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受欢迎的三角函数 >

2-2cos^2(x/2)=2cos^2(x)

解答

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

解答

x = π + 4 πn, x = 3 π + 4 πn, x = \frac{π}{3} + 4 πn, x = \frac{11 π}{3} + 4 πn, x = \frac{5 π}{3} + 4 πn, x = \frac{7 π}{3} + 4 πn

+1

度数

x = 18 0^{\circ} + 72 0^{\circ} n, x = 54 0^{\circ} + 72 0^{\circ} n, x = 6 0^{\circ} + 72 0^{\circ} n, x = 66 0^{\circ} + 72 0^{\circ} n, x = 30 0^{\circ} + 72 0^{\circ} n, x = 42 0^{\circ} + 72 0^{\circ} n

求解步骤

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

两边减去

2 cos^{2} (x)

2 - 2 cos^{2} (\frac{x}{2}) - 2 cos^{2} (x) = 0

令

: u = \frac{x}{2}

2 - 2 cos^{2} (u) - 2 cos^{2} (2 u) = 0

使用三角恒等式改写

2 - 2 cos^{2} (2 u) - 2 cos^{2} (u)

使用倍角公式:

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

= 2 - 2 (2 cos^{2} (u) - 1)^{2} - 2 cos^{2} (u)

化简

2 - 2 (2 cos^{2} (u) - 1)^{2} - 2 cos^{2} (u) : 6 cos^{2} (u) - 8 cos^{4} (u)

2 - 2 (2 cos^{2} (u) - 1)^{2} - 2 cos^{2} (u)

(2 cos^{2} (u) - 1)^{2} : 4 cos^{4} (u) - 4 cos^{2} (u) + 1

使用完全平方公式:

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

a = 2 cos^{2} (u), b = 1

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

化简

(2 cos^{2} (u))^{2} - 2 \cdot 2 cos^{2} (u) \cdot 1 + 1^{2} : 4 cos^{4} (u) - 4 cos^{2} (u) + 1

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

使用法则

1^{a} = 1

1^{2} = 1

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

(2 cos^{2} (u))^{2} = 4 cos^{4} (u)

(2 cos^{2} (u))^{2}

使用指数法则:

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

= 2^{2} (cos^{2} (u))^{2}

(cos^{2} (u))^{2} : cos^{4} (u)

使用指数法则:

(a^{b})^{c} = a^{b c}

= cos^{2 \cdot 2} (u)

数字相乘:

2 \cdot 2 = 4

= cos^{4} (u)

= 2^{2} cos^{4} (u)

2^{2} = 4

= 4 cos^{4} (u)

2 \cdot 2 cos^{2} (u) \cdot 1 = 4 cos^{2} (u)

2 \cdot 2 cos^{2} (u) \cdot 1

数字相乘:

2 \cdot 2 \cdot 1 = 4

= 4 cos^{2} (u)

= 4 cos^{4} (u) - 4 cos^{2} (u) + 1

= 4 cos^{4} (u) - 4 cos^{2} (u) + 1

= 2 - 2 (4 cos^{4} (u) - 4 cos^{2} (u) + 1) - 2 cos^{2} (u)

乘开

- 2 (4 cos^{4} (u) - 4 cos^{2} (u) + 1) : - 8 cos^{4} (u) + 8 cos^{2} (u) - 2

- 2 (4 cos^{4} (u) - 4 cos^{2} (u) + 1)

打开括号

= (- 2) \cdot 4 cos^{4} (u) + (- 2) (- 4 cos^{2} (u)) + (- 2) \cdot 1

使用加减运算法则

+ (- a) = - a, (- a) (- b) = ab

= - 2 \cdot 4 cos^{4} (u) + 2 \cdot 4 cos^{2} (u) - 2 \cdot 1

化简

- 2 \cdot 4 cos^{4} (u) + 2 \cdot 4 cos^{2} (u) - 2 \cdot 1 : - 8 cos^{4} (u) + 8 cos^{2} (u) - 2

- 2 \cdot 4 cos^{4} (u) + 2 \cdot 4 cos^{2} (u) - 2 \cdot 1

数字相乘:

2 \cdot 4 = 8

= - 8 cos^{4} (u) + 8 cos^{2} (u) - 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

= - 8 cos^{4} (u) + 8 cos^{2} (u) - 2

= - 8 cos^{4} (u) + 8 cos^{2} (u) - 2

= 2 - 8 cos^{4} (u) + 8 cos^{2} (u) - 2 - 2 cos^{2} (u)

化简

2 - 8 cos^{4} (u) + 8 cos^{2} (u) - 2 - 2 cos^{2} (u) : 6 cos^{2} (u) - 8 cos^{4} (u)

2 - 8 cos^{4} (u) + 8 cos^{2} (u) - 2 - 2 cos^{2} (u)

对同类项分组

= - 8 cos^{4} (u) + 8 cos^{2} (u) - 2 cos^{2} (u) + 2 - 2

同类项相加:

8 cos^{2} (u) - 2 cos^{2} (u) = 6 cos^{2} (u)

= - 8 cos^{4} (u) + 6 cos^{2} (u) + 2 - 2

2 - 2 = 0

= 6 cos^{2} (u) - 8 cos^{4} (u)

= 6 cos^{2} (u) - 8 cos^{4} (u)

= 6 cos^{2} (u) - 8 cos^{4} (u)

6 cos^{2} (u) - 8 cos^{4} (u) = 0

用替代法求解

6 cos^{2} (u) - 8 cos^{4} (u) = 0

令

: cos (u) = u

6 u^{2} - 8 u^{4} = 0

6 u^{2} - 8 u^{4} = 0 : u = 0, u = \frac{3}{2}, u = - \frac{3}{2}

6 u^{2} - 8 u^{4} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- 8 u^{4} + 6 u^{2} = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

- 8 v^{2} + 6 v = 0

解

- 8 v^{2} + 6 v = 0 : v = 0, v = \frac{3}{4}

- 8 v^{2} + 6 v = 0

使用求根公式求解

- 8 v^{2} + 6 v = 0

二次方程求根公式:

若

a = - 8, b = 6, c = 0

v_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 8 ) \cdot 0}{2 ( - 8 )}

v_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 8 ) \cdot 0}{2 ( - 8 )}

6^{2} - 4 (- 8) \cdot 0 = 6

6^{2} - 4 (- 8) \cdot 0

使用法则

- (- a) = a

= 6^{2} + 4 \cdot 8 \cdot 0

使用法则

0 \cdot a = 0

= 6^{2} + 0

6^{2} + 0 = 6^{2}

= 6^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 6

v_{1, 2} = \frac{- 6 \pm 6}{2 ( - 8 )}

将解分隔开

v_{1} = \frac{- 6 + 6}{2 ( - 8 )}, v_{2} = \frac{- 6 - 6}{2 ( - 8 )}

v = \frac{- 6 + 6}{2 ( - 8 )} : 0

\frac{- 6 + 6}{2 ( - 8 )}

去除括号:

(- a) = - a

= \frac{- 6 + 6}{- 2 \cdot 8}

数字相加/相减:

- 6 + 6 = 0

= \frac{0}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{0}{- 16}

使用分式法则:

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

= - \frac{0}{16}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

v = \frac{- 6 - 6}{2 ( - 8 )} : \frac{3}{4}

\frac{- 6 - 6}{2 ( - 8 )}

去除括号:

(- a) = - a

= \frac{- 6 - 6}{- 2 \cdot 8}

数字相减:

- 6 - 6 = - 12

= \frac{- 12}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 12}{- 16}

使用分式法则:

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

= \frac{12}{16}

约分:

4

= \frac{3}{4}

二次方程组的解是:

v = 0, v = \frac{3}{4}

v = 0, v = \frac{3}{4}

代回

v = u^{2}

，求解

u

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

u = 0

解

u^{2} = \frac{3}{4} : u = \frac{3}{2}, u = - \frac{3}{2}

u^{2} = \frac{3}{4}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = \frac{3}{4}, u = - \frac{3}{4}

\frac{3}{4} = \frac{3}{2}

\frac{3}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{3}{2}

- \frac{3}{4} = - \frac{3}{2}

- \frac{3}{4}

化简

\frac{3}{4} : \frac{3}{2}

\frac{3}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{3}{2}

= - \frac{3}{2}

u = \frac{3}{2}, u = - \frac{3}{2}

解为

u = 0, u = \frac{3}{2}, u = - \frac{3}{2}

u = cos (u)

代回

cos (u) = 0, cos (u) = \frac{3}{2}, cos (u) = - \frac{3}{2}

cos (u) = 0, cos (u) = \frac{3}{2}, cos (u) = - \frac{3}{2}

cos (u) = 0 : u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

cos (u) = 0

cos (u) = 0

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}

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

cos (u) = \frac{3}{2} : u = \frac{π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn

cos (u) = \frac{3}{2}

cos (u) = \frac{3}{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}

u = \frac{π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn

u = \frac{π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn

cos (u) = - \frac{3}{2} : u = \frac{5 π}{6} + 2 πn, u = \frac{7 π}{6} + 2 πn

cos (u) = - \frac{3}{2}

cos (u) = - \frac{3}{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}

u = \frac{5 π}{6} + 2 πn, u = \frac{7 π}{6} + 2 πn

u = \frac{5 π}{6} + 2 πn, u = \frac{7 π}{6} + 2 πn

合并所有解

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn, u = \frac{π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn, u = \frac{7 π}{6} + 2 πn

u = \frac{x}{2}

代回

\frac{x}{2} = \frac{π}{2} + 2 πn : x = π + 4 πn

\frac{x}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

化简

\frac{2 x}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{π}{2} + 2 \cdot 2 πn : π + 4 πn

2 \cdot \frac{π}{2} + 2 \cdot 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{2}

约分:

2

= π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= π + 4 πn

x = π + 4 πn

x = π + 4 πn

x = π + 4 πn

\frac{x}{2} = \frac{3 π}{2} + 2 πn : x = 3 π + 4 πn

\frac{x}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

化简

\frac{2 x}{2} = 2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn : 3 π + 4 πn

2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

2 \cdot \frac{3 π}{2} = 3 π

2 \cdot \frac{3 π}{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 π 2}{2}

约分:

2

= 3 π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= 3 π + 4 πn

x = 3 π + 4 πn

x = 3 π + 4 πn

x = 3 π + 4 πn

\frac{x}{2} = \frac{π}{6} + 2 πn : x = \frac{π}{3} + 4 πn

\frac{x}{2} = \frac{π}{6} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{π}{6} + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot \frac{π}{6} + 2 \cdot 2 πn

化简

\frac{2 x}{2} = 2 \cdot \frac{π}{6} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{π}{6} + 2 \cdot 2 πn : \frac{π}{3} + 4 πn

2 \cdot \frac{π}{6} + 2 \cdot 2 πn

2 \cdot \frac{π}{6} = \frac{π}{3}

2 \cdot \frac{π}{6}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{6}

约分:

2

= \frac{π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{π}{3} + 4 πn

x = \frac{π}{3} + 4 πn

x = \frac{π}{3} + 4 πn

x = \frac{π}{3} + 4 πn

\frac{x}{2} = \frac{11 π}{6} + 2 πn : x = \frac{11 π}{3} + 4 πn

\frac{x}{2} = \frac{11 π}{6} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{11 π}{6} + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn

化简

\frac{2 x}{2} = 2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn : \frac{11 π}{3} + 4 πn

2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{11 π}{6} = \frac{11 π}{3}

2 \cdot \frac{11 π}{6}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{11 π 2}{6}

数字相乘:

11 \cdot 2 = 22

= \frac{22 π}{6}

约分:

2

= \frac{11 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{11 π}{3} + 4 πn

x = \frac{11 π}{3} + 4 πn

x = \frac{11 π}{3} + 4 πn

x = \frac{11 π}{3} + 4 πn

\frac{x}{2} = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{3} + 4 πn

\frac{x}{2} = \frac{5 π}{6} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{5 π}{6} + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

化简

\frac{2 x}{2} = 2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn : \frac{5 π}{3} + 4 πn

2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{5 π}{6} = \frac{5 π}{3}

2 \cdot \frac{5 π}{6}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{5 π 2}{6}

数字相乘:

5 \cdot 2 = 10

= \frac{10 π}{6}

约分:

2

= \frac{5 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{5 π}{3} + 4 πn

x = \frac{5 π}{3} + 4 πn

x = \frac{5 π}{3} + 4 πn

x = \frac{5 π}{3} + 4 πn

\frac{x}{2} = \frac{7 π}{6} + 2 πn : x = \frac{7 π}{3} + 4 πn

\frac{x}{2} = \frac{7 π}{6} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{7 π}{6} + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot \frac{7 π}{6} + 2 \cdot 2 πn

化简

\frac{2 x}{2} = 2 \cdot \frac{7 π}{6} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{7 π}{6} + 2 \cdot 2 πn : \frac{7 π}{3} + 4 πn

2 \cdot \frac{7 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{7 π}{6} = \frac{7 π}{3}

2 \cdot \frac{7 π}{6}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{7 π 2}{6}

数字相乘:

7 \cdot 2 = 14

= \frac{14 π}{6}

约分:

2

= \frac{7 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{7 π}{3} + 4 πn

x = \frac{7 π}{3} + 4 πn

x = \frac{7 π}{3} + 4 πn

x = \frac{7 π}{3} + 4 πn

x = π + 4 πn, x = 3 π + 4 πn, x = \frac{π}{3} + 4 πn, x = \frac{11 π}{3} + 4 πn, x = \frac{5 π}{3} + 4 πn, x = \frac{7 π}{3} + 4 πn

作图

流行的例子

sec^2(θ)-2=tan^2(θ)

sec^{2} (θ) - 2 = tan^{2} (θ)

cos(x)=sin(x)+1

cos (x) = sin (x) + 1

2sin^2(θ)=3(1-cos(-θ))

2 sin^{2} (θ) = 3 (1 - cos (- θ))

sec(x)-sqrt(2)=0

sec (x) - 2 = 0

2sin^4(x)+sin^2(x)-1=0

2 sin^{4} (x) + sin^{2} (x) - 1 = 0