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受欢迎的三角函数 >

cot(θ)+2csc(θ)=4,0<= θ<= 360

解答

cot (θ) + 2 csc (θ) = 4, 0^{\circ} \leq θ \leq 36 0^{\circ}

解答

θ = 0.75142 \dots, θ = 2.88012 \dots

求解步骤

cot (θ) + 2 csc (θ) = 4, 0^{\circ} \leq θ \leq 36 0^{\circ}

两边减去

4

cot (θ) + 2 csc (θ) - 4 = 0

用 sin, cos 表示

\frac{cos ( θ )}{sin ( θ )} + 2 \cdot \frac{1}{sin ( θ )} - 4 = 0

化简

\frac{cos ( θ )}{sin ( θ )} + 2 \cdot \frac{1}{sin ( θ )} - 4 : \frac{cos ( θ ) + 2 - 4 sin ( θ )}{sin ( θ )}

\frac{cos ( θ )}{sin ( θ )} + 2 \cdot \frac{1}{sin ( θ )} - 4

2 \cdot \frac{1}{sin ( θ )} = \frac{2}{sin ( θ )}

2 \cdot \frac{1}{sin ( θ )}

分式相乘:

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

= \frac{1 \cdot 2}{sin ( θ )}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )} + \frac{2}{sin ( θ )} - 4

合并分式

\frac{cos ( θ )}{sin ( θ )} + \frac{2}{sin ( θ )} : \frac{cos ( θ ) + 2}{sin ( θ )}

使用法则

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

= \frac{cos ( θ ) + 2}{sin ( θ )}

= \frac{cos ( θ ) + 2}{sin ( θ )} - 4

将项转换为分式:

4 = \frac{4 sin ( θ )}{sin ( θ )}

= \frac{cos ( θ ) + 2}{sin ( θ )} - \frac{4 sin ( θ )}{sin ( θ )}

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 ( θ ) + 2 - 4 sin ( θ )}{sin ( θ )}

\frac{cos ( θ ) + 2 - 4 sin ( θ )}{sin ( θ )} = 0

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

cos (θ) + 2 - 4 sin (θ) = 0

两边加上

4 sin (θ)

cos (θ) + 2 = 4 sin (θ)

两边进行平方

(cos (θ) + 2)^{2} = (4 sin (θ))^{2}

两边减去

(4 sin (θ))^{2}

(cos (θ) + 2)^{2} - 16 sin^{2} (θ) = 0

使用三角恒等式改写

(2 + cos (θ))^{2} - 16 sin^{2} (θ)

使用毕达哥拉斯恒等式:

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

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

= (2 + cos (θ))^{2} - 16 (1 - cos^{2} (θ))

化简

(2 + cos (θ))^{2} - 16 (1 - cos^{2} (θ)) : 17 cos^{2} (θ) + 4 cos (θ) - 12

(2 + cos (θ))^{2} - 16 (1 - cos^{2} (θ))

(2 + cos (θ))^{2} : 4 + 4 cos (θ) + cos^{2} (θ)

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 2, b = cos (θ)

= 2^{2} + 2 \cdot 2 cos (θ) + cos^{2} (θ)

化简

2^{2} + 2 \cdot 2 cos (θ) + cos^{2} (θ) : 4 + 4 cos (θ) + cos^{2} (θ)

2^{2} + 2 \cdot 2 cos (θ) + cos^{2} (θ)

2^{2} = 4

= 4 + 2 \cdot 2 cos (θ) + cos^{2} (θ)

数字相乘:

2 \cdot 2 = 4

= 4 + 4 cos (θ) + cos^{2} (θ)

= 4 + 4 cos (θ) + cos^{2} (θ)

= 4 + 4 cos (θ) + cos^{2} (θ) - 16 (1 - cos^{2} (θ))

乘开

- 16 (1 - cos^{2} (θ)) : - 16 + 16 cos^{2} (θ)

- 16 (1 - cos^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 16, b = 1, c = cos^{2} (θ)

= - 16 \cdot 1 - (- 16) cos^{2} (θ)

使用加减运算法则

- (- a) = a

= - 16 \cdot 1 + 16 cos^{2} (θ)

数字相乘:

16 \cdot 1 = 16

= - 16 + 16 cos^{2} (θ)

= 4 + 4 cos (θ) + cos^{2} (θ) - 16 + 16 cos^{2} (θ)

化简

4 + 4 cos (θ) + cos^{2} (θ) - 16 + 16 cos^{2} (θ) : 17 cos^{2} (θ) + 4 cos (θ) - 12

4 + 4 cos (θ) + cos^{2} (θ) - 16 + 16 cos^{2} (θ)

对同类项分组

= 4 cos (θ) + cos^{2} (θ) + 16 cos^{2} (θ) + 4 - 16

同类项相加:

cos^{2} (θ) + 16 cos^{2} (θ) = 17 cos^{2} (θ)

= 4 cos (θ) + 17 cos^{2} (θ) + 4 - 16

数字相加/相减:

4 - 16 = - 12

= 17 cos^{2} (θ) + 4 cos (θ) - 12

= 17 cos^{2} (θ) + 4 cos (θ) - 12

= 17 cos^{2} (θ) + 4 cos (θ) - 12

- 12 + 17 cos^{2} (θ) + 4 cos (θ) = 0

用替代法求解

- 12 + 17 cos^{2} (θ) + 4 cos (θ) = 0

令

: cos (θ) = u

- 12 + 17 u^{2} + 4 u = 0

- 12 + 17 u^{2} + 4 u = 0 : u = \frac{2 ( 2 13 - 1 )}{17}, u = - \frac{2 ( 1 + 2 13 )}{17}

- 12 + 17 u^{2} + 4 u = 0

改写成标准形式

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

17 u^{2} + 4 u - 12 = 0

使用求根公式求解

17 u^{2} + 4 u - 12 = 0

二次方程求根公式:

若

a = 17, b = 4, c = - 12

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 17 ( - 12 )}{2 \cdot 17}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 17 ( - 12 )}{2 \cdot 17}

4^{2} - 4 \cdot 17 (- 12) = 813

4^{2} - 4 \cdot 17 (- 12)

使用法则

- (- a) = a

= 4^{2} + 4 \cdot 17 \cdot 12

数字相乘:

4 \cdot 17 \cdot 12 = 816

= 4^{2} + 816

4^{2} = 16

= 16 + 816

数字相加:

16 + 816 = 832

= 832

832

质因数分解:

2^{6} \cdot 13

832

832

除以

2832 = 416 \cdot 2

= 2 \cdot 416

416

除以

2416 = 208 \cdot 2

= 2 \cdot 2 \cdot 208

208

除以

2208 = 104 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 104

104

除以

2104 = 52 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 52

52

除以

252 = 26 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 26

26

除以

226 = 13 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 13

2, 13

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 13

= 2^{6} \cdot 13

= 2^{6} \cdot 13

使用根式运算法则:

n ab = n a n b

= 13 2^{6}

使用根式运算法则:

n a^{m} = a^{\frac{m}{n}}

2^{6} = 2^{\frac{6}{2}} = 2^{3}

= 2^{3} 13

整理后得

= 813

u_{1, 2} = \frac{- 4 \pm 8 13}{2 \cdot 17}

将解分隔开

u_{1} = \frac{- 4 + 8 13}{2 \cdot 17}, u_{2} = \frac{- 4 - 8 13}{2 \cdot 17}

u = \frac{- 4 + 8 13}{2 \cdot 17} : \frac{2 ( 2 13 - 1 )}{17}

\frac{- 4 + 8 13}{2 \cdot 17}

数字相乘:

2 \cdot 17 = 34

= \frac{- 4 + 8 13}{34}

分解

- 4 + 813 : 4 (- 1 + 213)

- 4 + 813

改写为

= - 4 \cdot 1 + 4 \cdot 213

因式分解出通项

4

= 4 (- 1 + 213)

= \frac{4 ( - 1 + 2 13 )}{34}

约分:

2

= \frac{2 ( 2 13 - 1 )}{17}

u = \frac{- 4 - 8 13}{2 \cdot 17} : - \frac{2 ( 1 + 2 13 )}{17}

\frac{- 4 - 8 13}{2 \cdot 17}

数字相乘:

2 \cdot 17 = 34

= \frac{- 4 - 8 13}{34}

分解

- 4 - 813 : - 4 (1 + 213)

- 4 - 813

改写为

= - 4 \cdot 1 - 4 \cdot 213

因式分解出通项

4

= - 4 (1 + 213)

= - \frac{4 ( 1 + 2 13 )}{34}

约分:

2

= - \frac{2 ( 1 + 2 13 )}{17}

二次方程组的解是:

u = \frac{2 ( 2 13 - 1 )}{17}, u = - \frac{2 ( 1 + 2 13 )}{17}

u = cos (θ)

代回

cos (θ) = \frac{2 ( 2 13 - 1 )}{17}, cos (θ) = - \frac{2 ( 1 + 2 13 )}{17}

cos (θ) = \frac{2 ( 2 13 - 1 )}{17}, cos (θ) = - \frac{2 ( 1 + 2 13 )}{17}

cos (θ) = \frac{2 ( 2 13 - 1 )}{17}, 0 \leq θ \leq 36 0^{\circ} : θ = arccos (\frac{2 ( 2 13 - 1 )}{17}), θ = 36 0^{\circ} - arccos (\frac{2 ( 2 13 - 1 )}{17})

cos (θ) = \frac{2 ( 2 13 - 1 )}{17}, 0 \leq θ \leq 36 0^{\circ}

使用反三角函数性质

cos (θ) = \frac{2 ( 2 13 - 1 )}{17}

cos (θ) = \frac{2 ( 2 13 - 1 )}{17}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

θ = arccos (\frac{2 ( 2 13 - 1 )}{17}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 36 0^{\circ} n

θ = arccos (\frac{2 ( 2 13 - 1 )}{17}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{2 ( 2 13 - 1 )}{17}) + 36 0^{\circ} n

在

0 \leq θ \leq 36 0^{\circ}

范围内的解

θ = arccos (\frac{2 ( 2 13 - 1 )}{17}), θ = 36 0^{\circ} - arccos (\frac{2 ( 2 13 - 1 )}{17})

cos (θ) = - \frac{2 ( 1 + 2 13 )}{17}, 0 \leq θ \leq 36 0^{\circ} : θ = arccos (- \frac{2 ( 1 + 2 13 )}{17}), θ = - arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 36 0^{\circ}

cos (θ) = - \frac{2 ( 1 + 2 13 )}{17}, 0 \leq θ \leq 36 0^{\circ}

使用反三角函数性质

cos (θ) = - \frac{2 ( 1 + 2 13 )}{17}

cos (θ) = - \frac{2 ( 1 + 2 13 )}{17}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 36 0^{\circ} n, x = - arccos (- a) + 36 0^{\circ} n

θ = arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 36 0^{\circ} n, θ = - arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 36 0^{\circ} n

θ = arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 36 0^{\circ} n, θ = - arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 36 0^{\circ} n

在

0 \leq θ \leq 36 0^{\circ}

范围内的解

θ = arccos (- \frac{2 ( 1 + 2 13 )}{17}), θ = - arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 36 0^{\circ}

合并所有解

θ = arccos (\frac{2 ( 2 13 - 1 )}{17}), θ = 36 0^{\circ} - arccos (\frac{2 ( 2 13 - 1 )}{17}), θ = arccos (- \frac{2 ( 1 + 2 13 )}{17}), θ = - arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 36 0^{\circ}

将解代入原方程进行验证

将它们代入

cot (θ) + 2 csc (θ) = 4

检验解是否符合

去除与方程不符的解。

检验

arccos (\frac{2 ( 2 13 - 1 )}{17})

的解:

真

arccos (\frac{2 ( 2 13 - 1 )}{17})

代入

n = 1

arccos (\frac{2 ( 2 13 - 1 )}{17})

对于

cot (θ) + 2 csc (θ) = 4

代入

θ = arccos (\frac{2 ( 2 13 - 1 )}{17})

cot (arccos (\frac{2 ( 2 13 - 1 )}{17})) + 2 csc (arccos (\frac{2 ( 2 13 - 1 )}{17})) = 4

整理后得

4 = 4

\Rightarrow 真

检验

36 0^{\circ} - arccos (\frac{2 ( 2 13 - 1 )}{17})

的解:

假

36 0^{\circ} - arccos (\frac{2 ( 2 13 - 1 )}{17})

代入

n = 1

36 0^{\circ} - arccos (\frac{2 ( 2 13 - 1 )}{17})

对于

cot (θ) + 2 csc (θ) = 4

代入

θ = 36 0^{\circ} - arccos (\frac{2 ( 2 13 - 1 )}{17})

cot (36 0^{\circ} - arccos (\frac{2 ( 2 13 - 1 )}{17})) + 2 csc (36 0^{\circ} - arccos (\frac{2 ( 2 13 - 1 )}{17})) = 4

整理后得

- 4 = 4

\Rightarrow 假

检验

arccos (- \frac{2 ( 1 + 2 13 )}{17})

的解:

真

arccos (- \frac{2 ( 1 + 2 13 )}{17})

代入

n = 1

arccos (- \frac{2 ( 1 + 2 13 )}{17})

对于

cot (θ) + 2 csc (θ) = 4

代入

θ = arccos (- \frac{2 ( 1 + 2 13 )}{17})

cot (arccos (- \frac{2 ( 1 + 2 13 )}{17})) + 2 csc (arccos (- \frac{2 ( 1 + 2 13 )}{17})) = 4

整理后得

4 = 4

\Rightarrow 真

检验

- arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 36 0^{\circ}

的解:

假

- arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 36 0^{\circ}

代入

n = 1

- arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 36 0^{\circ}

对于

cot (θ) + 2 csc (θ) = 4

代入

θ = - arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 36 0^{\circ}

cot (- arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 36 0^{\circ}) + 2 csc (- arccos (- \frac{2 ( 1 + 2 13 )}{17}) + 36 0^{\circ}) = 4

整理后得

- 4 = 4

\Rightarrow 假

θ = arccos (\frac{2 ( 2 13 - 1 )}{17}), θ = arccos (- \frac{2 ( 1 + 2 13 )}{17})

以小数形式表示解

θ = 0.75142 \dots, θ = 2.88012 \dots

作图

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