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受欢迎的三角函数 >

cos(θ)(1+cos(θ))=sin^2(θ)

解答

cos (θ) (1 + cos (θ)) = sin^{2} (θ)

解答

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = π + 2 πn

+1

度数

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

cos (θ) (1 + cos (θ)) = sin^{2} (θ)

两边减去

sin^{2} (θ)

cos (θ) (1 + cos (θ)) - sin^{2} (θ) = 0

使用三角恒等式改写

- sin^{2} (θ) + (1 + cos (θ)) cos (θ)

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)) cos (θ)

化简

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

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

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

乘开

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

cos (θ) (1 + cos (θ))

使用分配律:

a (b + c) = ab + a c

a = cos (θ), b = 1, c = cos (θ)

= cos (θ) \cdot 1 + cos (θ) cos (θ)

= 1 \cdot cos (θ) + cos (θ) cos (θ)

化简

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

1 \cdot cos (θ) + cos (θ) cos (θ)

1 \cdot cos (θ) = cos (θ)

1 \cdot cos (θ)

乘以:

1 \cdot cos (θ) = cos (θ)

= cos (θ)

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

数字相加:

1 + 1 = 2

= cos^{2} (θ)

= cos (θ) + cos^{2} (θ)

= cos (θ) + cos^{2} (θ)

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

化简

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

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

对同类项分组

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

同类项相加:

cos^{2} (θ) + cos^{2} (θ) = 2 cos^{2} (θ)

= 2 cos^{2} (θ) + cos (θ) - 1

= 2 cos^{2} (θ) + cos (θ) - 1

= 2 cos^{2} (θ) + cos (θ) - 1

- 1 + cos (θ) + 2 cos^{2} (θ) = 0

用替代法求解

- 1 + cos (θ) + 2 cos^{2} (θ) = 0

令

: cos (θ) = u

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

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

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

改写成标准形式

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

2 u^{2} + u - 1 = 0

使用求根公式求解

2 u^{2} + u - 1 = 0

二次方程求根公式:

若

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

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

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 1)

使用法则

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

3^{2} = 3

= 3

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

将解分隔开

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

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

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

数字相加/相减:

- 1 + 3 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

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

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

数字相减:

- 1 - 3 = - 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 4}{4}

使用分式法则:

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

= - \frac{4}{4}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = \frac{1}{2}, u = - 1

u = cos (θ)

代回

cos (θ) = \frac{1}{2}, cos (θ) = - 1

cos (θ) = \frac{1}{2}, cos (θ) = - 1

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = \frac{1}{2}

cos (θ) = \frac{1}{2}

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 = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = - 1 : θ = π + 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 = π + 2 πn

θ = π + 2 πn

合并所有解

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = π + 2 πn

作图

流行的例子

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