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受欢迎的三角函数 >

sin(A)-0.1*cos(A)=(6.94)/(9.8)

解答

sin (A) - 0.1 \cdot cos (A) = \frac{6.94}{9.8}

解答

A = 2.45933 \dots + 2 πn, A = 0.88159 \dots + 2 πn

+1

度数

A = 140.90941 \dots^{\circ} + 36 0^{\circ} n, A = 50.51177 \dots^{\circ} + 36 0^{\circ} n

求解步骤

sin (A) - 0.1 cos (A) = \frac{6.94}{9.8}

两边加上

0.1 cos (A)

sin (A) = 0.70816 \dots + 0.1 cos (A)

两边进行平方

sin^{2} (A) = (0.70816 \dots + 0.1 cos (A))^{2}

两边减去

(0.70816 \dots + 0.1 cos (A))^{2}

sin^{2} (A) - 0.50149 \dots - 0.14163 \dots cos (A) - 0.01 cos^{2} (A) = 0

使用三角恒等式改写

- 0.50149 \dots + sin^{2} (A) - 0.01 cos^{2} (A) - 0.14163 \dots cos (A)

使用毕达哥拉斯恒等式:

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

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

= - 0.50149 \dots + 1 - cos^{2} (A) - 0.01 cos^{2} (A) - 0.14163 \dots cos (A)

化简

- 0.50149 \dots + 1 - cos^{2} (A) - 0.01 cos^{2} (A) - 0.14163 \dots cos (A) : - 1.01 cos^{2} (A) - 0.14163 \dots cos (A) + 0.49850 \dots

- 0.50149 \dots + 1 - cos^{2} (A) - 0.01 cos^{2} (A) - 0.14163 \dots cos (A)

同类项相加:

- cos^{2} (A) - 0.01 cos^{2} (A) = - 1.01 cos^{2} (A)

= - 0.50149 \dots + 1 - 1.01 cos^{2} (A) - 0.14163 \dots cos (A)

数字相加/相减:

- 0.50149 \dots + 1 = 0.49850 \dots

= - 1.01 cos^{2} (A) - 0.14163 \dots cos (A) + 0.49850 \dots

= - 1.01 cos^{2} (A) - 0.14163 \dots cos (A) + 0.49850 \dots

0.49850 \dots - 0.14163 \dots cos (A) - 1.01 cos^{2} (A) = 0

用替代法求解

0.49850 \dots - 0.14163 \dots cos (A) - 1.01 cos^{2} (A) = 0

令

: cos (A) = u

0.49850 \dots - 0.14163 \dots u - 1.01 u^{2} = 0

0.49850 \dots - 0.14163 \dots u - 1.01 u^{2} = 0 : u = - \frac{0.14163 \dots + 2.03401 \dots}{2.02}, u = \frac{2.03401 \dots - 0.14163 \dots}{2.02}

0.49850 \dots - 0.14163 \dots u - 1.01 u^{2} = 0

改写成标准形式

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

- 1.01 u^{2} - 0.14163 \dots u + 0.49850 \dots = 0

使用求根公式求解

- 1.01 u^{2} - 0.14163 \dots u + 0.49850 \dots = 0

二次方程求根公式:

若

a = - 1.01, b = - 0.14163 \dots, c = 0.49850 \dots

u_{1, 2} = \frac{- ( - 0.14163 \dots ) \pm ( - 0.14163 \dots ) ^{2} - 4 ( - 1.01 ) \cdot 0.49850 \dots}{2 ( - 1.01 )}

u_{1, 2} = \frac{- ( - 0.14163 \dots ) \pm ( - 0.14163 \dots ) ^{2} - 4 ( - 1.01 ) \cdot 0.49850 \dots}{2 ( - 1.01 )}

(- 0.14163 \dots)^{2} - 4 (- 1.01) \cdot 0.49850 \dots = 2.03401 \dots

(- 0.14163 \dots)^{2} - 4 (- 1.01) \cdot 0.49850 \dots

使用法则

- (- a) = a

= (- 0.14163 \dots)^{2} + 4 \cdot 1.01 \cdot 0.49850 \dots

使用指数法则:

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

若

n

是偶数

(- 0.14163 \dots)^{2} = 0.14163 \dots^{2}

= 0.14163 \dots^{2} + 4 \cdot 0.49850 \dots \cdot 1.01

数字相乘:

4 \cdot 1.01 \cdot 0.49850 \dots = 2.01395 \dots

= 0.14163 \dots^{2} + 2.01395 \dots

0.14163 \dots^{2} = 0.02005 \dots

= 0.02005 \dots + 2.01395 \dots

数字相加:

0.02005 \dots + 2.01395 \dots = 2.03401 \dots

= 2.03401 \dots

u_{1, 2} = \frac{- ( - 0.14163 \dots ) \pm 2.03401 \dots}{2 ( - 1.01 )}

将解分隔开

u_{1} = \frac{- ( - 0.14163 \dots ) + 2.03401 \dots}{2 ( - 1.01 )}, u_{2} = \frac{- ( - 0.14163 \dots ) - 2.03401 \dots}{2 ( - 1.01 )}

u = \frac{- ( - 0.14163 \dots ) + 2.03401 \dots}{2 ( - 1.01 )} : - \frac{0.14163 \dots + 2.03401 \dots}{2.02}

\frac{- ( - 0.14163 \dots ) + 2.03401 \dots}{2 ( - 1.01 )}

去除括号:

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

= \frac{0.14163 \dots + 2.03401 \dots}{- 2 \cdot 1.01}

数字相乘:

2 \cdot 1.01 = 2.02

= \frac{0.14163 \dots + 2.03401 \dots}{- 2.02}

使用分式法则:

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

= - \frac{0.14163 \dots + 2.03401 \dots}{2.02}

u = \frac{- ( - 0.14163 \dots ) - 2.03401 \dots}{2 ( - 1.01 )} : \frac{2.03401 \dots - 0.14163 \dots}{2.02}

\frac{- ( - 0.14163 \dots ) - 2.03401 \dots}{2 ( - 1.01 )}

去除括号:

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

= \frac{0.14163 \dots - 2.03401 \dots}{- 2 \cdot 1.01}

数字相乘:

2 \cdot 1.01 = 2.02

= \frac{0.14163 \dots - 2.03401 \dots}{- 2.02}

使用分式法则:

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

0.14163 \dots - 2.03401 \dots = - (2.03401 \dots - 0.14163 \dots)

= \frac{2.03401 \dots - 0.14163 \dots}{2.02}

二次方程组的解是:

u = - \frac{0.14163 \dots + 2.03401 \dots}{2.02}, u = \frac{2.03401 \dots - 0.14163 \dots}{2.02}

u = cos (A)

代回

cos (A) = - \frac{0.14163 \dots + 2.03401 \dots}{2.02}, cos (A) = \frac{2.03401 \dots - 0.14163 \dots}{2.02}

cos (A) = - \frac{0.14163 \dots + 2.03401 \dots}{2.02}, cos (A) = \frac{2.03401 \dots - 0.14163 \dots}{2.02}

cos (A) = - \frac{0.14163 \dots + 2.03401 \dots}{2.02} : A = arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 πn, A = - arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 πn

cos (A) = - \frac{0.14163 \dots + 2.03401 \dots}{2.02}

使用反三角函数性质

cos (A) = - \frac{0.14163 \dots + 2.03401 \dots}{2.02}

cos (A) = - \frac{0.14163 \dots + 2.03401 \dots}{2.02}

的通解

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

A = arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 πn, A = - arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 πn

A = arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 πn, A = - arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 πn

cos (A) = \frac{2.03401 \dots - 0.14163 \dots}{2.02} : A = arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 πn, A = 2 π - arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 πn

cos (A) = \frac{2.03401 \dots - 0.14163 \dots}{2.02}

使用反三角函数性质

cos (A) = \frac{2.03401 \dots - 0.14163 \dots}{2.02}

cos (A) = \frac{2.03401 \dots - 0.14163 \dots}{2.02}

的通解

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

A = arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 πn, A = 2 π - arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 πn

A = arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 πn, A = 2 π - arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 πn

合并所有解

A = arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 πn, A = - arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 πn, A = arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 πn, A = 2 π - arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 πn

将解代入原方程进行验证

将它们代入

sin (A) - 0.1 cos (A) = \frac{6.94}{9.8}

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 πn

的解:

真

arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 πn

代入

n = 1

arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 π 1

对于

sin (A) - 0.1 cos (A) = \frac{6.94}{9.8}

代入

A = arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 π 1

sin (arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 π 1) - 0.1 cos (arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 π 1) = \frac{6.94}{9.8}

整理后得

0.70816 \dots = 0.70816 \dots

\Rightarrow 真

检验

- arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 πn

的解:

假

- arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 πn

代入

n = 1

- arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 π 1

对于

sin (A) - 0.1 cos (A) = \frac{6.94}{9.8}

代入

A = - arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 π 1

sin (- arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 π 1) - 0.1 cos (- arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 π 1) = \frac{6.94}{9.8}

整理后得

- 0.55293 \dots = 0.70816 \dots

\Rightarrow 假

检验

arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 πn

的解:

真

arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 πn

代入

n = 1

arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 π 1

对于

sin (A) - 0.1 cos (A) = \frac{6.94}{9.8}

代入

A = arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 π 1

sin (arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 π 1) - 0.1 cos (arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 π 1) = \frac{6.94}{9.8}

整理后得

0.70816 \dots = 0.70816 \dots

\Rightarrow 真

检验

2 π - arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 πn

的解:

假

2 π - arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 πn

代入

n = 1

2 π - arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 π 1

对于

sin (A) - 0.1 cos (A) = \frac{6.94}{9.8}

代入

A = 2 π - arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 π 1

sin (2 π - arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 π 1) - 0.1 cos (2 π - arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 π 1) = \frac{6.94}{9.8}

整理后得

- 0.83534 \dots = 0.70816 \dots

\Rightarrow 假

A = arccos (- \frac{0.14163 \dots + 2.03401 \dots}{2.02}) + 2 πn, A = arccos (\frac{2.03401 \dots - 0.14163 \dots}{2.02}) + 2 πn

以小数形式表示解

A = 2.45933 \dots + 2 πn, A = 0.88159 \dots + 2 πn

作图

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