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受欢迎的三角函数 >

(tan(x))/(sec(x))=cot(x)

解答

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

解答

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

+1

度数

x = 51.82729 \dots^{\circ} + 36 0^{\circ} n, x = 308.17270 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

cot (x)

\frac{tan ( x )}{sec ( x )} - cot (x) = 0

化简

\frac{tan ( x )}{sec ( x )} - cot (x) : \frac{tan ( x ) - cot ( x ) sec ( x )}{sec ( x )}

\frac{tan ( x )}{sec ( x )} - cot (x)

将项转换为分式:

cot (x) = \frac{cot ( x ) sec ( x )}{sec ( x )}

= \frac{tan ( x )}{sec ( x )} - \frac{cot ( x ) sec ( x )}{sec ( x )}

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 ( x ) - cot ( x ) sec ( x )}{sec ( x )}

\frac{tan ( x ) - cot ( x ) sec ( x )}{sec ( x )} = 0

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

tan (x) - cot (x) sec (x) = 0

用 sin, cos 表示

\frac{sin ( x )}{cos ( x )} - \frac{cos ( x )}{sin ( x )} \cdot \frac{1}{cos ( x )} = 0

化简

\frac{sin ( x )}{cos ( x )} - \frac{cos ( x )}{sin ( x )} \cdot \frac{1}{cos ( x )} : \frac{sin ^{2} ( x ) - cos ( x )}{cos ( x ) sin ( x )}

\frac{sin ( x )}{cos ( x )} - \frac{cos ( x )}{sin ( x )} \cdot \frac{1}{cos ( x )}

\frac{cos ( x )}{sin ( x )} \cdot \frac{1}{cos ( x )} = \frac{1}{sin ( x )}

\frac{cos ( x )}{sin ( x )} \cdot \frac{1}{cos ( x )}

分式相乘:

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

= \frac{cos ( x ) \cdot 1}{sin ( x ) cos ( x )}

约分:

cos (x)

= \frac{1}{sin ( x )}

= \frac{sin ( x )}{cos ( x )} - \frac{1}{sin ( x )}

cos (x), sin (x)

的最小公倍数:

cos (x) sin (x)

cos (x), sin (x)

最小公倍数 (LCM)

计算出由出现在

cos (x)

或

sin (x)

中的因子组成的表达式

= cos (x) sin (x)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

cos (x) sin (x)

对于

\frac{sin ( x )}{cos ( x )} :

将分母和分子乘以

sin (x)

\frac{sin ( x )}{cos ( x )} = \frac{sin ( x ) sin ( x )}{cos ( x ) sin ( x )} = \frac{sin ^{2} ( x )}{cos ( x ) sin ( x )}

对于

\frac{1}{sin ( x )} :

将分母和分子乘以

cos (x)

\frac{1}{sin ( x )} = \frac{1 \cdot cos ( x )}{sin ( x ) cos ( x )} = \frac{cos ( x )}{cos ( x ) sin ( x )}

= \frac{sin ^{2} ( x )}{cos ( x ) sin ( x )} - \frac{cos ( x )}{cos ( x ) sin ( x )}

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{sin ^{2} ( x ) - cos ( x )}{cos ( x ) sin ( x )}

\frac{sin ^{2} ( x ) - cos ( x )}{cos ( x ) sin ( x )} = 0

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

sin^{2} (x) - cos (x) = 0

两边加上

cos (x)

sin^{2} (x) = cos (x)

两边进行平方

(sin^{2} (x))^{2} = cos^{2} (x)

两边减去

cos^{2} (x)

sin^{4} (x) - cos^{2} (x) = 0

分解

sin^{4} (x) - cos^{2} (x) : (sin^{2} (x) + cos (x)) (sin^{2} (x) - cos (x))

sin^{4} (x) - cos^{2} (x)

使用指数法则:

a^{b c} = (a^{b})^{c}

sin^{4} (x) = (sin^{2} (x))^{2}

= (sin^{2} (x))^{2} - cos^{2} (x)

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(sin^{2} (x))^{2} - cos^{2} (x) = (sin^{2} (x) + cos (x)) (sin^{2} (x) - cos (x))

= (sin^{2} (x) + cos (x)) (sin^{2} (x) - cos (x))

(sin^{2} (x) + cos (x)) (sin^{2} (x) - cos (x)) = 0

分别求解每个部分

sin^{2} (x) + cos (x) = 0 or sin^{2} (x) - cos (x) = 0

sin^{2} (x) + cos (x) = 0 : x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

sin^{2} (x) + cos (x) = 0

使用三角恒等式改写

cos (x) + sin^{2} (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= cos (x) + 1 - cos^{2} (x)

1 + cos (x) - cos^{2} (x) = 0

用替代法求解

1 + cos (x) - cos^{2} (x) = 0

令

: cos (x) = u

1 + u - u^{2} = 0

1 + u - u^{2} = 0 : u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

1 + u - u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

1^{2} - 4 (- 1) \cdot 1 = 5

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 1

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- 1 + 5}{2 ( - 1 )}, u_{2} = \frac{- 1 - 5}{2 ( - 1 )}

u = \frac{- 1 + 5}{2 ( - 1 )} : - \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 + 5}{- 2}

使用分式法则:

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

= - \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 ( - 1 )} : \frac{1 + 5}{2}

\frac{- 1 - 5}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 1 - 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 - 5}{- 2}

使用分式法则:

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

- 1 - 5 = - (1 + 5)

= \frac{1 + 5}{2}

二次方程组的解是:

u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

u = cos (x)

代回

cos (x) = - \frac{- 1 + 5}{2}, cos (x) = \frac{1 + 5}{2}

cos (x) = - \frac{- 1 + 5}{2}, cos (x) = \frac{1 + 5}{2}

cos (x) = - \frac{- 1 + 5}{2} : x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (x) = - \frac{- 1 + 5}{2}

使用反三角函数性质

cos (x) = - \frac{- 1 + 5}{2}

cos (x) = - \frac{- 1 + 5}{2}

的通解

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

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (x) = \frac{1 + 5}{2} :

无解

cos (x) = \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

sin^{2} (x) - cos (x) = 0 : x = arccos (\frac{5 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{5 - 1}{2}) + 2 πn

sin^{2} (x) - cos (x) = 0

使用三角恒等式改写

- cos (x) + sin^{2} (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= - cos (x) + 1 - cos^{2} (x)

1 - cos (x) - cos^{2} (x) = 0

用替代法求解

1 - cos (x) - cos^{2} (x) = 0

令

: cos (x) = u

1 - u - u^{2} = 0

1 - u - u^{2} = 0 : u = - \frac{1 + 5}{2}, u = \frac{5 - 1}{2}

1 - u - u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

(- 1)^{2} - 4 (- 1) \cdot 1 = 5

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

使用法则

- (- a) = a

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

(- 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 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 5}{2 ( - 1 )}, u_{2} = \frac{- ( - 1 ) - 5}{2 ( - 1 )}

u = \frac{- ( - 1 ) + 5}{2 ( - 1 )} : - \frac{1 + 5}{2}

\frac{- ( - 1 ) + 5}{2 ( - 1 )}

去除括号:

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

= \frac{1 + 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 + 5}{- 2}

使用分式法则:

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

= - \frac{1 + 5}{2}

u = \frac{- ( - 1 ) - 5}{2 ( - 1 )} : \frac{5 - 1}{2}

\frac{- ( - 1 ) - 5}{2 ( - 1 )}

去除括号:

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

= \frac{1 - 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 - 5}{- 2}

使用分式法则:

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

1 - 5 = - (5 - 1)

= \frac{5 - 1}{2}

二次方程组的解是:

u = - \frac{1 + 5}{2}, u = \frac{5 - 1}{2}

u = cos (x)

代回

cos (x) = - \frac{1 + 5}{2}, cos (x) = \frac{5 - 1}{2}

cos (x) = - \frac{1 + 5}{2}, cos (x) = \frac{5 - 1}{2}

cos (x) = - \frac{1 + 5}{2} :

无解

cos (x) = - \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

无解

cos (x) = \frac{5 - 1}{2} : x = arccos (\frac{5 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{5 - 1}{2}) + 2 πn

cos (x) = \frac{5 - 1}{2}

使用反三角函数性质

cos (x) = \frac{5 - 1}{2}

cos (x) = \frac{5 - 1}{2}

的通解

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

x = arccos (\frac{5 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{5 - 1}{2}) + 2 πn

x = arccos (\frac{5 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{5 - 1}{2}) + 2 πn

合并所有解

x = arccos (\frac{5 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{5 - 1}{2}) + 2 πn

合并所有解

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = arccos (\frac{5 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{5 - 1}{2}) + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{- 1 + 5}{2}) + 2 πn

的解:

假

arccos (- \frac{- 1 + 5}{2}) + 2 πn

代入

n = 1

arccos (- \frac{- 1 + 5}{2}) + 2 π 1

对于

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

代入

x = arccos (- \frac{- 1 + 5}{2}) + 2 π 1

\frac{tan ( arccos ( - \frac{- 1 + 5}{2} ) + 2 π 1 )}{sec ( arccos ( - \frac{- 1 + 5}{2} ) + 2 π 1 )} = cot (arccos (- \frac{- 1 + 5}{2}) + 2 π 1)

整理后得

0.78615 \dots = - 0.78615 \dots

\Rightarrow 假

检验

- arccos (- \frac{- 1 + 5}{2}) + 2 πn

的解:

假

- arccos (- \frac{- 1 + 5}{2}) + 2 πn

代入

n = 1

- arccos (- \frac{- 1 + 5}{2}) + 2 π 1

对于

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

代入

x = - arccos (- \frac{- 1 + 5}{2}) + 2 π 1

\frac{tan ( - arccos ( - \frac{- 1 + 5}{2} ) + 2 π 1 )}{sec ( - arccos ( - \frac{- 1 + 5}{2} ) + 2 π 1 )} = cot (- arccos (- \frac{- 1 + 5}{2}) + 2 π 1)

整理后得

- 0.78615 \dots = 0.78615 \dots

\Rightarrow 假

检验

arccos (\frac{5 - 1}{2}) + 2 πn

的解:

真

arccos (\frac{5 - 1}{2}) + 2 πn

代入

n = 1

arccos (\frac{5 - 1}{2}) + 2 π 1

对于

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

代入

x = arccos (\frac{5 - 1}{2}) + 2 π 1

\frac{tan ( arccos ( \frac{5 - 1}{2} ) + 2 π 1 )}{sec ( arccos ( \frac{5 - 1}{2} ) + 2 π 1 )} = cot (arccos (\frac{5 - 1}{2}) + 2 π 1)

整理后得

0.78615 \dots = 0.78615 \dots

\Rightarrow 真

检验

2 π - arccos (\frac{5 - 1}{2}) + 2 πn

的解:

真

2 π - arccos (\frac{5 - 1}{2}) + 2 πn

代入

n = 1

2 π - arccos (\frac{5 - 1}{2}) + 2 π 1

对于

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

代入

x = 2 π - arccos (\frac{5 - 1}{2}) + 2 π 1

\frac{tan ( 2 π - arccos ( \frac{5 - 1}{2} ) + 2 π 1 )}{sec ( 2 π - arccos ( \frac{5 - 1}{2} ) + 2 π 1 )} = cot (2 π - arccos (\frac{5 - 1}{2}) + 2 π 1)

整理后得

- 0.78615 \dots = - 0.78615 \dots

\Rightarrow 真

x = arccos (\frac{5 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{5 - 1}{2}) + 2 πn

以小数形式表示解

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

作图

流行的例子

cos(7x)+cos(3x)=0

cos (7 x) + cos (3 x) = 0

5sin^2(x)+2sin(x)=0

5 sin^{2} (x) + 2 sin (x) = 0

cos(2x)-11cos(x)+6=0

cos (2 x) - 11 cos (x) + 6 = 0

sin (x) = - 0.45

sin (x) = - 0.59