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受欢迎的三角函数 >

tan(u)-cot(u)=2

解答

tan (u) - cot (u) = 2

解答

u = 2.74889 \dots + πn, u = 1.17809 \dots + πn

+1

度数

u = 157. 5^{\circ} + 18 0^{\circ} n, u = 67. 5^{\circ} + 18 0^{\circ} n

求解步骤

tan (u) - cot (u) = 2

两边减去

2

tan (u) - cot (u) - 2 = 0

使用三角恒等式改写

- 2 - cot (u) + tan (u)

使用基本三角恒等式:

tan (x) = \frac{1}{cot ( x )}

= - 2 - cot (u) + \frac{1}{cot ( u )}

- 2 - cot (u) + \frac{1}{cot ( u )} = 0

用替代法求解

- 2 - cot (u) + \frac{1}{cot ( u )} = 0

令

: cot (u) = v

- 2 - v + \frac{1}{v} = 0

- 2 - v + \frac{1}{v} = 0 : v = - 1 - 2, v = 2 - 1

- 2 - v + \frac{1}{v} = 0

在两边乘以

v

- 2 - v + \frac{1}{v} = 0

在两边乘以

v

- 2 v - vv + \frac{1}{v} v = 0 \cdot v

化简

- 2 v - vv + \frac{1}{v} v = 0 \cdot v

化简

- vv : - v^{2}

- vv

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

vv = v^{1 + 1}

= - v^{1 + 1}

数字相加:

1 + 1 = 2

= - v^{2}

化简

\frac{1}{v} v : 1

\frac{1}{v} v

分式相乘:

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

= \frac{1 \cdot v}{v}

约分:

v

= 1

化简

0 \cdot v : 0

0 \cdot v

使用法则

0 \cdot a = 0

= 0

- 2 v - v^{2} + 1 = 0

- 2 v - v^{2} + 1 = 0

- 2 v - v^{2} + 1 = 0

解

- 2 v - v^{2} + 1 = 0 : v = - 1 - 2, v = 2 - 1

- 2 v - v^{2} + 1 = 0

改写成标准形式

a x^{2} + b x + c = 0

- v^{2} - 2 v + 1 = 0

使用求根公式求解

- v^{2} - 2 v + 1 = 0

二次方程求根公式:

若

a = - 1, b = - 2, c = 1

v_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 1 ) \cdot 1}{2 ( - 1 )}

v_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 1 ) \cdot 1}{2 ( - 1 )}

(- 2)^{2} - 4 (- 1) \cdot 1 = 22

(- 2)^{2} - 4 (- 1) \cdot 1

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 1 \cdot 1

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 2)^{2} = 2^{2}

= 2^{2} + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

数字相加:

4 + 4 = 8

= 8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

使用根式运算法则:

= 2 2^{2}

使用根式运算法则:

2^{2} = 2

= 22

v_{1, 2} = \frac{- ( - 2 ) \pm 2 2}{2 ( - 1 )}

将解分隔开

v_{1} = \frac{- ( - 2 ) + 2 2}{2 ( - 1 )}, v_{2} = \frac{- ( - 2 ) - 2 2}{2 ( - 1 )}

v = \frac{- ( - 2 ) + 2 2}{2 ( - 1 )} : - 1 - 2

\frac{- ( - 2 ) + 2 2}{2 ( - 1 )}

去除括号:

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

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

数字相乘:

2 \cdot 1 = 2

= \frac{2 + 2 2}{- 2}

使用分式法则:

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

= - \frac{2 + 2 2}{2}

消掉

\frac{2 + 2 2}{2} : 1 + 2

\frac{2 + 2 2}{2}

分解

2 + 22 : 2 (1 + 2)

2 + 22

改写为

= 2 \cdot 1 + 22

因式分解出通项

2

= 2 (1 + 2)

= \frac{2 ( 1 + 2 )}{2}

数字相除:

\frac{2}{2} = 1

= 1 + 2

= - (1 + 2)

打开括号

= - (1) - (2)

使用加减运算法则

+ (- a) = - a

= - 1 - 2

v = \frac{- ( - 2 ) - 2 2}{2 ( - 1 )} : 2 - 1

\frac{- ( - 2 ) - 2 2}{2 ( - 1 )}

去除括号:

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

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

数字相乘:

2 \cdot 1 = 2

= \frac{2 - 2 2}{- 2}

使用分式法则:

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

2 - 22 = - (22 - 2)

= \frac{2 2 - 2}{2}

分解

22 - 2 : 2 (2 - 1)

22 - 2

改写为

= 22 - 2 \cdot 1

因式分解出通项

2

= 2 (2 - 1)

= \frac{2 ( 2 - 1 )}{2}

数字相除:

\frac{2}{2} = 1

= 2 - 1

二次方程组的解是:

v = - 1 - 2, v = 2 - 1

v = - 1 - 2, v = 2 - 1

验证解

找到无定义的点（奇点）:

v = 0

取

- 2 - v + \frac{1}{v}

的分母，令其等于零

v = 0

以下点无定义

v = 0

将不在定义域的点与解相综合:

v = - 1 - 2, v = 2 - 1

v = cot (u)

代回

cot (u) = - 1 - 2, cot (u) = 2 - 1

cot (u) = - 1 - 2, cot (u) = 2 - 1

cot (u) = - 1 - 2 : u = arccot (- 1 - 2) + πn

cot (u) = - 1 - 2

使用反三角函数性质

cot (u) = - 1 - 2

cot (u) = - 1 - 2

的通解

cot (x) = - a \Rightarrow x = arccot (- a) + πn

u = arccot (- 1 - 2) + πn

u = arccot (- 1 - 2) + πn

cot (u) = 2 - 1 : u = arccot (2 - 1) + πn

cot (u) = 2 - 1

使用反三角函数性质

cot (u) = 2 - 1

cot (u) = 2 - 1

的通解

cot (x) = a \Rightarrow x = arccot (a) + πn

u = arccot (2 - 1) + πn

u = arccot (2 - 1) + πn

合并所有解

u = arccot (- 1 - 2) + πn, u = arccot (2 - 1) + πn

以小数形式表示解

u = 2.74889 \dots + πn, u = 1.17809 \dots + πn

作图

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