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受欢迎的三角函数 >

solvefor a,90=180-2*arccos((2a)/c)

解答

求解

a, 9 0^{\circ} = 18 0^{\circ} - 2 \cdot arccos (\frac{2 a}{c})

解答

a = \frac{c}{2 2}

求解步骤

9 0^{\circ} = 18 0^{\circ} - 2 arccos (\frac{2 a}{c})

交换两边

π - 2 arccos (\frac{2 a}{c}) = \frac{π}{2}

用替代法求解

π - 2 arccos (\frac{2 a}{c}) = \frac{π}{2}

令

: arccos (\frac{2 a}{c}) = u

18 0^{\circ} - 2 u = 9 0^{\circ}

18 0^{\circ} - 2 u = 9 0^{\circ} : u = 4 5^{\circ}

18 0^{\circ} - 2 u = 9 0^{\circ}

将

18 0^{\circ}

到右边

18 0^{\circ} - 2 u = 9 0^{\circ}

两边减去

18 0^{\circ}

18 0^{\circ} - 2 u - 18 0^{\circ} = 9 0^{\circ} - 18 0^{\circ}

化简

- 2 u = 9 0^{\circ} - 18 0^{\circ}

- 2 u = 9 0^{\circ} - 18 0^{\circ}

两边除以

- 2

- 2 u = 9 0^{\circ} - 18 0^{\circ}

两边除以

- 2

\frac{- 2 u}{- 2} = \frac{9 0 ^{\circ}}{- 2} - \frac{18 0 ^{\circ}}{- 2}

化简

\frac{- 2 u}{- 2} = \frac{9 0 ^{\circ}}{- 2} - \frac{18 0 ^{\circ}}{- 2}

化简

\frac{- 2 u}{- 2} : u

\frac{- 2 u}{- 2}

使用分式法则:

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

= \frac{2 u}{2}

数字相除:

\frac{2}{2} = 1

= u

化简

\frac{9 0 ^{\circ}}{- 2} - \frac{18 0 ^{\circ}}{- 2} : 4 5^{\circ}

\frac{9 0 ^{\circ}}{- 2} - \frac{18 0 ^{\circ}}{- 2}

使用法则

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

= \frac{9 0 ^{\circ} - 18 0 ^{\circ}}{- 2}

使用分式法则:

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

9 0^{\circ} - 18 0^{\circ} = - (18 0^{\circ} - 9 0^{\circ})

= \frac{18 0 ^{\circ} - 9 0 ^{\circ}}{2}

化简

18 0^{\circ} - 9 0^{\circ} : 9 0^{\circ}

18 0^{\circ} - 9 0^{\circ}

将项转换为分式:

18 0^{\circ} = 18 0^{\circ}

= 18 0^{\circ} - 9 0^{\circ}

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{18 0 ^{\circ} 2 - 18 0 ^{\circ}}{2}

同类项相加:

36 0^{\circ} - 18 0^{\circ} = 18 0^{\circ}

= 9 0^{\circ}

= \frac{9 0 ^{\circ}}{2}

使用分式法则:

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

= \frac{18 0 ^{\circ}}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= 4 5^{\circ}

u = 4 5^{\circ}

u = 4 5^{\circ}

u = 4 5^{\circ}

arccos (\frac{2 a}{c}) = \frac{π}{4}

u = arccos (\frac{2 a}{c})

代回

arccos (\frac{2 a}{c}) = \frac{π}{4}

arccos (\frac{2 a}{c}) = \frac{π}{4}

两边减去

9 0^{\circ}

π - 2 arccos (\frac{2 a}{c}) - \frac{π}{2} = 0

化简

18 0^{\circ} - 2 arccos (\frac{2 a}{c}) - 9 0^{\circ} : \frac{18 0 ^{\circ} - 4 arccos ( \frac{2 a}{c} )}{2}

π - 2 arccos (\frac{2 a}{c}) - \frac{π}{2}

将项转换为分式:

18 0^{\circ} = 18 0^{\circ}, 2 arccos (\frac{2 a}{c}) = \frac{2 arccos ( \frac{2 a}{c} ) 2}{2}

= \frac{π 2}{2} - \frac{2 arccos ( \frac{2 a}{c} ) \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{π 2 - 2 arccos ( \frac{2 a}{c} ) \cdot 2 - π}{2}

18 0^{\circ} 2 - 2 arccos (\frac{2 a}{c}) \cdot 2 - 18 0^{\circ} = 18 0^{\circ} - 4 arccos (\frac{2 a}{c})

π 2 - 2 arccos (\frac{2 a}{c}) \cdot 2 - π

同类项相加:

36 0^{\circ} - 18 0^{\circ} = 18 0^{\circ}

= π - 2 \cdot 2 arccos (\frac{2 a}{c})

数字相乘:

2 \cdot 2 = 4

= π - 4 arccos (\frac{2 a}{c})

= \frac{π - 4 arccos ( \frac{2 a}{c} )}{2}

\frac{π - 4 arccos ( \frac{2 a}{c} )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

π - 4 arccos (\frac{2 a}{c}) = 0

用替代法求解

π - 4 arccos (\frac{2 a}{c}) = 0

令

: arccos (\frac{2 a}{c}) = u

18 0^{\circ} - 4 u = 0

18 0^{\circ} - 4 u = 0 : u = 4 5^{\circ}

18 0^{\circ} - 4 u = 0

将

18 0^{\circ}

到右边

18 0^{\circ} - 4 u = 0

两边减去

18 0^{\circ}

18 0^{\circ} - 4 u - 18 0^{\circ} = 0 - 18 0^{\circ}

化简

- 4 u = - 18 0^{\circ}

- 4 u = - 18 0^{\circ}

两边除以

- 4

- 4 u = - 18 0^{\circ}

两边除以

- 4

\frac{- 4 u}{- 4} = \frac{- 18 0 ^{\circ}}{- 4}

化简

u = 4 5^{\circ}

u = 4 5^{\circ}

arccos (\frac{2 a}{c}) = \frac{π}{4}

u = arccos (\frac{2 a}{c})

代回

arccos (\frac{2 a}{c}) = \frac{π}{4}

arccos (\frac{2 a}{c}) = \frac{π}{4}

令

: u = \frac{2 a}{c}

18 0^{\circ} - 4 arccos (u) = 0

将

18 0^{\circ}

到右边

18 0^{\circ} - 4 arccos (u) = 0

两边减去

18 0^{\circ}

18 0^{\circ} - 4 arccos (u) - 18 0^{\circ} = 0 - 18 0^{\circ}

化简

- 4 arccos (u) = - 18 0^{\circ}

- 4 arccos (u) = - 18 0^{\circ}

两边除以

- 4

- 4 arccos (u) = - 18 0^{\circ}

两边除以

- 4

\frac{- 4 arccos ( u )}{- 4} = \frac{- 18 0 ^{\circ}}{- 4}

化简

arccos (u) = 4 5^{\circ}

arccos (u) = 4 5^{\circ}

使用反三角函数性质

arccos (u) = 4 5^{\circ}

arccos (x) = a \Rightarrow x = cos (a)

u = cos (4 5^{\circ})

cos (4 5^{\circ}) = \frac{2}{2}

cos (4 5^{\circ})

使用以下普通恒等式:

cos (4 5^{\circ}) = \frac{2}{2}

cos (4 5^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

= \frac{2}{2}

u = \frac{2}{2}

u = \frac{2}{2}

u = \frac{2 a}{c}

代回

\frac{2 a}{c} = \frac{2}{2} : a = \frac{c}{2 2}; c \neq = 0

\frac{2 a}{c} = \frac{2}{2}

使用根式运算法则:

a = a a

2 = 22

\frac{2 a}{c} = \frac{2}{2 2}

约分:

2

\frac{2 a}{c} = \frac{1}{2}

在两边乘以

c

\frac{2 a}{c} = \frac{1}{2}

在两边乘以

c

\frac{2 a c}{c} = \frac{1 \cdot c}{2}; c \neq = 0

化简

2 a = \frac{1 \cdot c}{2}; c \neq = 0

2 a = \frac{1 \cdot c}{2}; c \neq = 0

两边除以

2; c \neq = 0

2 a = \frac{1 \cdot c}{2}; c \neq = 0

两边除以

2; c \neq = 0

\frac{2 a}{2} = \frac{\frac{1 \cdot c}{2}}{2}; c \neq = 0

化简

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

化简

\frac{2 a}{2} : a

\frac{2 a}{2}

约分:

2

= a

化简

\frac{\frac{1 \cdot c}{2}}{2} : \frac{c}{2 2}

\frac{\frac{1 \cdot c}{2}}{2}

使用法则:

1 \cdot a = a

= \frac{\frac{c}{2}}{2}

使用分式法则:

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

= \frac{c}{2 \cdot 2}

= \frac{c}{2 2}

a = \frac{c}{2 2}; c \neq = 0

a = \frac{c}{2 2}; c \neq = 0

a = \frac{c}{2 2}; c \neq = 0

a = \frac{c}{2 2}

作图

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