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受欢迎的三角函数 >

3(1-cos(θ))=sin^2(θ)

解答

3 (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

3 (1 - cos (θ)) = sin^{2} (θ)

两边减去

sin^{2} (θ)

3 (1 - cos (θ)) - sin^{2} (θ) = 0

使用三角恒等式改写

- sin^{2} (θ) + (1 - cos (θ)) \cdot 3

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 3

化简

- (1 - cos^{2} (θ)) + (1 - cos (θ)) \cdot 3 : cos^{2} (θ) - 3 cos (θ) + 2

- (1 - cos^{2} (θ)) + (1 - cos (θ)) \cdot 3

= - (1 - cos^{2} (θ)) + 3 (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 3

乘开

3 (1 - cos (θ)) : 3 - 3 cos (θ)

3 (1 - cos (θ))

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = cos (θ)

= 3 \cdot 1 - 3 cos (θ)

数字相乘:

3 \cdot 1 = 3

= 3 - 3 cos (θ)

= - 1 + cos^{2} (θ) + 3 - 3 cos (θ)

化简

- 1 + cos^{2} (θ) + 3 - 3 cos (θ) : cos^{2} (θ) - 3 cos (θ) + 2

- 1 + cos^{2} (θ) + 3 - 3 cos (θ)

对同类项分组

= cos^{2} (θ) - 3 cos (θ) - 1 + 3

数字相加/相减:

- 1 + 3 = 2

= cos^{2} (θ) - 3 cos (θ) + 2

= cos^{2} (θ) - 3 cos (θ) + 2

= cos^{2} (θ) - 3 cos (θ) + 2

2 + cos^{2} (θ) - 3 cos (θ) = 0

用替代法求解

2 + cos^{2} (θ) - 3 cos (θ) = 0

令

: cos (θ) = u

2 + u^{2} - 3 u = 0

2 + u^{2} - 3 u = 0 : u = 2, u = 1

2 + u^{2} - 3 u = 0

改写成标准形式

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

u^{2} - 3 u + 2 = 0

使用求根公式求解

u^{2} - 3 u + 2 = 0

二次方程求根公式:

若

a = 1, b = - 3, c = 2

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot 2 = 1

(- 3)^{2} - 4 \cdot 1 \cdot 2

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

= 3^{2} - 4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数字相减:

9 - 8 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- ( - 3 ) \pm 1}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 1}{2 \cdot 1}, u_{2} = \frac{- ( - 3 ) - 1}{2 \cdot 1}

u = \frac{- ( - 3 ) + 1}{2 \cdot 1} : 2

\frac{- ( - 3 ) + 1}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{3 + 1}{2 \cdot 1}

数字相加:

3 + 1 = 4

= \frac{4}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

u = \frac{- ( - 3 ) - 1}{2 \cdot 1} : 1

\frac{- ( - 3 ) - 1}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{3 - 1}{2 \cdot 1}

数字相减:

3 - 1 = 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 = 2, u = 1

u = cos (θ)

代回

cos (θ) = 2, cos (θ) = 1

cos (θ) = 2, cos (θ) = 1

cos (θ) = 2 :

无解

cos (θ) = 2

- 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

流行的例子

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29 cos (x) = - 17 - cos (x)

2cos(x)+2cos(2x)=0

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