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受欢迎的三角函数 >

cos(x)= 1/(cot(x))

解答

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

解答

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

+1

度数

x = 38.17270 \dots^{\circ} + 36 0^{\circ} n, x = 141.82729 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

\frac{1}{cot ( x )}

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

化简

cos (x) - \frac{1}{cot ( x )} : \frac{cos ( x ) cot ( x ) - 1}{cot ( x )}

cos (x) - \frac{1}{cot ( x )}

将项转换为分式:

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

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

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

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

cos (x) cot (x) - 1 = 0

使用三角恒等式改写

- 1 + cos (x) cot (x)

使用基本三角恒等式:

cot (x) = \frac{cos ( x )}{sin ( x )}

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

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

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

分式相乘:

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

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

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

cos (x) cos (x)

使用指数法则:

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

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

= cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= cos^{2} (x)

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

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

使用毕达哥拉斯恒等式:

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

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

= - 1 + \frac{1 - sin ^{2} ( x )}{sin ( x )}

- 1 + \frac{1 - sin ^{2} ( x )}{sin ( x )} = 0

用替代法求解

- 1 + \frac{1 - sin ^{2} ( x )}{sin ( x )} = 0

令

: sin (x) = u

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

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

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

在两边乘以

u

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

在两边乘以

u

- 1 \cdot u + \frac{1 - u ^{2}}{u} u = 0 \cdot u

化简

- 1 \cdot u + \frac{1 - u ^{2}}{u} u = 0 \cdot u

化简

- 1 \cdot u : - u

- 1 \cdot u

乘以:

1 \cdot u = u

= - u

化简

\frac{1 - u ^{2}}{u} u : 1 - u^{2}

\frac{1 - u ^{2}}{u} u

分式相乘:

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

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

约分:

u

= 1 - u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

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

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

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

解

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

- 1 + \frac{1 - u ^{2}}{u}

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

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

u = sin (x)

代回

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

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

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

无解

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

- 1 \leq sin (x) \leq 1

无解

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

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

使用反三角函数性质

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

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

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

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

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

合并所有解

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

以小数形式表示解

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

作图

流行的例子

2 - 3 cos (θ) = 0

(12)/(sin(120))= 5/(sin(x))

\frac{12}{sin ( 12 0 ^{\circ} )} = \frac{5}{sin ( x )}

3 - tan^{2} (b) = 0

tan (2 θ) = - \frac{2}{5}

cos(2x)-3cos(x)=-2

cos (2 x) - 3 cos (x) = - 2