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受欢迎的三角函数 >

0= pi/2-2arctan((pi/2-2-c)/(pi/2-c))

解答

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

解答

c \in R 无解

求解步骤

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

交换两边

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

将

\frac{π}{2}

到右边

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

两边减去

\frac{π}{2}

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

化简

- 2 arctan (\frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c}) = - \frac{π}{2}

- 2 arctan (\frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c}) = - \frac{π}{2}

两边除以

- 2

- 2 arctan (\frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c}) = - \frac{π}{2}

两边除以

- 2

\frac{- 2 arctan ( \frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c} )}{- 2} = \frac{- \frac{π}{2}}{- 2}

化简

\frac{- 2 arctan ( \frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c} )}{- 2} = \frac{- \frac{π}{2}}{- 2}

化简

\frac{- 2 arctan ( \frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c} )}{- 2} : arctan (\frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c})

\frac{- 2 arctan ( \frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c} )}{- 2}

使用分式法则:

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

= \frac{2 arctan ( \frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c} )}{2}

数字相除:

\frac{2}{2} = 1

= arctan (\frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c})

化简

\frac{- \frac{π}{2}}{- 2} : \frac{π}{4}

\frac{- \frac{π}{2}}{- 2}

使用分式法则:

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

= \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}

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

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

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

使用反三角函数性质

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

arctan (x) = a \Rightarrow x = tan (a)

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

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

使用以下普通恒等式:

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

= 1

\frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c} = 1

\frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c} = 1

解

\frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c} = 1 : c \in R

无解

\frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c} = 1

化简

\frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c} : \frac{π - 4 - 2 c}{π - 2 c}

\frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c}

化简

\frac{π}{2} - c : \frac{π - 2 c}{2}

\frac{π}{2} - c

将项转换为分式:

c = \frac{c 2}{2}

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

= \frac{\frac{π}{2} - 2 - c}{\frac{π - 2 c}{2}}

化简

\frac{π}{2} - 2 - c : \frac{π - 4 - 2 c}{2}

\frac{π}{2} - 2 - c

将项转换为分式:

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

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

数字相乘:

2 \cdot 2 = 4

= \frac{π - 4 - 2 c}{2}

= \frac{\frac{π - 4 - 2 c}{2}}{\frac{π - 2 c}{2}}

分式相除:

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

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

约分:

2

= \frac{π - 4 - c \cdot 2}{π - c \cdot 2}

\frac{π - 4 - 2 c}{π - 2 c} = 1

在两边乘以

π - 2 c

\frac{π - 4 - 2 c}{π - 2 c} = 1

在两边乘以

π - 2 c

\frac{π - 4 - 2 c}{π - 2 c} (π - 2 c) = 1 \cdot (π - 2 c)

化简

\frac{π - 4 - 2 c}{π - 2 c} (π - 2 c) = 1 \cdot (π - 2 c)

化简

\frac{π - 4 - 2 c}{π - 2 c} (π - 2 c) : π - 4 - 2 c

\frac{π - 4 - 2 c}{π - 2 c} (π - 2 c)

分式相乘:

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

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

约分:

π - 2 c

= π - 4 - 2 c

化简

1 \cdot (π - 2 c) : π - 2 c

1 \cdot (π - 2 c)

乘以:

1 \cdot (π - 2 c) = (π - 2 c)

= (π - 2 c)

去除括号:

(a) = a

= π - 2 c

π - 4 - 2 c = π - 2 c

π - 4 - 2 c = π - 2 c

π - 4 - 2 c = π - 2 c

解

π - 4 - 2 c = π - 2 c :

无解

π - 4 - 2 c = π - 2 c

两边减去

π - 2 c

π - 4 - 2 c - (π - 2 c) = π - 2 c - (π - 2 c)

化简

- 4 = 0

两侧不相等

无解

c \in R 无解

c \in R 无解

作图

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