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受欢迎的三角函数 >

sin^4(x)=1-cos^4(x)

解答

sin^{4} (x) = 1 - cos^{4} (x)

解答

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

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

sin^{4} (x) = 1 - cos^{4} (x)

两边减去

1 - cos^{4} (x)

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

使用指数法则:

a^{b} = a^{2} a^{b - 2}

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

使用三角恒等式改写

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

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

化简

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

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

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

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

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

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

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

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

使用完全平方公式:

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

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

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

2 \cdot 1 \cdot cos^{2} (x) = 2 cos^{2} (x)

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

数字相乘:

2 \cdot 1 = 2

= 2 cos^{2} (x)

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

(cos^{2} (x))^{2}

使用指数法则:

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

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

数字相乘:

2 \cdot 2 = 4

= cos^{4} (x)

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

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

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

化简

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

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

对同类项分组

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

同类项相加:

cos^{4} (x) + cos^{4} (x) = 2 cos^{4} (x)

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

- 1 + 1 = 0

= 2 cos^{4} (x) - 2 cos^{2} (x)

= 2 cos^{4} (x) - 2 cos^{2} (x)

= 2 cos^{4} (x) - 2 cos^{2} (x)

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

用替代法求解

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

令

: cos (x) = u

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

- 2 u^{2} + 2 u^{4} = 0 : u = 1, u = - 1, u = 0

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

改写成标准形式

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

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

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

2 v^{2} - 2 v = 0

解

2 v^{2} - 2 v = 0 : v = 1, v = 0

2 v^{2} - 2 v = 0

使用求根公式求解

2 v^{2} - 2 v = 0

二次方程求根公式:

若

a = 2, b = - 2, c = 0

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

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

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

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

使用指数法则:

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

若

n

是偶数

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

= 2^{2} - 4 \cdot 2 \cdot 0

使用法则

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 2

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

2 + 2 = 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

v = \frac{- ( - 2 ) - 2}{2 \cdot 2} : 0

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

使用法则

- (- a) = a

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

数字相减:

2 - 2 = 0

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

数字相乘:

2 \cdot 2 = 4

= \frac{0}{4}

使用法则

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

= 0

二次方程组的解是:

v = 1, v = 0

v = 1, v = 0

代回

v = u^{2}

，求解

u

解

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

对于

x^{2} = f (a)

解为

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

u = 1, u = - 1

1 = 1

1

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

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

u = 0

解为

u = 1, u = - 1, u = 0

u = cos (x)

代回

cos (x) = 1, cos (x) = - 1, cos (x) = 0

cos (x) = 1, cos (x) = - 1, cos (x) = 0

cos (x) = 1 : x = 2 πn

cos (x) = 1

cos (x) = 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}

x = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

cos (x) = - 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}

x = π + 2 πn

x = π + 2 πn

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

cos (x) = 0

cos (x) = 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}

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

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

合并所有解

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

作图

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