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受欢迎的三角函数 >

(-2+sqrt(3))/2 =-1+cos(4θ+135)

解答

\frac{- 2 + 3}{2} = - 1 + cos (4 θ + 135)

解答

θ = \frac{π}{24} - \frac{135}{4} + \frac{πn}{2}, θ = \frac{π}{2} - \frac{π}{24} - \frac{135}{4} + \frac{πn}{2}

+1

度数

θ = - 1926.23255 \dots^{\circ} + 9 0^{\circ} n, θ = - 1851.23255 \dots^{\circ} + 9 0^{\circ} n

求解步骤

\frac{- 2 + 3}{2} = - 1 + cos (4 θ + 135)

交换两边

- 1 + cos (4 θ + 135) = \frac{- 2 + 3}{2}

将

1

到右边

- 1 + cos (4 θ + 135) = \frac{- 2 + 3}{2}

两边加上

1

- 1 + cos (4 θ + 135) + 1 = \frac{- 2 + 3}{2} + 1

化简

cos (4 θ + 135) = \frac{- 2 + 3}{2} + 1

cos (4 θ + 135) = \frac{- 2 + 3}{2} + 1

使用反三角函数性质

cos (4 θ + 135) = \frac{- 2 + 3}{2} + 1

cos (4 θ + 135) = \frac{- 2 + 3}{2} + 1

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

4 θ + 135 = arccos (\frac{- 2 + 3}{2} + 1) + 2 πn, 4 θ + 135 = 2 π - arccos (\frac{- 2 + 3}{2} + 1) + 2 πn

4 θ + 135 = arccos (\frac{- 2 + 3}{2} + 1) + 2 πn, 4 θ + 135 = 2 π - arccos (\frac{- 2 + 3}{2} + 1) + 2 πn

解

4 θ + 135 = arccos (\frac{- 2 + 3}{2} + 1) + 2 πn : θ = \frac{π}{24} - \frac{135}{4} + \frac{πn}{2}

4 θ + 135 = arccos (\frac{- 2 + 3}{2} + 1) + 2 πn

化简

arccos (\frac{- 2 + 3}{2} + 1) + 2 πn : \frac{π}{6} + 2 πn

arccos (\frac{- 2 + 3}{2} + 1) + 2 πn

使用以下普通恒等式:

arccos (\frac{3 - 2}{2} + 1) = \frac{π}{6}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{6} + 2 πn

4 θ + 135 = \frac{π}{6} + 2 πn

将

135

到右边

4 θ + 135 = \frac{π}{6} + 2 πn

两边减去

135

4 θ + 135 - 135 = \frac{π}{6} + 2 πn - 135

化简

4 θ = \frac{π}{6} + 2 πn - 135

4 θ = \frac{π}{6} + 2 πn - 135

两边除以

4

4 θ = \frac{π}{6} + 2 πn - 135

两边除以

4

\frac{4 θ}{4} = \frac{\frac{π}{6}}{4} + \frac{2 πn}{4} - \frac{135}{4}

化简

\frac{4 θ}{4} = \frac{\frac{π}{6}}{4} + \frac{2 πn}{4} - \frac{135}{4}

化简

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

数字相除:

\frac{4}{4} = 1

= θ

化简

\frac{\frac{π}{6}}{4} + \frac{2 πn}{4} - \frac{135}{4} : \frac{π}{24} - \frac{135}{4} + \frac{πn}{2}

\frac{\frac{π}{6}}{4} + \frac{2 πn}{4} - \frac{135}{4}

对同类项分组

= - \frac{135}{4} + \frac{2 πn}{4} + \frac{\frac{π}{6}}{4}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

\frac{\frac{π}{6}}{4} = \frac{π}{24}

\frac{\frac{π}{6}}{4}

使用分式法则:

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

= \frac{π}{6 \cdot 4}

数字相乘:

6 \cdot 4 = 24

= \frac{π}{24}

= - \frac{135}{4} + \frac{πn}{2} + \frac{π}{24}

对同类项分组

= \frac{π}{24} - \frac{135}{4} + \frac{πn}{2}

θ = \frac{π}{24} - \frac{135}{4} + \frac{πn}{2}

θ = \frac{π}{24} - \frac{135}{4} + \frac{πn}{2}

θ = \frac{π}{24} - \frac{135}{4} + \frac{πn}{2}

解

4 θ + 135 = 2 π - arccos (\frac{- 2 + 3}{2} + 1) + 2 πn : θ = \frac{π}{2} - \frac{π}{24} - \frac{135}{4} + \frac{πn}{2}

4 θ + 135 = 2 π - arccos (\frac{- 2 + 3}{2} + 1) + 2 πn

化简

2 π - arccos (\frac{- 2 + 3}{2} + 1) + 2 πn : 2 π - \frac{π}{6} + 2 πn

2 π - arccos (\frac{- 2 + 3}{2} + 1) + 2 πn

使用以下普通恒等式:

arccos (\frac{3 - 2}{2} + 1) = \frac{π}{6}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{π}{6} + 2 πn

4 θ + 135 = 2 π - \frac{π}{6} + 2 πn

将

135

到右边

4 θ + 135 = 2 π - \frac{π}{6} + 2 πn

两边减去

135

4 θ + 135 - 135 = 2 π - \frac{π}{6} + 2 πn - 135

化简

4 θ = 2 π - \frac{π}{6} + 2 πn - 135

4 θ = 2 π - \frac{π}{6} + 2 πn - 135

两边除以

4

4 θ = 2 π - \frac{π}{6} + 2 πn - 135

两边除以

4

\frac{4 θ}{4} = \frac{2 π}{4} - \frac{\frac{π}{6}}{4} + \frac{2 πn}{4} - \frac{135}{4}

化简

\frac{4 θ}{4} = \frac{2 π}{4} - \frac{\frac{π}{6}}{4} + \frac{2 πn}{4} - \frac{135}{4}

化简

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

数字相除:

\frac{4}{4} = 1

= θ

化简

\frac{2 π}{4} - \frac{\frac{π}{6}}{4} + \frac{2 πn}{4} - \frac{135}{4} : \frac{π}{2} - \frac{π}{24} - \frac{135}{4} + \frac{πn}{2}

\frac{2 π}{4} - \frac{\frac{π}{6}}{4} + \frac{2 πn}{4} - \frac{135}{4}

对同类项分组

= \frac{2 π}{4} - \frac{135}{4} + \frac{2 πn}{4} - \frac{\frac{π}{6}}{4}

\frac{2 π}{4} = \frac{π}{2}

\frac{2 π}{4}

约分:

2

= \frac{π}{2}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

\frac{\frac{π}{6}}{4} = \frac{π}{24}

\frac{\frac{π}{6}}{4}

使用分式法则:

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

= \frac{π}{6 \cdot 4}

数字相乘:

6 \cdot 4 = 24

= \frac{π}{24}

= \frac{π}{2} - \frac{135}{4} + \frac{πn}{2} - \frac{π}{24}

对同类项分组

= \frac{π}{2} - \frac{π}{24} - \frac{135}{4} + \frac{πn}{2}

θ = \frac{π}{2} - \frac{π}{24} - \frac{135}{4} + \frac{πn}{2}

θ = \frac{π}{2} - \frac{π}{24} - \frac{135}{4} + \frac{πn}{2}

θ = \frac{π}{2} - \frac{π}{24} - \frac{135}{4} + \frac{πn}{2}

θ = \frac{π}{24} - \frac{135}{4} + \frac{πn}{2}, θ = \frac{π}{2} - \frac{π}{24} - \frac{135}{4} + \frac{πn}{2}

作图

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