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受欢迎的三角函数 >

cos(θ)-sin(θ)=sqrt(2)

解答

cos (θ) - sin (θ) = 2

解答

θ = \frac{7 π}{4} + 2 πn

+1

度数

θ = 31 5^{\circ} + 36 0^{\circ} n

求解步骤

cos (θ) - sin (θ) = 2

两边减去

2

cos (θ) - sin (θ) - 2 = 0

使用三角恒等式改写

cos (θ) - sin (θ) - 2

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

= cos (θ) - cos (\frac{π}{2} - θ) - 2

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

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

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

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

\frac{θ + \frac{π}{2} - θ}{2}

θ + \frac{π}{2} - θ = \frac{π}{2}

θ + \frac{π}{2} - θ

对同类项分组

= θ - θ + \frac{π}{2}

同类项相加:

θ - θ = 0

= \frac{π}{2}

= \frac{\frac{π}{2}}{2}

使用分式法则:

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

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

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

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

\frac{θ - ( \frac{π}{2} - θ )}{2}

乘开

θ - (\frac{π}{2} - θ) : 2 θ - \frac{π}{2}

θ - (\frac{π}{2} - θ)

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

- (\frac{π}{2} - θ)

打开括号

= - (\frac{π}{2}) - (- θ)

使用加减运算法则

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

= - \frac{π}{2} + θ

= θ - \frac{π}{2} + θ

化简

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

θ - \frac{π}{2} + θ

对同类项分组

= θ + θ - \frac{π}{2}

同类项相加:

θ + θ = 2 θ

= 2 θ - \frac{π}{2}

= 2 θ - \frac{π}{2}

= \frac{2 θ - \frac{π}{2}}{2}

化简

2 θ - \frac{π}{2} : \frac{4 θ - π}{2}

2 θ - \frac{π}{2}

将项转换为分式:

2 θ = \frac{2 θ 2}{2}

= \frac{2 θ \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 θ \cdot 2 - π}{2}

数字相乘:

2 \cdot 2 = 4

= \frac{4 θ - π}{2}

= \frac{\frac{4 θ - π}{2}}{2}

使用分式法则:

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

= \frac{4 θ - π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{4 θ - π}{4}

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

化简

sin (\frac{π}{4}) : \frac{2}{2}

sin (\frac{π}{4})

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{2}{2}

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}{2}

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

分式相乘:

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

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

约分:

2

= 2 sin (\frac{4 θ - π}{4})

= - 2 - 2 sin (\frac{4 θ - π}{4})

- 2 - 2 sin (\frac{4 θ - π}{4}) = 0

将

2

到右边

- 2 - 2 sin (\frac{4 θ - π}{4}) = 0

两边加上

2

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

化简

- 2 sin (\frac{4 θ - π}{4}) = 2

- 2 sin (\frac{4 θ - π}{4}) = 2

两边除以

- 2

- 2 sin (\frac{4 θ - π}{4}) = 2

两边除以

- 2

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

化简

sin (\frac{4 θ - π}{4}) = - 1

sin (\frac{4 θ - π}{4}) = - 1

sin (\frac{4 θ - π}{4}) = - 1

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{4 θ - π}{4} = \frac{3 π}{2} + 2 πn

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

解

\frac{4 θ - π}{4} = \frac{3 π}{2} + 2 πn : θ = \frac{7 π}{4} + 2 πn

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

在两边乘以

4

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

在两边乘以

4

\frac{4 ( 4 θ - π )}{4} = 4 \cdot \frac{3 π}{2} + 4 \cdot 2 πn

化简

\frac{4 ( 4 θ - π )}{4} = 4 \cdot \frac{3 π}{2} + 4 \cdot 2 πn

化简

\frac{4 ( 4 θ - π )}{4} : 4 θ - π

\frac{4 ( 4 θ - π )}{4}

数字相除:

\frac{4}{4} = 1

= 4 θ - π

化简

4 \cdot \frac{3 π}{2} + 4 \cdot 2 πn : 6 π + 8 πn

4 \cdot \frac{3 π}{2} + 4 \cdot 2 πn

4 \cdot \frac{3 π}{2} = 6 π

4 \cdot \frac{3 π}{2}

分式相乘:

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

= \frac{3 π 4}{2}

数字相乘:

3 \cdot 4 = 12

= \frac{12 π}{2}

数字相除:

\frac{12}{2} = 6

= 6 π

4 \cdot 2 πn = 8 πn

4 \cdot 2 πn

数字相乘:

4 \cdot 2 = 8

= 8 πn

= 6 π + 8 πn

4 θ - π = 6 π + 8 πn

4 θ - π = 6 π + 8 πn

4 θ - π = 6 π + 8 πn

将

π

到右边

4 θ - π = 6 π + 8 πn

两边加上

π

4 θ - π + π = 6 π + 8 πn + π

化简

4 θ = 7 π + 8 πn

4 θ = 7 π + 8 πn

两边除以

4

4 θ = 7 π + 8 πn

两边除以

4

\frac{4 θ}{4} = \frac{7 π}{4} + \frac{8 πn}{4}

化简

θ = \frac{7 π}{4} + 2 πn

θ = \frac{7 π}{4} + 2 πn

θ = \frac{7 π}{4} + 2 πn

作图

流行的例子

3sin(2x+30)=tan(2x+30)

3 sin (2 x + 30) = tan (2 x + 30)

16sin(t)cos(t)=4sin(t)

16 sin (t) cos (t) = 4 sin (t)

4cos(x)=4-4cos(x)

4 cos (x) = 4 - 4 cos (x)

cos(2x)-cos(6x)-sin(4x)=0,0<= x<= pi

cos (2 x) - cos (6 x) - sin (4 x) = 0, 0 \leq x \leq π

0 = 3 sin (θ)