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受欢迎的三角函数 >

-9cos^2(θ)+9=17sin(θ)-8

解答

- 9 cos^{2} (θ) + 9 = 17 sin (θ) - 8

解答

θ = \frac{π}{2} + 2 πn, θ = 1.09491 \dots + 2 πn, θ = π - 1.09491 \dots + 2 πn

+1

度数

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 62.73395 \dots^{\circ} + 36 0^{\circ} n, θ = 117.26604 \dots^{\circ} + 36 0^{\circ} n

求解步骤

- 9 cos^{2} (θ) + 9 = 17 sin (θ) - 8

两边减去

17 sin (θ) - 8

- 9 cos^{2} (θ) - 17 sin (θ) + 17 = 0

使用三角恒等式改写

17 - 17 sin (θ) - 9 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)

= 17 - 17 sin (θ) - 9 (1 - sin^{2} (θ))

化简

17 - 17 sin (θ) - 9 (1 - sin^{2} (θ)) : 9 sin^{2} (θ) - 17 sin (θ) + 8

17 - 17 sin (θ) - 9 (1 - sin^{2} (θ))

乘开

- 9 (1 - sin^{2} (θ)) : - 9 + 9 sin^{2} (θ)

- 9 (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 9, b = 1, c = sin^{2} (θ)

= - 9 \cdot 1 - (- 9) sin^{2} (θ)

使用加减运算法则

- (- a) = a

= - 9 \cdot 1 + 9 sin^{2} (θ)

数字相乘:

9 \cdot 1 = 9

= - 9 + 9 sin^{2} (θ)

= 17 - 17 sin (θ) - 9 + 9 sin^{2} (θ)

化简

17 - 17 sin (θ) - 9 + 9 sin^{2} (θ) : 9 sin^{2} (θ) - 17 sin (θ) + 8

17 - 17 sin (θ) - 9 + 9 sin^{2} (θ)

对同类项分组

= - 17 sin (θ) + 9 sin^{2} (θ) + 17 - 9

数字相加/相减:

17 - 9 = 8

= 9 sin^{2} (θ) - 17 sin (θ) + 8

= 9 sin^{2} (θ) - 17 sin (θ) + 8

= 9 sin^{2} (θ) - 17 sin (θ) + 8

8 - 17 sin (θ) + 9 sin^{2} (θ) = 0

用替代法求解

8 - 17 sin (θ) + 9 sin^{2} (θ) = 0

令

: sin (θ) = u

8 - 17 u + 9 u^{2} = 0

8 - 17 u + 9 u^{2} = 0 : u = 1, u = \frac{8}{9}

8 - 17 u + 9 u^{2} = 0

改写成标准形式

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

9 u^{2} - 17 u + 8 = 0

使用求根公式求解

9 u^{2} - 17 u + 8 = 0

二次方程求根公式:

若

a = 9, b = - 17, c = 8

u_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 9 \cdot 8}{2 \cdot 9}

u_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 9 \cdot 8}{2 \cdot 9}

(- 17)^{2} - 4 \cdot 9 \cdot 8 = 1

(- 17)^{2} - 4 \cdot 9 \cdot 8

使用指数法则:

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

若

n

是偶数

(- 17)^{2} = 1 7^{2}

= 1 7^{2} - 4 \cdot 9 \cdot 8

数字相乘:

4 \cdot 9 \cdot 8 = 288

= 1 7^{2} - 288

1 7^{2} = 289

= 289 - 288

数字相减:

289 - 288 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- ( - 17 ) \pm 1}{2 \cdot 9}

将解分隔开

u_{1} = \frac{- ( - 17 ) + 1}{2 \cdot 9}, u_{2} = \frac{- ( - 17 ) - 1}{2 \cdot 9}

u = \frac{- ( - 17 ) + 1}{2 \cdot 9} : 1

\frac{- ( - 17 ) + 1}{2 \cdot 9}

使用法则

- (- a) = a

= \frac{17 + 1}{2 \cdot 9}

数字相加:

17 + 1 = 18

= \frac{18}{2 \cdot 9}

数字相乘:

2 \cdot 9 = 18

= \frac{18}{18}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 17 ) - 1}{2 \cdot 9} : \frac{8}{9}

\frac{- ( - 17 ) - 1}{2 \cdot 9}

使用法则

- (- a) = a

= \frac{17 - 1}{2 \cdot 9}

数字相减:

17 - 1 = 16

= \frac{16}{2 \cdot 9}

数字相乘:

2 \cdot 9 = 18

= \frac{16}{18}

约分:

2

= \frac{8}{9}

二次方程组的解是:

u = 1, u = \frac{8}{9}

u = sin (θ)

代回

sin (θ) = 1, sin (θ) = \frac{8}{9}

sin (θ) = 1, sin (θ) = \frac{8}{9}

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = 1

sin (θ) = 1

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

sin (θ) = \frac{8}{9} : θ = arcsin (\frac{8}{9}) + 2 πn, θ = π - arcsin (\frac{8}{9}) + 2 πn

sin (θ) = \frac{8}{9}

使用反三角函数性质

sin (θ) = \frac{8}{9}

sin (θ) = \frac{8}{9}

的通解

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

θ = arcsin (\frac{8}{9}) + 2 πn, θ = π - arcsin (\frac{8}{9}) + 2 πn

θ = arcsin (\frac{8}{9}) + 2 πn, θ = π - arcsin (\frac{8}{9}) + 2 πn

合并所有解

θ = \frac{π}{2} + 2 πn, θ = arcsin (\frac{8}{9}) + 2 πn, θ = π - arcsin (\frac{8}{9}) + 2 πn

以小数形式表示解

θ = \frac{π}{2} + 2 πn, θ = 1.09491 \dots + 2 πn, θ = π - 1.09491 \dots + 2 πn

作图

流行的例子

4 tan^{2} (θ) - 9 = 0

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\frac{sin ( 9 0 ^{\circ} )}{3.37} = \frac{sin ( x )}{0.7}

cos(x)=sqrt((1-cos(x))/2)

cos (x) = \frac{1 - cos ( x )}{2}

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