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受欢迎的三角函数 >

49.55cos(θ)-30sin(θ)=1.225

解答

49.55 cos (θ) - 30 sin (θ) = 1.225

解答

θ = π + 1.04752 \dots + 2 πn, θ = 1.00522 \dots + 2 πn

+1

度数

θ = 240.01903 \dots^{\circ} + 36 0^{\circ} n, θ = 57.59542 \dots^{\circ} + 36 0^{\circ} n

求解步骤

49.55 cos (θ) - 30 sin (θ) = 1.225

两边加上

30 sin (θ)

49.55 cos (θ) = 1.225 + 30 sin (θ)

两边进行平方

(49.55 cos (θ))^{2} = (1.225 + 30 sin (θ))^{2}

两边减去

(1.225 + 30 sin (θ))^{2}

2455.2025 cos^{2} (θ) - 1.500625 - 73.5 sin (θ) - 900 sin^{2} (θ) = 0

使用三角恒等式改写

- 1.500625 + 2455.2025 cos^{2} (θ) - 73.5 sin (θ) - 900 sin^{2} (θ)

使用毕达哥拉斯恒等式:

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

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

= - 1.500625 + 2455.2025 (1 - sin^{2} (θ)) - 73.5 sin (θ) - 900 sin^{2} (θ)

化简

- 1.500625 + 2455.2025 (1 - sin^{2} (θ)) - 73.5 sin (θ) - 900 sin^{2} (θ) : - 3355.2025 sin^{2} (θ) - 73.5 sin (θ) + 2453.701875

- 1.500625 + 2455.2025 (1 - sin^{2} (θ)) - 73.5 sin (θ) - 900 sin^{2} (θ)

乘开

2455.2025 (1 - sin^{2} (θ)) : 2455.2025 - 2455.2025 sin^{2} (θ)

2455.2025 (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 2455.2025, b = 1, c = sin^{2} (θ)

= 2455.2025 \cdot 1 - 2455.2025 sin^{2} (θ)

= 1 \cdot 2455.2025 - 2455.2025 sin^{2} (θ)

数字相乘:

1 \cdot 2455.2025 = 2455.2025

= 2455.2025 - 2455.2025 sin^{2} (θ)

= - 1.500625 + 2455.2025 - 2455.2025 sin^{2} (θ) - 73.5 sin (θ) - 900 sin^{2} (θ)

化简

- 1.500625 + 2455.2025 - 2455.2025 sin^{2} (θ) - 73.5 sin (θ) - 900 sin^{2} (θ) : - 3355.2025 sin^{2} (θ) - 73.5 sin (θ) + 2453.701875

- 1.500625 + 2455.2025 - 2455.2025 sin^{2} (θ) - 73.5 sin (θ) - 900 sin^{2} (θ)

对同类项分组

= - 2455.2025 sin^{2} (θ) - 73.5 sin (θ) - 900 sin^{2} (θ) - 1.500625 + 2455.2025

同类项相加:

- 2455.2025 sin^{2} (θ) - 900 sin^{2} (θ) = - 3355.2025 sin^{2} (θ)

= - 3355.2025 sin^{2} (θ) - 73.5 sin (θ) - 1.500625 + 2455.2025

数字相加/相减:

- 1.500625 + 2455.2025 = 2453.701875

= - 3355.2025 sin^{2} (θ) - 73.5 sin (θ) + 2453.701875

= - 3355.2025 sin^{2} (θ) - 73.5 sin (θ) + 2453.701875

= - 3355.2025 sin^{2} (θ) - 73.5 sin (θ) + 2453.701875

2453.701875 - 3355.2025 sin^{2} (θ) - 73.5 sin (θ) = 0

用替代法求解

2453.701875 - 3355.2025 sin^{2} (θ) - 73.5 sin (θ) = 0

令

: sin (θ) = u

2453.701875 - 3355.2025 u^{2} - 73.5 u = 0

2453.701875 - 3355.2025 u^{2} - 73.5 u = 0 : u = - \frac{73.5 + 32936068.91101 \dots}{6710.405}, u = \frac{32936068.91101 \dots - 73.5}{6710.405}

2453.701875 - 3355.2025 u^{2} - 73.5 u = 0

改写成标准形式

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

- 3355.2025 u^{2} - 73.5 u + 2453.701875 = 0

使用求根公式求解

- 3355.2025 u^{2} - 73.5 u + 2453.701875 = 0

二次方程求根公式:

若

a = - 3355.2025, b = - 73.5, c = 2453.701875

u_{1, 2} = \frac{- ( - 73.5 ) \pm ( - 73.5 ) ^{2} - 4 ( - 3355.2025 ) \cdot 2453.701875}{2 ( - 3355.2025 )}

u_{1, 2} = \frac{- ( - 73.5 ) \pm ( - 73.5 ) ^{2} - 4 ( - 3355.2025 ) \cdot 2453.701875}{2 ( - 3355.2025 )}

(- 73.5)^{2} - 4 (- 3355.2025) \cdot 2453.701875 = 32936068.91101 \dots

(- 73.5)^{2} - 4 (- 3355.2025) \cdot 2453.701875

使用法则

- (- a) = a

= (- 73.5)^{2} + 4 \cdot 3355.2025 \cdot 2453.701875

使用指数法则:

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

若

n

是偶数

(- 73.5)^{2} = 73. 5^{2}

= 73. 5^{2} + 4 \cdot 2453.701875 \cdot 3355.2025

数字相乘:

4 \cdot 3355.2025 \cdot 2453.701875 = 32930666.66101 \dots

= 73. 5^{2} + 32930666.66101 \dots

73. 5^{2} = 5402.25

= 5402.25 + 32930666.66101 \dots

数字相加:

5402.25 + 32930666.66101 \dots = 32936068.91101 \dots

= 32936068.91101 \dots

u_{1, 2} = \frac{- ( - 73.5 ) \pm 32936068.91101 \dots}{2 ( - 3355.2025 )}

将解分隔开

u_{1} = \frac{- ( - 73.5 ) + 32936068.91101 \dots}{2 ( - 3355.2025 )}, u_{2} = \frac{- ( - 73.5 ) - 32936068.91101 \dots}{2 ( - 3355.2025 )}

u = \frac{- ( - 73.5 ) + 32936068.91101 \dots}{2 ( - 3355.2025 )} : - \frac{73.5 + 32936068.91101 \dots}{6710.405}

\frac{- ( - 73.5 ) + 32936068.91101 \dots}{2 ( - 3355.2025 )}

去除括号:

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

= \frac{73.5 + 32936068.91101 \dots}{- 2 \cdot 3355.2025}

数字相乘:

2 \cdot 3355.2025 = 6710.405

= \frac{73.5 + 32936068.91101 \dots}{- 6710.405}

使用分式法则:

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

= - \frac{73.5 + 32936068.91101 \dots}{6710.405}

u = \frac{- ( - 73.5 ) - 32936068.91101 \dots}{2 ( - 3355.2025 )} : \frac{32936068.91101 \dots - 73.5}{6710.405}

\frac{- ( - 73.5 ) - 32936068.91101 \dots}{2 ( - 3355.2025 )}

去除括号:

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

= \frac{73.5 - 32936068.91101 \dots}{- 2 \cdot 3355.2025}

数字相乘:

2 \cdot 3355.2025 = 6710.405

= \frac{73.5 - 32936068.91101 \dots}{- 6710.405}

使用分式法则:

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

73.5 - 32936068.91101 \dots = - (32936068.91101 \dots - 73.5)

= \frac{32936068.91101 \dots - 73.5}{6710.405}

二次方程组的解是:

u = - \frac{73.5 + 32936068.91101 \dots}{6710.405}, u = \frac{32936068.91101 \dots - 73.5}{6710.405}

u = sin (θ)

代回

sin (θ) = - \frac{73.5 + 32936068.91101 \dots}{6710.405}, sin (θ) = \frac{32936068.91101 \dots - 73.5}{6710.405}

sin (θ) = - \frac{73.5 + 32936068.91101 \dots}{6710.405}, sin (θ) = \frac{32936068.91101 \dots - 73.5}{6710.405}

sin (θ) = - \frac{73.5 + 32936068.91101 \dots}{6710.405} : θ = arcsin (- \frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 πn, θ = π + arcsin (\frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 πn

sin (θ) = - \frac{73.5 + 32936068.91101 \dots}{6710.405}

使用反三角函数性质

sin (θ) = - \frac{73.5 + 32936068.91101 \dots}{6710.405}

sin (θ) = - \frac{73.5 + 32936068.91101 \dots}{6710.405}

的通解

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

θ = arcsin (- \frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 πn, θ = π + arcsin (\frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 πn

θ = arcsin (- \frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 πn, θ = π + arcsin (\frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 πn

sin (θ) = \frac{32936068.91101 \dots - 73.5}{6710.405} : θ = arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 πn, θ = π - arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 πn

sin (θ) = \frac{32936068.91101 \dots - 73.5}{6710.405}

使用反三角函数性质

sin (θ) = \frac{32936068.91101 \dots - 73.5}{6710.405}

sin (θ) = \frac{32936068.91101 \dots - 73.5}{6710.405}

的通解

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

θ = arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 πn, θ = π - arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 πn

θ = arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 πn, θ = π - arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 πn

合并所有解

θ = arcsin (- \frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 πn, θ = π + arcsin (\frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 πn, θ = arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 πn, θ = π - arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 πn

将解代入原方程进行验证

将它们代入

49.55 cos (θ) - 30 sin (θ) = 1.225

检验解是否符合

去除与方程不符的解。

检验

arcsin (- \frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 πn

的解:

假

arcsin (- \frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 πn

代入

n = 1

arcsin (- \frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 π 1

对于

49.55 cos (θ) - 30 sin (θ) = 1.225

代入

θ = arcsin (- \frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 π 1

49.55 cos (arcsin (- \frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 π 1) - 30 sin (arcsin (- \frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 π 1) = 1.225

整理后得

50.74648 \dots = 1.225

\Rightarrow 假

检验

π + arcsin (\frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 πn

的解:

真

π + arcsin (\frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 πn

代入

n = 1

π + arcsin (\frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 π 1

对于

49.55 cos (θ) - 30 sin (θ) = 1.225

代入

θ = π + arcsin (\frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 π 1

49.55 cos (π + arcsin (\frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 π 1) - 30 sin (π + arcsin (\frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 π 1) = 1.225

整理后得

1.22499 \dots = 1.225

\Rightarrow 真

检验

arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 πn

的解:

真

arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 πn

代入

n = 1

arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 π 1

对于

49.55 cos (θ) - 30 sin (θ) = 1.225

代入

θ = arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 π 1

49.55 cos (arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 π 1) - 30 sin (arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 π 1) = 1.225

整理后得

1.225 = 1.225

\Rightarrow 真

检验

π - arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 πn

的解:

假

π - arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 πn

代入

n = 1

π - arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 π 1

对于

49.55 cos (θ) - 30 sin (θ) = 1.225

代入

θ = π - arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 π 1

49.55 cos (π - arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 π 1) - 30 sin (π - arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 π 1) = 1.225

整理后得

- 51.88211 \dots = 1.225

\Rightarrow 假

θ = π + arcsin (\frac{73.5 + 32936068.91101 \dots}{6710.405}) + 2 πn, θ = arcsin (\frac{32936068.91101 \dots - 73.5}{6710.405}) + 2 πn

以小数形式表示解

θ = π + 1.04752 \dots + 2 πn, θ = 1.00522 \dots + 2 πn

作图

流行的例子

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tan (2 x) + sin (2 x) + cos (2 x) = sec (2 x)

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cos(2x)=sin^2(x)

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

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sec^{2} (x) - 3 tan (x) = 5

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