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受欢迎的三角函数 >

14(1-cos(θ))=sin^2(θ)

解答

14 (1 - cos (θ)) = sin^{2} (θ)

解答

θ = 2 πn

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n

求解步骤

14 (1 - cos (θ)) = sin^{2} (θ)

两边减去

sin^{2} (θ)

14 (1 - cos (θ)) - sin^{2} (θ) = 0

使用三角恒等式改写

- sin^{2} (θ) + (1 - cos (θ)) \cdot 14

使用毕达哥拉斯恒等式:

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

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

= - (1 - cos^{2} (θ)) + (1 - cos (θ)) \cdot 14

化简

- (1 - cos^{2} (θ)) + (1 - cos (θ)) \cdot 14 : cos^{2} (θ) - 14 cos (θ) + 13

- (1 - cos^{2} (θ)) + (1 - cos (θ)) \cdot 14

= - (1 - cos^{2} (θ)) + 14 (1 - cos (θ))

- (1 - cos^{2} (θ)) : - 1 + cos^{2} (θ)

- (1 - cos^{2} (θ))

打开括号

= - (1) - (- cos^{2} (θ))

使用加减运算法则

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

= - 1 + cos^{2} (θ)

= - 1 + cos^{2} (θ) + (1 - cos (θ)) \cdot 14

乘开

14 (1 - cos (θ)) : 14 - 14 cos (θ)

14 (1 - cos (θ))

使用分配律:

a (b - c) = ab - a c

a = 14, b = 1, c = cos (θ)

= 14 \cdot 1 - 14 cos (θ)

数字相乘:

14 \cdot 1 = 14

= 14 - 14 cos (θ)

= - 1 + cos^{2} (θ) + 14 - 14 cos (θ)

化简

- 1 + cos^{2} (θ) + 14 - 14 cos (θ) : cos^{2} (θ) - 14 cos (θ) + 13

- 1 + cos^{2} (θ) + 14 - 14 cos (θ)

对同类项分组

= cos^{2} (θ) - 14 cos (θ) - 1 + 14

数字相加/相减:

- 1 + 14 = 13

= cos^{2} (θ) - 14 cos (θ) + 13

= cos^{2} (θ) - 14 cos (θ) + 13

= cos^{2} (θ) - 14 cos (θ) + 13

13 + cos^{2} (θ) - 14 cos (θ) = 0

用替代法求解

13 + cos^{2} (θ) - 14 cos (θ) = 0

令

: cos (θ) = u

13 + u^{2} - 14 u = 0

13 + u^{2} - 14 u = 0 : u = 13, u = 1

13 + u^{2} - 14 u = 0

改写成标准形式

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

u^{2} - 14 u + 13 = 0

使用求根公式求解

u^{2} - 14 u + 13 = 0

二次方程求根公式:

若

a = 1, b = - 14, c = 13

u_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot 13}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot 13}{2 \cdot 1}

(- 14)^{2} - 4 \cdot 1 \cdot 13 = 12

(- 14)^{2} - 4 \cdot 1 \cdot 13

使用指数法则:

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

若

n

是偶数

(- 14)^{2} = 1 4^{2}

= 1 4^{2} - 4 \cdot 1 \cdot 13

数字相乘:

4 \cdot 1 \cdot 13 = 52

= 1 4^{2} - 52

1 4^{2} = 196

= 196 - 52

数字相减:

196 - 52 = 144

= 144

因式分解数字:

144 = 1 2^{2}

= 1 2^{2}

使用根式运算法则:

n a^{n} = a

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- ( - 14 ) \pm 12}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 14 ) + 12}{2 \cdot 1}, u_{2} = \frac{- ( - 14 ) - 12}{2 \cdot 1}

u = \frac{- ( - 14 ) + 12}{2 \cdot 1} : 13

\frac{- ( - 14 ) + 12}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{14 + 12}{2 \cdot 1}

数字相加:

14 + 12 = 26

= \frac{26}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{26}{2}

数字相除:

\frac{26}{2} = 13

= 13

u = \frac{- ( - 14 ) - 12}{2 \cdot 1} : 1

\frac{- ( - 14 ) - 12}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{14 - 12}{2 \cdot 1}

数字相减:

14 - 12 = 2

= \frac{2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = 13, u = 1

u = cos (θ)

代回

cos (θ) = 13, cos (θ) = 1

cos (θ) = 13, cos (θ) = 1

cos (θ) = 13 :

无解

cos (θ) = 13

- 1 \leq cos (x) \leq 1

无解

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

cos (θ) = 1

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = 0 + 2 πn

θ = 0 + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

合并所有解

θ = 2 πn

作图

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