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受欢迎的三角函数 >

sin(θ)-0.1cos(θ)=(8.87)/(9.8)

解答

sin (θ) - 0.1 cos (θ) = \frac{8.87}{9.8}

解答

θ = 2.12008 \dots + 2 πn, θ = 1.22084 \dots + 2 πn

+1

度数

θ = 121.47220 \dots^{\circ} + 36 0^{\circ} n, θ = 69.94898 \dots^{\circ} + 36 0^{\circ} n

求解步骤

sin (θ) - 0.1 cos (θ) = \frac{8.87}{9.8}

两边加上

0.1 cos (θ)

sin (θ) = 0.90510 \dots + 0.1 cos (θ)

两边进行平方

sin^{2} (θ) = (0.90510 \dots + 0.1 cos (θ))^{2}

两边减去

(0.90510 \dots + 0.1 cos (θ))^{2}

sin^{2} (θ) - 0.81920 \dots - 0.18102 \dots cos (θ) - 0.01 cos^{2} (θ) = 0

使用三角恒等式改写

- 0.81920 \dots + sin^{2} (θ) - 0.01 cos^{2} (θ) - 0.18102 \dots cos (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)

= - 0.81920 \dots + 1 - cos^{2} (θ) - 0.01 cos^{2} (θ) - 0.18102 \dots cos (θ)

化简

- 0.81920 \dots + 1 - cos^{2} (θ) - 0.01 cos^{2} (θ) - 0.18102 \dots cos (θ) : - 1.01 cos^{2} (θ) - 0.18102 \dots cos (θ) + 0.18079 \dots

- 0.81920 \dots + 1 - cos^{2} (θ) - 0.01 cos^{2} (θ) - 0.18102 \dots cos (θ)

同类项相加:

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

= - 0.81920 \dots + 1 - 1.01 cos^{2} (θ) - 0.18102 \dots cos (θ)

数字相加/相减:

- 0.81920 \dots + 1 = 0.18079 \dots

= - 1.01 cos^{2} (θ) - 0.18102 \dots cos (θ) + 0.18079 \dots

= - 1.01 cos^{2} (θ) - 0.18102 \dots cos (θ) + 0.18079 \dots

0.18079 \dots - 0.18102 \dots cos (θ) - 1.01 cos^{2} (θ) = 0

用替代法求解

0.18079 \dots - 0.18102 \dots cos (θ) - 1.01 cos^{2} (θ) = 0

令

: cos (θ) = u

0.18079 \dots - 0.18102 \dots u - 1.01 u^{2} = 0

0.18079 \dots - 0.18102 \dots u - 1.01 u^{2} = 0 : u = - \frac{0.18102 \dots + 0.76316 \dots}{2.02}, u = \frac{0.76316 \dots - 0.18102 \dots}{2.02}

0.18079 \dots - 0.18102 \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} - 0.18102 \dots u + 0.18079 \dots = 0

使用求根公式求解

- 1.01 u^{2} - 0.18102 \dots u + 0.18079 \dots = 0

二次方程求根公式:

若

a = - 1.01, b = - 0.18102 \dots, c = 0.18079 \dots

u_{1, 2} = \frac{- ( - 0.18102 \dots ) \pm ( - 0.18102 \dots ) ^{2} - 4 ( - 1.01 ) \cdot 0.18079 \dots}{2 ( - 1.01 )}

u_{1, 2} = \frac{- ( - 0.18102 \dots ) \pm ( - 0.18102 \dots ) ^{2} - 4 ( - 1.01 ) \cdot 0.18079 \dots}{2 ( - 1.01 )}

(- 0.18102 \dots)^{2} - 4 (- 1.01) \cdot 0.18079 \dots = 0.76316 \dots

(- 0.18102 \dots)^{2} - 4 (- 1.01) \cdot 0.18079 \dots

使用法则

- (- a) = a

= (- 0.18102 \dots)^{2} + 4 \cdot 1.01 \cdot 0.18079 \dots

使用指数法则:

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

若

n

是偶数

(- 0.18102 \dots)^{2} = 0.18102 \dots^{2}

= 0.18102 \dots^{2} + 4 \cdot 0.18079 \dots \cdot 1.01

数字相乘:

4 \cdot 1.01 \cdot 0.18079 \dots = 0.73039 \dots

= 0.18102 \dots^{2} + 0.73039 \dots

0.18102 \dots^{2} = 0.03276 \dots

= 0.03276 \dots + 0.73039 \dots

数字相加:

0.03276 \dots + 0.73039 \dots = 0.76316 \dots

= 0.76316 \dots

u_{1, 2} = \frac{- ( - 0.18102 \dots ) \pm 0.76316 \dots}{2 ( - 1.01 )}

将解分隔开

u_{1} = \frac{- ( - 0.18102 \dots ) + 0.76316 \dots}{2 ( - 1.01 )}, u_{2} = \frac{- ( - 0.18102 \dots ) - 0.76316 \dots}{2 ( - 1.01 )}

u = \frac{- ( - 0.18102 \dots ) + 0.76316 \dots}{2 ( - 1.01 )} : - \frac{0.18102 \dots + 0.76316 \dots}{2.02}

\frac{- ( - 0.18102 \dots ) + 0.76316 \dots}{2 ( - 1.01 )}

去除括号:

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

= \frac{0.18102 \dots + 0.76316 \dots}{- 2 \cdot 1.01}

数字相乘:

2 \cdot 1.01 = 2.02

= \frac{0.18102 \dots + 0.76316 \dots}{- 2.02}

使用分式法则:

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

= - \frac{0.18102 \dots + 0.76316 \dots}{2.02}

u = \frac{- ( - 0.18102 \dots ) - 0.76316 \dots}{2 ( - 1.01 )} : \frac{0.76316 \dots - 0.18102 \dots}{2.02}

\frac{- ( - 0.18102 \dots ) - 0.76316 \dots}{2 ( - 1.01 )}

去除括号:

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

= \frac{0.18102 \dots - 0.76316 \dots}{- 2 \cdot 1.01}

数字相乘:

2 \cdot 1.01 = 2.02

= \frac{0.18102 \dots - 0.76316 \dots}{- 2.02}

使用分式法则:

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

0.18102 \dots - 0.76316 \dots = - (0.76316 \dots - 0.18102 \dots)

= \frac{0.76316 \dots - 0.18102 \dots}{2.02}

二次方程组的解是:

u = - \frac{0.18102 \dots + 0.76316 \dots}{2.02}, u = \frac{0.76316 \dots - 0.18102 \dots}{2.02}

u = cos (θ)

代回

cos (θ) = - \frac{0.18102 \dots + 0.76316 \dots}{2.02}, cos (θ) = \frac{0.76316 \dots - 0.18102 \dots}{2.02}

cos (θ) = - \frac{0.18102 \dots + 0.76316 \dots}{2.02}, cos (θ) = \frac{0.76316 \dots - 0.18102 \dots}{2.02}

cos (θ) = - \frac{0.18102 \dots + 0.76316 \dots}{2.02} : θ = arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 πn, θ = - arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 πn

cos (θ) = - \frac{0.18102 \dots + 0.76316 \dots}{2.02}

使用反三角函数性质

cos (θ) = - \frac{0.18102 \dots + 0.76316 \dots}{2.02}

cos (θ) = - \frac{0.18102 \dots + 0.76316 \dots}{2.02}

的通解

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

θ = arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 πn, θ = - arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 πn

θ = arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 πn, θ = - arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 πn

cos (θ) = \frac{0.76316 \dots - 0.18102 \dots}{2.02} : θ = arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 πn, θ = 2 π - arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 πn

cos (θ) = \frac{0.76316 \dots - 0.18102 \dots}{2.02}

使用反三角函数性质

cos (θ) = \frac{0.76316 \dots - 0.18102 \dots}{2.02}

cos (θ) = \frac{0.76316 \dots - 0.18102 \dots}{2.02}

的通解

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

θ = arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 πn, θ = 2 π - arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 πn

θ = arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 πn, θ = 2 π - arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 πn

合并所有解

θ = arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 πn, θ = - arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 πn, θ = arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 πn, θ = 2 π - arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 πn

将解代入原方程进行验证

将它们代入

sin (θ) - 0.1 cos (θ) = \frac{8.87}{9.8}

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 πn

的解:

真

arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 πn

代入

n = 1

arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 π 1

对于

sin (θ) - 0.1 cos (θ) = \frac{8.87}{9.8}

代入

θ = arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 π 1

sin (arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 π 1) - 0.1 cos (arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 π 1) = \frac{8.87}{9.8}

整理后得

0.90510 \dots = 0.90510 \dots

\Rightarrow 真

检验

- arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 πn

的解:

假

- arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 πn

代入

n = 1

- arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 π 1

对于

sin (θ) - 0.1 cos (θ) = \frac{8.87}{9.8}

代入

θ = - arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 π 1

sin (- arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 π 1) - 0.1 cos (- arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 π 1) = \frac{8.87}{9.8}

整理后得

- 0.80068 \dots = 0.90510 \dots

\Rightarrow 假

检验

arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 πn

的解:

真

arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 πn

代入

n = 1

arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 π 1

对于

sin (θ) - 0.1 cos (θ) = \frac{8.87}{9.8}

代入

θ = arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 π 1

sin (arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 π 1) - 0.1 cos (arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 π 1) = \frac{8.87}{9.8}

整理后得

0.90510 \dots = 0.90510 \dots

\Rightarrow 真

检验

2 π - arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 πn

的解:

假

2 π - arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 πn

代入

n = 1

2 π - arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 π 1

对于

sin (θ) - 0.1 cos (θ) = \frac{8.87}{9.8}

代入

θ = 2 π - arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 π 1

sin (2 π - arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 π 1) - 0.1 cos (2 π - arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 π 1) = \frac{8.87}{9.8}

整理后得

- 0.97367 \dots = 0.90510 \dots

\Rightarrow 假

θ = arccos (- \frac{0.18102 \dots + 0.76316 \dots}{2.02}) + 2 πn, θ = arccos (\frac{0.76316 \dots - 0.18102 \dots}{2.02}) + 2 πn

以小数形式表示解

θ = 2.12008 \dots + 2 πn, θ = 1.22084 \dots + 2 πn

作图

流行的例子

(cos(x))/2 =(sqrt(3))/2

\frac{cos ( x )}{2} = \frac{3}{2}

2 cot (2 x) + 2 = 0

sec^2(θ)tan(θ)=2tan(θ),0<= θ<= 360

sec^{2} (θ) tan (θ) = 2 tan (θ), 0^{\circ} \leq θ \leq 36 0^{\circ}

arccos (x) = 18 0^{\circ}

sin(a)=(29.7)/(54.5)

sin (a) = \frac{29.7}{54.5}