sin (2 θ) - 2 sin (θ) = 0

cot(2x+pi/3)-sqrt(3)=0,-pi<x<pi

cot (2 x + \frac{π}{3}) - 3 = 0, - π < x < π

cos^{4} (x) = \frac{1}{2}

tan(x)=tan(2x-30)

tan (x) = tan (2 x - 30)