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12cos(β)-5sin(β)=4.7

解答

12 cos (β) - 5 sin (β) = 4.7

解答

β = π + 1.54592 \dots + 2 πn, β = 0.80608 \dots + 2 πn

+1

度数

β = 268.57484 \dots^{\circ} + 36 0^{\circ} n, β = 46.18542 \dots^{\circ} + 36 0^{\circ} n

求解步骤

12 cos (β) - 5 sin (β) = 4.7

两边加上

5 sin (β)

12 cos (β) = 4.7 + 5 sin (β)

两边进行平方

(12 cos (β))^{2} = (4.7 + 5 sin (β))^{2}

两边减去

(4.7 + 5 sin (β))^{2}

144 cos^{2} (β) - 22.09 - 47 sin (β) - 25 sin^{2} (β) = 0

使用三角恒等式改写

- 22.09 + 144 cos^{2} (β) - 25 sin^{2} (β) - 47 sin (β)

使用毕达哥拉斯恒等式:

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

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

= - 22.09 + 144 (1 - sin^{2} (β)) - 25 sin^{2} (β) - 47 sin (β)

化简

- 22.09 + 144 (1 - sin^{2} (β)) - 25 sin^{2} (β) - 47 sin (β) : - 169 sin^{2} (β) - 47 sin (β) + 121.91

- 22.09 + 144 (1 - sin^{2} (β)) - 25 sin^{2} (β) - 47 sin (β)

乘开

144 (1 - sin^{2} (β)) : 144 - 144 sin^{2} (β)

144 (1 - sin^{2} (β))

使用分配律:

a (b - c) = ab - a c

a = 144, b = 1, c = sin^{2} (β)

= 144 \cdot 1 - 144 sin^{2} (β)

数字相乘:

144 \cdot 1 = 144

= 144 - 144 sin^{2} (β)

= - 22.09 + 144 - 144 sin^{2} (β) - 25 sin^{2} (β) - 47 sin (β)

化简

- 22.09 + 144 - 144 sin^{2} (β) - 25 sin^{2} (β) - 47 sin (β) : - 169 sin^{2} (β) - 47 sin (β) + 121.91

- 22.09 + 144 - 144 sin^{2} (β) - 25 sin^{2} (β) - 47 sin (β)

同类项相加:

- 144 sin^{2} (β) - 25 sin^{2} (β) = - 169 sin^{2} (β)

= - 22.09 + 144 - 169 sin^{2} (β) - 47 sin (β)

数字相加/相减:

- 22.09 + 144 = 121.91

= - 169 sin^{2} (β) - 47 sin (β) + 121.91

= - 169 sin^{2} (β) - 47 sin (β) + 121.91

= - 169 sin^{2} (β) - 47 sin (β) + 121.91

121.91 - 169 sin^{2} (β) - 47 sin (β) = 0

用替代法求解

121.91 - 169 sin^{2} (β) - 47 sin (β) = 0

令

: sin (β) = u

121.91 - 169 u^{2} - 47 u = 0

121.91 - 169 u^{2} - 47 u = 0 : u = - \frac{4700 + 846201600}{33800}, u = \frac{846201600 - 4700}{33800}

121.91 - 169 u^{2} - 47 u = 0

在两边乘以

100

121.91 - 169 u^{2} - 47 u = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are

2

digits to the right of the decimal point, therefore multiply by

100

121.91 \cdot 100 - 169 u^{2} \cdot 100 - 47 u \cdot 100 = 0 \cdot 100

整理后得

12191 - 16900 u^{2} - 4700 u = 0

12191 - 16900 u^{2} - 4700 u = 0

改写成标准形式

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

- 16900 u^{2} - 4700 u + 12191 = 0

使用求根公式求解

- 16900 u^{2} - 4700 u + 12191 = 0

二次方程求根公式:

若

a = - 16900, b = - 4700, c = 12191

u_{1, 2} = \frac{- ( - 4700 ) \pm ( - 4700 ) ^{2} - 4 ( - 16900 ) \cdot 12191}{2 ( - 16900 )}

u_{1, 2} = \frac{- ( - 4700 ) \pm ( - 4700 ) ^{2} - 4 ( - 16900 ) \cdot 12191}{2 ( - 16900 )}

(- 4700)^{2} - 4 (- 16900) \cdot 12191 = 846201600

(- 4700)^{2} - 4 (- 16900) \cdot 12191

使用法则

- (- a) = a

= (- 4700)^{2} + 4 \cdot 16900 \cdot 12191

使用指数法则:

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

若

n

是偶数

(- 4700)^{2} = 470 0^{2}

= 470 0^{2} + 4 \cdot 16900 \cdot 12191

数字相乘:

4 \cdot 16900 \cdot 12191 = 824111600

= 470 0^{2} + 824111600

470 0^{2} = 22090000

= 22090000 + 824111600

数字相加:

22090000 + 824111600 = 846201600

= 846201600

u_{1, 2} = \frac{- ( - 4700 ) \pm 846201600}{2 ( - 16900 )}

将解分隔开

u_{1} = \frac{- ( - 4700 ) + 846201600}{2 ( - 16900 )}, u_{2} = \frac{- ( - 4700 ) - 846201600}{2 ( - 16900 )}

u = \frac{- ( - 4700 ) + 846201600}{2 ( - 16900 )} : - \frac{4700 + 846201600}{33800}

\frac{- ( - 4700 ) + 846201600}{2 ( - 16900 )}

去除括号:

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

= \frac{4700 + 846201600}{- 2 \cdot 16900}

数字相乘:

2 \cdot 16900 = 33800

= \frac{4700 + 846201600}{- 33800}

使用分式法则:

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

= - \frac{4700 + 846201600}{33800}

u = \frac{- ( - 4700 ) - 846201600}{2 ( - 16900 )} : \frac{846201600 - 4700}{33800}

\frac{- ( - 4700 ) - 846201600}{2 ( - 16900 )}

去除括号:

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

= \frac{4700 - 846201600}{- 2 \cdot 16900}

数字相乘:

2 \cdot 16900 = 33800

= \frac{4700 - 846201600}{- 33800}

使用分式法则:

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

4700 - 846201600 = - (846201600 - 4700)

= \frac{846201600 - 4700}{33800}

二次方程组的解是:

u = - \frac{4700 + 846201600}{33800}, u = \frac{846201600 - 4700}{33800}

u = sin (β)

代回

sin (β) = - \frac{4700 + 846201600}{33800}, sin (β) = \frac{846201600 - 4700}{33800}

sin (β) = - \frac{4700 + 846201600}{33800}, sin (β) = \frac{846201600 - 4700}{33800}

sin (β) = - \frac{4700 + 846201600}{33800} : β = arcsin (- \frac{4700 + 846201600}{33800}) + 2 πn, β = π + arcsin (\frac{4700 + 846201600}{33800}) + 2 πn

sin (β) = - \frac{4700 + 846201600}{33800}

使用反三角函数性质

sin (β) = - \frac{4700 + 846201600}{33800}

sin (β) = - \frac{4700 + 846201600}{33800}

的通解

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

β = arcsin (- \frac{4700 + 846201600}{33800}) + 2 πn, β = π + arcsin (\frac{4700 + 846201600}{33800}) + 2 πn

β = arcsin (- \frac{4700 + 846201600}{33800}) + 2 πn, β = π + arcsin (\frac{4700 + 846201600}{33800}) + 2 πn

sin (β) = \frac{846201600 - 4700}{33800} : β = arcsin (\frac{846201600 - 4700}{33800}) + 2 πn, β = π - arcsin (\frac{846201600 - 4700}{33800}) + 2 πn

sin (β) = \frac{846201600 - 4700}{33800}

使用反三角函数性质

sin (β) = \frac{846201600 - 4700}{33800}

sin (β) = \frac{846201600 - 4700}{33800}

的通解

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

β = arcsin (\frac{846201600 - 4700}{33800}) + 2 πn, β = π - arcsin (\frac{846201600 - 4700}{33800}) + 2 πn

β = arcsin (\frac{846201600 - 4700}{33800}) + 2 πn, β = π - arcsin (\frac{846201600 - 4700}{33800}) + 2 πn

合并所有解

β = arcsin (- \frac{4700 + 846201600}{33800}) + 2 πn, β = π + arcsin (\frac{4700 + 846201600}{33800}) + 2 πn, β = arcsin (\frac{846201600 - 4700}{33800}) + 2 πn, β = π - arcsin (\frac{846201600 - 4700}{33800}) + 2 πn

将解代入原方程进行验证

将它们代入

12 cos (β) - 5 sin (β) = 4.7

检验解是否符合

去除与方程不符的解。

检验

arcsin (- \frac{4700 + 846201600}{33800}) + 2 πn

的解:

假

arcsin (- \frac{4700 + 846201600}{33800}) + 2 πn

代入

n = 1

arcsin (- \frac{4700 + 846201600}{33800}) + 2 π 1

对于

12 cos (β) - 5 sin (β) = 4.7

代入

β = arcsin (- \frac{4700 + 846201600}{33800}) + 2 π 1

12 cos (arcsin (- \frac{4700 + 846201600}{33800}) + 2 π 1) - 5 sin (arcsin (- \frac{4700 + 846201600}{33800}) + 2 π 1) = 4.7

整理后得

5.29690 \dots = 4.7

\Rightarrow 假

检验

π + arcsin (\frac{4700 + 846201600}{33800}) + 2 πn

的解:

真

π + arcsin (\frac{4700 + 846201600}{33800}) + 2 πn

代入

n = 1

π + arcsin (\frac{4700 + 846201600}{33800}) + 2 π 1

对于

12 cos (β) - 5 sin (β) = 4.7

代入

β = π + arcsin (\frac{4700 + 846201600}{33800}) + 2 π 1

12 cos (π + arcsin (\frac{4700 + 846201600}{33800}) + 2 π 1) - 5 sin (π + arcsin (\frac{4700 + 846201600}{33800}) + 2 π 1) = 4.7

整理后得

4.7 = 4.7

\Rightarrow 真

检验

arcsin (\frac{846201600 - 4700}{33800}) + 2 πn

的解:

真

arcsin (\frac{846201600 - 4700}{33800}) + 2 πn

代入

n = 1

arcsin (\frac{846201600 - 4700}{33800}) + 2 π 1

对于

12 cos (β) - 5 sin (β) = 4.7

代入

β = arcsin (\frac{846201600 - 4700}{33800}) + 2 π 1

12 cos (arcsin (\frac{846201600 - 4700}{33800}) + 2 π 1) - 5 sin (arcsin (\frac{846201600 - 4700}{33800}) + 2 π 1) = 4.7

整理后得

4.7 = 4.7

\Rightarrow 真

检验

π - arcsin (\frac{846201600 - 4700}{33800}) + 2 πn

的解:

假

π - arcsin (\frac{846201600 - 4700}{33800}) + 2 πn

代入

n = 1

π - arcsin (\frac{846201600 - 4700}{33800}) + 2 π 1

对于

12 cos (β) - 5 sin (β) = 4.7

代入

β = π - arcsin (\frac{846201600 - 4700}{33800}) + 2 π 1

12 cos (π - arcsin (\frac{846201600 - 4700}{33800}) + 2 π 1) - 5 sin (π - arcsin (\frac{846201600 - 4700}{33800}) + 2 π 1) = 4.7

整理后得

- 11.91584 \dots = 4.7

\Rightarrow 假

β = π + arcsin (\frac{4700 + 846201600}{33800}) + 2 πn, β = arcsin (\frac{846201600 - 4700}{33800}) + 2 πn

以小数形式表示解

β = π + 1.54592 \dots + 2 πn, β = 0.80608 \dots + 2 πn

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