zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

117.72sin(θ)-35.316cos(θ)-12.5=0

解答

117.72 sin (θ) - 35.316 cos (θ) - 12.5 = 0

解答

θ = 0.39333 \dots + 2 πn, θ = π + 0.18957 \dots + 2 πn

+1

度数

θ = 22.53666 \dots^{\circ} + 36 0^{\circ} n, θ = 190.86182 \dots^{\circ} + 36 0^{\circ} n

求解步骤

117.72 sin (θ) - 35.316 cos (θ) - 12.5 = 0

两边加上

35.316 cos (θ)

117.72 sin (θ) - 12.5 = 35.316 cos (θ)

两边进行平方

(117.72 sin (θ) - 12.5)^{2} = (35.316 cos (θ))^{2}

两边减去

(35.316 cos (θ))^{2}

(117.72 sin (θ) - 12.5)^{2} - 1247.219856 cos^{2} (θ) = 0

使用三角恒等式改写

(- 12.5 + 117.72 sin (θ))^{2} - 1247.219856 cos^{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)

= (- 12.5 + 117.72 sin (θ))^{2} - 1247.219856 (1 - sin^{2} (θ))

化简

(- 12.5 + 117.72 sin (θ))^{2} - 1247.219856 (1 - sin^{2} (θ)) : 15105.218256 sin^{2} (θ) - 2943 sin (θ) - 1090.969856

(- 12.5 + 117.72 sin (θ))^{2} - 1247.219856 (1 - sin^{2} (θ))

(- 12.5 + 117.72 sin (θ))^{2} : 156.25 - 2943 sin (θ) + 13857.9984 sin^{2} (θ)

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = - 12.5, b = 117.72 sin (θ)

= (- 12.5)^{2} + 2 (- 12.5) \cdot 117.72 sin (θ) + (117.72 sin (θ))^{2}

化简

(- 12.5)^{2} + 2 (- 12.5) \cdot 117.72 sin (θ) + (117.72 sin (θ))^{2} : 156.25 - 2943 sin (θ) + 13857.9984 sin^{2} (θ)

(- 12.5)^{2} + 2 (- 12.5) \cdot 117.72 sin (θ) + (117.72 sin (θ))^{2}

去除括号:

(- a) = - a

= (- 12.5)^{2} - 2 \cdot 12.5 \cdot 117.72 sin (θ) + (117.72 sin (θ))^{2}

(- 12.5)^{2} = 156.25

(- 12.5)^{2}

使用指数法则:

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

若

n

是偶数

(- 12.5)^{2} = 12. 5^{2}

= 12. 5^{2}

12. 5^{2} = 156.25

= 156.25

2 \cdot 12.5 \cdot 117.72 sin (θ) = 2943 sin (θ)

2 \cdot 12.5 \cdot 117.72 sin (θ)

数字相乘:

2 \cdot 12.5 \cdot 117.72 = 2943

= 2943 sin (θ)

(117.72 sin (θ))^{2} = 13857.9984 sin^{2} (θ)

(117.72 sin (θ))^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 117.7 2^{2} sin^{2} (θ)

117.7 2^{2} = 13857.9984

= 13857.9984 sin^{2} (θ)

= 156.25 - 2943 sin (θ) + 13857.9984 sin^{2} (θ)

= 156.25 - 2943 sin (θ) + 13857.9984 sin^{2} (θ)

= 156.25 - 2943 sin (θ) + 13857.9984 sin^{2} (θ) - 1247.219856 (1 - sin^{2} (θ))

乘开

- 1247.219856 (1 - sin^{2} (θ)) : - 1247.219856 + 1247.219856 sin^{2} (θ)

- 1247.219856 (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 1247.219856, b = 1, c = sin^{2} (θ)

= - 1247.219856 \cdot 1 - (- 1247.219856) sin^{2} (θ)

使用加减运算法则

- (- a) = a

= - 1 \cdot 1247.219856 + 1247.219856 sin^{2} (θ)

数字相乘:

1 \cdot 1247.219856 = 1247.219856

= - 1247.219856 + 1247.219856 sin^{2} (θ)

= 156.25 - 2943 sin (θ) + 13857.9984 sin^{2} (θ) - 1247.219856 + 1247.219856 sin^{2} (θ)

化简

156.25 - 2943 sin (θ) + 13857.9984 sin^{2} (θ) - 1247.219856 + 1247.219856 sin^{2} (θ) : 15105.218256 sin^{2} (θ) - 2943 sin (θ) - 1090.969856

156.25 - 2943 sin (θ) + 13857.9984 sin^{2} (θ) - 1247.219856 + 1247.219856 sin^{2} (θ)

对同类项分组

= - 2943 sin (θ) + 13857.9984 sin^{2} (θ) + 1247.219856 sin^{2} (θ) + 156.25 - 1247.219856

同类项相加:

13857.9984 sin^{2} (θ) + 1247.219856 sin^{2} (θ) = 15105.218256 sin^{2} (θ)

= - 2943 sin (θ) + 15105.218256 sin^{2} (θ) + 156.25 - 1247.219856

数字相加/相减:

156.25 - 1247.219856 = - 1090.969856

= 15105.218256 sin^{2} (θ) - 2943 sin (θ) - 1090.969856

= 15105.218256 sin^{2} (θ) - 2943 sin (θ) - 1090.969856

= 15105.218256 sin^{2} (θ) - 2943 sin (θ) - 1090.969856

- 1090.969856 + 15105.218256 sin^{2} (θ) - 2943 sin (θ) = 0

用替代法求解

- 1090.969856 + 15105.218256 sin^{2} (θ) - 2943 sin (θ) = 0

令

: sin (θ) = u

- 1090.969856 + 15105.218256 u^{2} - 2943 u = 0

- 1090.969856 + 15105.218256 u^{2} - 2943 u = 0 : u = \frac{0.19483 \dots + 0.32685 \dots}{2}, u = \frac{0.19483 \dots - 0.32685 \dots}{2}

- 1090.969856 + 15105.218256 u^{2} - 2943 u = 0

两边除以

15105.218256

- \frac{1090.969856}{15105.218256} + \frac{15105.218256 u ^{2}}{15105.218256} - \frac{2943 u}{15105.218256} = \frac{0}{15105.218256}

改写成标准形式

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

u^{2} - 0.19483 \dots u - 0.07222 \dots = 0

使用求根公式求解

u^{2} - 0.19483 \dots u - 0.07222 \dots = 0

二次方程求根公式:

若

a = 1, b = - 0.19483 \dots, c = - 0.07222 \dots

u_{1, 2} = \frac{- ( - 0.19483 \dots ) \pm ( - 0.19483 \dots ) ^{2} - 4 \cdot 1 \cdot ( - 0.07222 \dots )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 0.19483 \dots ) \pm ( - 0.19483 \dots ) ^{2} - 4 \cdot 1 \cdot ( - 0.07222 \dots )}{2 \cdot 1}

(- 0.19483 \dots)^{2} - 4 \cdot 1 \cdot (- 0.07222 \dots) = 0.32685 \dots

(- 0.19483 \dots)^{2} - 4 \cdot 1 \cdot (- 0.07222 \dots)

使用法则

- (- a) = a

= (- 0.19483 \dots)^{2} + 4 \cdot 1 \cdot 0.07222 \dots

使用指数法则:

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

若

n

是偶数

(- 0.19483 \dots)^{2} = 0.19483 \dots^{2}

= 0.19483 \dots^{2} + 4 \cdot 1 \cdot 0.07222 \dots

数字相乘:

4 \cdot 1 \cdot 0.07222 \dots = 0.28889 \dots

= 0.19483 \dots^{2} + 0.28889 \dots

0.19483 \dots^{2} = 0.03796 \dots

= 0.03796 \dots + 0.28889 \dots

数字相加:

0.03796 \dots + 0.28889 \dots = 0.32685 \dots

= 0.32685 \dots

u_{1, 2} = \frac{- ( - 0.19483 \dots ) \pm 0.32685 \dots}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 0.19483 \dots ) + 0.32685 \dots}{2 \cdot 1}, u_{2} = \frac{- ( - 0.19483 \dots ) - 0.32685 \dots}{2 \cdot 1}

u = \frac{- ( - 0.19483 \dots ) + 0.32685 \dots}{2 \cdot 1} : \frac{0.19483 \dots + 0.32685 \dots}{2}

\frac{- ( - 0.19483 \dots ) + 0.32685 \dots}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{0.19483 \dots + 0.32685 \dots}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0.19483 \dots + 0.32685 \dots}{2}

u = \frac{- ( - 0.19483 \dots ) - 0.32685 \dots}{2 \cdot 1} : \frac{0.19483 \dots - 0.32685 \dots}{2}

\frac{- ( - 0.19483 \dots ) - 0.32685 \dots}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{0.19483 \dots - 0.32685 \dots}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0.19483 \dots - 0.32685 \dots}{2}

二次方程组的解是:

u = \frac{0.19483 \dots + 0.32685 \dots}{2}, u = \frac{0.19483 \dots - 0.32685 \dots}{2}

u = sin (θ)

代回

sin (θ) = \frac{0.19483 \dots + 0.32685 \dots}{2}, sin (θ) = \frac{0.19483 \dots - 0.32685 \dots}{2}

sin (θ) = \frac{0.19483 \dots + 0.32685 \dots}{2}, sin (θ) = \frac{0.19483 \dots - 0.32685 \dots}{2}

sin (θ) = \frac{0.19483 \dots + 0.32685 \dots}{2} : θ = arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 πn, θ = π - arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 πn

sin (θ) = \frac{0.19483 \dots + 0.32685 \dots}{2}

使用反三角函数性质

sin (θ) = \frac{0.19483 \dots + 0.32685 \dots}{2}

sin (θ) = \frac{0.19483 \dots + 0.32685 \dots}{2}

的通解

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

θ = arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 πn, θ = π - arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 πn

θ = arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 πn, θ = π - arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 πn

sin (θ) = \frac{0.19483 \dots - 0.32685 \dots}{2} : θ = arcsin (\frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 πn, θ = π + arcsin (- \frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 πn

sin (θ) = \frac{0.19483 \dots - 0.32685 \dots}{2}

使用反三角函数性质

sin (θ) = \frac{0.19483 \dots - 0.32685 \dots}{2}

sin (θ) = \frac{0.19483 \dots - 0.32685 \dots}{2}

的通解

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

θ = arcsin (\frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 πn, θ = π + arcsin (- \frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 πn

θ = arcsin (\frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 πn, θ = π + arcsin (- \frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 πn

合并所有解

θ = arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 πn, θ = π - arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 πn, θ = arcsin (\frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 πn, θ = π + arcsin (- \frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 πn

将解代入原方程进行验证

将它们代入

117.72 sin (θ) - 35.316 cos (θ) - 12.5 = 0

检验解是否符合

去除与方程不符的解。

检验

arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 πn

的解:

真

arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 πn

代入

n = 1

arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 π 1

对于

117.72 sin (θ) - 35.316 cos (θ) - 12.5 = 0

代入

θ = arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 π 1

117.72 sin (arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 π 1) - 35.316 cos (arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 π 1) - 12.5 = 0

整理后得

0 = 0

\Rightarrow 真

检验

π - arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 πn

的解:

假

π - arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 πn

代入

n = 1

π - arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 π 1

对于

117.72 sin (θ) - 35.316 cos (θ) - 12.5 = 0

代入

θ = π - arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 π 1

117.72 sin (π - arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 π 1) - 35.316 cos (π - arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 π 1) - 12.5 = 0

整理后得

65.23815 \dots = 0

\Rightarrow 假

检验

arcsin (\frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 πn

的解:

假

arcsin (\frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 πn

代入

n = 1

arcsin (\frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 π 1

对于

117.72 sin (θ) - 35.316 cos (θ) - 12.5 = 0

代入

θ = arcsin (\frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 π 1

117.72 sin (arcsin (\frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 π 1) - 35.316 cos (arcsin (\frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 π 1) - 12.5 = 0

整理后得

- 69.36659 \dots = 0

\Rightarrow 假

检验

π + arcsin (- \frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 πn

的解:

真

π + arcsin (- \frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 πn

代入

n = 1

π + arcsin (- \frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 π 1

对于

117.72 sin (θ) - 35.316 cos (θ) - 12.5 = 0

代入

θ = π + arcsin (- \frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 π 1

117.72 sin (π + arcsin (- \frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 π 1) - 35.316 cos (π + arcsin (- \frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 π 1) - 12.5 = 0

整理后得

0 = 0

\Rightarrow 真

θ = arcsin (\frac{0.19483 \dots + 0.32685 \dots}{2}) + 2 πn, θ = π + arcsin (- \frac{0.19483 \dots - 0.32685 \dots}{2}) + 2 πn

以小数形式表示解

θ = 0.39333 \dots + 2 πn, θ = π + 0.18957 \dots + 2 πn

作图

流行的例子

1/(cot^2(x))+sqrt(3)tan(x)=0

\frac{1}{cot ^{2} ( x )} + 3 tan (x) = 0

tan (x) = 0.23

tan (x) = 0.45

2cos(x)-1=sec(x)

2 cos (x) - 1 = sec (x)

cos (2 x) = - 0.32