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受欢迎的三角函数 >

solvefor u,tan(u)+cot(u)= 1/f

解答

求解

u, tan (u) + cot (u) = \frac{1}{f}

解答

u = arccot (\frac{1 + 1 - 4 f ^{2}}{2 f}) + πn, u = arccot (\frac{1 - 1 - 4 f ^{2}}{2 f}) + πn

求解步骤

tan (u) + cot (u) = \frac{1}{f}

两边减去

\frac{1}{f}

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

化简

tan (u) + cot (u) - \frac{1}{f} : \frac{f tan ( u ) + f cot ( u ) - 1}{f}

tan (u) + cot (u) - \frac{1}{f}

将项转换为分式:

tan (u) = \frac{tan ( u ) f}{f}, cot (u) = \frac{cot ( u ) f}{f}

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

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{tan ( u ) f + cot ( u ) f - 1}{f}

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

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

f tan (u) + f cot (u) - 1 = 0

使用三角恒等式改写

- 1 + cot (u) f + tan (u) f

使用基本三角恒等式:

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

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

\frac{1}{cot ( u )} f = \frac{f}{cot ( u )}

\frac{1}{cot ( u )} f

分式相乘:

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

= \frac{1 \cdot f}{cot ( u )}

乘以:

1 \cdot f = f

= \frac{f}{cot ( u )}

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

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

用替代法求解

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

令

: cot (u) = v

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

- 1 + \frac{f}{v} + v f = 0 : v = \frac{1 + 1 - 4 f ^{2}}{2 f}, v = \frac{1 - 1 - 4 f ^{2}}{2 f}; f \neq = 0

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

在两边乘以

v

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

在两边乘以

v

- 1 \cdot v + \frac{f}{v} v + v f v = 0 \cdot v

化简

- 1 \cdot v + \frac{f}{v} v + v f v = 0 \cdot v

化简

- 1 \cdot v : - v

- 1 \cdot v

乘以:

1 \cdot v = v

= - v

化简

\frac{f}{v} v : f

\frac{f}{v} v

分式相乘:

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

= \frac{f v}{v}

约分:

v

= f

化简

v f v : f v^{2}

v f v

使用指数法则:

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

vv = v^{1 + 1}

= f v^{1 + 1}

数字相加:

1 + 1 = 2

= f v^{2}

化简

0 \cdot v : 0

0 \cdot v

使用法则

0 \cdot a = 0

= 0

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

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

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

解

- v + f + f v^{2} = 0 : v = \frac{1 + 1 - 4 f ^{2}}{2 f}, v = \frac{1 - 1 - 4 f ^{2}}{2 f}; f \neq = 0

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = f, b = - 1, c = f

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

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

化简

(- 1)^{2} - 4 ff : 1 - 4 f^{2}

(- 1)^{2} - 4 ff

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 ff = 4 f^{2}

4 ff

使用指数法则:

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

ff = f^{1 + 1}

= 4 f^{1 + 1}

数字相加:

1 + 1 = 2

= 4 f^{2}

= 1 - 4 f^{2}

v_{1, 2} = \frac{- ( - 1 ) \pm 1 - 4 f ^{2}}{2 f}; f \neq = 0

将解分隔开

v_{1} = \frac{- ( - 1 ) + 1 - 4 f ^{2}}{2 f}, v_{2} = \frac{- ( - 1 ) - 1 - 4 f ^{2}}{2 f}

v = \frac{- ( - 1 ) + 1 - 4 f ^{2}}{2 f} : \frac{1 + 1 - 4 f ^{2}}{2 f}

\frac{- ( - 1 ) + 1 - 4 f ^{2}}{2 f}

使用法则

- (- a) = a

= \frac{1 + 1 - 4 f ^{2}}{2 f}

v = \frac{- ( - 1 ) - 1 - 4 f ^{2}}{2 f} : \frac{1 - 1 - 4 f ^{2}}{2 f}

\frac{- ( - 1 ) - 1 - 4 f ^{2}}{2 f}

使用法则

- (- a) = a

= \frac{1 - 1 - 4 f ^{2}}{2 f}

二次方程组的解是:

v = \frac{1 + 1 - 4 f ^{2}}{2 f}, v = \frac{1 - 1 - 4 f ^{2}}{2 f}; f \neq = 0

v = \frac{1 + 1 - 4 f ^{2}}{2 f}, v = \frac{1 - 1 - 4 f ^{2}}{2 f}; f \neq = 0

v = cot (u)

代回

cot (u) = \frac{1 + 1 - 4 f ^{2}}{2 f}, cot (u) = \frac{1 - 1 - 4 f ^{2}}{2 f}; f \neq = 0

cot (u) = \frac{1 + 1 - 4 f ^{2}}{2 f}, cot (u) = \frac{1 - 1 - 4 f ^{2}}{2 f}; f \neq = 0

cot (u) = \frac{1 + 1 - 4 f ^{2}}{2 f} : u = arccot (\frac{1 + 1 - 4 f ^{2}}{2 f}) + πn

cot (u) = \frac{1 + 1 - 4 f ^{2}}{2 f}

使用反三角函数性质

cot (u) = \frac{1 + 1 - 4 f ^{2}}{2 f}

cot (u) = \frac{1 + 1 - 4 f ^{2}}{2 f}

的通解

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

u = arccot (\frac{1 + 1 - 4 f ^{2}}{2 f}) + πn

u = arccot (\frac{1 + 1 - 4 f ^{2}}{2 f}) + πn

cot (u) = \frac{1 - 1 - 4 f ^{2}}{2 f} : u = arccot (\frac{1 - 1 - 4 f ^{2}}{2 f}) + πn

cot (u) = \frac{1 - 1 - 4 f ^{2}}{2 f}

使用反三角函数性质

cot (u) = \frac{1 - 1 - 4 f ^{2}}{2 f}

cot (u) = \frac{1 - 1 - 4 f ^{2}}{2 f}

的通解

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

u = arccot (\frac{1 - 1 - 4 f ^{2}}{2 f}) + πn

u = arccot (\frac{1 - 1 - 4 f ^{2}}{2 f}) + πn

合并所有解

u = arccot (\frac{1 + 1 - 4 f ^{2}}{2 f}) + πn, u = arccot (\frac{1 - 1 - 4 f ^{2}}{2 f}) + πn

作图

流行的例子

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