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受欢迎的三角函数 >

sin(θ)-0.2cos(θ)=0.6377

解答

sin (θ) - 0.2 cos (θ) = 0.6377

解答

θ = 2.66345 \dots + 2 πn, θ = 0.87293 \dots + 2 πn

+1

度数

θ = 152.60452 \dots^{\circ} + 36 0^{\circ} n, θ = 50.01533 \dots^{\circ} + 36 0^{\circ} n

求解步骤

sin (θ) - 0.2 cos (θ) = 0.6377

两边加上

0.2 cos (θ)

sin (θ) = 0.6377 + 0.2 cos (θ)

两边进行平方

sin^{2} (θ) = (0.6377 + 0.2 cos (θ))^{2}

两边减去

(0.6377 + 0.2 cos (θ))^{2}

sin^{2} (θ) - 0.40666129 - 0.25508 cos (θ) - 0.04 cos^{2} (θ) = 0

使用三角恒等式改写

- 0.40666129 + sin^{2} (θ) - 0.04 cos^{2} (θ) - 0.25508 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)

= - 0.40666129 + 1 - cos^{2} (θ) - 0.04 cos^{2} (θ) - 0.25508 cos (θ)

化简

- 0.40666129 + 1 - cos^{2} (θ) - 0.04 cos^{2} (θ) - 0.25508 cos (θ) : - 1.04 cos^{2} (θ) - 0.25508 cos (θ) + 0.59333871

- 0.40666129 + 1 - cos^{2} (θ) - 0.04 cos^{2} (θ) - 0.25508 cos (θ)

同类项相加:

- cos^{2} (θ) - 0.04 cos^{2} (θ) = - 1.04 cos^{2} (θ)

= - 0.40666129 + 1 - 1.04 cos^{2} (θ) - 0.25508 cos (θ)

数字相加/相减:

- 0.40666129 + 1 = 0.59333871

= - 1.04 cos^{2} (θ) - 0.25508 cos (θ) + 0.59333871

= - 1.04 cos^{2} (θ) - 0.25508 cos (θ) + 0.59333871

0.59333871 - 0.25508 cos (θ) - 1.04 cos^{2} (θ) = 0

用替代法求解

0.59333871 - 0.25508 cos (θ) - 1.04 cos^{2} (θ) = 0

令

: cos (θ) = u

0.59333871 - 0.25508 u - 1.04 u^{2} = 0

0.59333871 - 0.25508 u - 1.04 u^{2} = 0 : u = - \frac{0.25508 + 2.53335484}{2.08}, u = \frac{2.53335484 - 0.25508}{2.08}

0.59333871 - 0.25508 u - 1.04 u^{2} = 0

改写成标准形式

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

- 1.04 u^{2} - 0.25508 u + 0.59333871 = 0

使用求根公式求解

- 1.04 u^{2} - 0.25508 u + 0.59333871 = 0

二次方程求根公式:

若

a = - 1.04, b = - 0.25508, c = 0.59333871

u_{1, 2} = \frac{- ( - 0.25508 ) \pm ( - 0.25508 ) ^{2} - 4 ( - 1.04 ) \cdot 0.59333871}{2 ( - 1.04 )}

u_{1, 2} = \frac{- ( - 0.25508 ) \pm ( - 0.25508 ) ^{2} - 4 ( - 1.04 ) \cdot 0.59333871}{2 ( - 1.04 )}

(- 0.25508)^{2} - 4 (- 1.04) \cdot 0.59333871 = 2.53335484

(- 0.25508)^{2} - 4 (- 1.04) \cdot 0.59333871

使用法则

- (- a) = a

= (- 0.25508)^{2} + 4 \cdot 1.04 \cdot 0.59333871

使用指数法则:

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

若

n

是偶数

(- 0.25508)^{2} = 0.2550 8^{2}

= 0.2550 8^{2} + 4 \cdot 0.59333871 \cdot 1.04

数字相乘:

4 \cdot 1.04 \cdot 0.59333871 = 2.46828 \dots

= 0.2550 8^{2} + 2.46828 \dots

0.2550 8^{2} = 0.0650658064

= 0.0650658064 + 2.46828 \dots

数字相加:

0.0650658064 + 2.46828 \dots = 2.53335484

= 2.53335484

u_{1, 2} = \frac{- ( - 0.25508 ) \pm 2.53335484}{2 ( - 1.04 )}

将解分隔开

u_{1} = \frac{- ( - 0.25508 ) + 2.53335484}{2 ( - 1.04 )}, u_{2} = \frac{- ( - 0.25508 ) - 2.53335484}{2 ( - 1.04 )}

u = \frac{- ( - 0.25508 ) + 2.53335484}{2 ( - 1.04 )} : - \frac{0.25508 + 2.53335484}{2.08}

\frac{- ( - 0.25508 ) + 2.53335484}{2 ( - 1.04 )}

去除括号:

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

= \frac{0.25508 + 2.53335484}{- 2 \cdot 1.04}

数字相乘:

2 \cdot 1.04 = 2.08

= \frac{0.25508 + 2.53335484}{- 2.08}

使用分式法则:

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

= - \frac{0.25508 + 2.53335484}{2.08}

u = \frac{- ( - 0.25508 ) - 2.53335484}{2 ( - 1.04 )} : \frac{2.53335484 - 0.25508}{2.08}

\frac{- ( - 0.25508 ) - 2.53335484}{2 ( - 1.04 )}

去除括号:

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

= \frac{0.25508 - 2.53335484}{- 2 \cdot 1.04}

数字相乘:

2 \cdot 1.04 = 2.08

= \frac{0.25508 - 2.53335484}{- 2.08}

使用分式法则:

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

0.25508 - 2.53335484 = - (2.53335484 - 0.25508)

= \frac{2.53335484 - 0.25508}{2.08}

二次方程组的解是:

u = - \frac{0.25508 + 2.53335484}{2.08}, u = \frac{2.53335484 - 0.25508}{2.08}

u = cos (θ)

代回

cos (θ) = - \frac{0.25508 + 2.53335484}{2.08}, cos (θ) = \frac{2.53335484 - 0.25508}{2.08}

cos (θ) = - \frac{0.25508 + 2.53335484}{2.08}, cos (θ) = \frac{2.53335484 - 0.25508}{2.08}

cos (θ) = - \frac{0.25508 + 2.53335484}{2.08} : θ = arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 πn, θ = - arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 πn

cos (θ) = - \frac{0.25508 + 2.53335484}{2.08}

使用反三角函数性质

cos (θ) = - \frac{0.25508 + 2.53335484}{2.08}

cos (θ) = - \frac{0.25508 + 2.53335484}{2.08}

的通解

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

θ = arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 πn, θ = - arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 πn

θ = arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 πn, θ = - arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 πn

cos (θ) = \frac{2.53335484 - 0.25508}{2.08} : θ = arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 πn, θ = 2 π - arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 πn

cos (θ) = \frac{2.53335484 - 0.25508}{2.08}

使用反三角函数性质

cos (θ) = \frac{2.53335484 - 0.25508}{2.08}

cos (θ) = \frac{2.53335484 - 0.25508}{2.08}

的通解

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

θ = arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 πn, θ = 2 π - arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 πn

θ = arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 πn, θ = 2 π - arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 πn

合并所有解

θ = arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 πn, θ = - arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 πn, θ = arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 πn, θ = 2 π - arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 πn

将解代入原方程进行验证

将它们代入

sin (θ) - 0.2 cos (θ) = 0.6377

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 πn

的解:

真

arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 πn

代入

n = 1

arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 π 1

对于

sin (θ) - 0.2 cos (θ) = 0.6377

代入

θ = arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 π 1

sin (arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 π 1) - 0.2 cos (arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 π 1) = 0.6377

整理后得

0.6377 = 0.6377

\Rightarrow 真

检验

- arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 πn

的解:

假

- arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 πn

代入

n = 1

- arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 π 1

对于

sin (θ) - 0.2 cos (θ) = 0.6377

代入

θ = - arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 π 1

sin (- arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 π 1) - 0.2 cos (- arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 π 1) = 0.6377

整理后得

- 0.28255 \dots = 0.6377

\Rightarrow 假

检验

arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 πn

的解:

真

arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 πn

代入

n = 1

arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 π 1

对于

sin (θ) - 0.2 cos (θ) = 0.6377

代入

θ = arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 π 1

sin (arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 π 1) - 0.2 cos (arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 π 1) = 0.6377

整理后得

0.6377 = 0.6377

\Rightarrow 真

检验

2 π - arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 πn

的解:

假

2 π - arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 πn

代入

n = 1

2 π - arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 π 1

对于

sin (θ) - 0.2 cos (θ) = 0.6377

代入

θ = 2 π - arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 π 1

sin (2 π - arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 π 1) - 0.2 cos (2 π - arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 π 1) = 0.6377

整理后得

- 0.89473 \dots = 0.6377

\Rightarrow 假

θ = arccos (- \frac{0.25508 + 2.53335484}{2.08}) + 2 πn, θ = arccos (\frac{2.53335484 - 0.25508}{2.08}) + 2 πn

以小数形式表示解

θ = 2.66345 \dots + 2 πn, θ = 0.87293 \dots + 2 πn

作图

流行的例子

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