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受欢迎的三角函数 >

cos^4(a)=8cos^4(a)-8cos^2(a)+1

解答

cos^{4} (a) = 8 cos^{4} (a) - 8 cos^{2} (a) + 1

解答

a = 2 πn, a = π + 2 πn, a = 1.18319 \dots + 2 πn, a = 2 π - 1.18319 \dots + 2 πn, a = 1.95839 \dots + 2 πn, a = - 1.95839 \dots + 2 πn

+1

度数

a = 0^{\circ} + 36 0^{\circ} n, a = 18 0^{\circ} + 36 0^{\circ} n, a = 67.79234 \dots^{\circ} + 36 0^{\circ} n, a = 292.20765 \dots^{\circ} + 36 0^{\circ} n, a = 112.20765 \dots^{\circ} + 36 0^{\circ} n, a = - 112.20765 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cos^{4} (a) = 8 cos^{4} (a) - 8 cos^{2} (a) + 1

用替代法求解

cos^{4} (a) = 8 cos^{4} (a) - 8 cos^{2} (a) + 1

令

: cos (a) = u

u^{4} = 8 u^{4} - 8 u^{2} + 1

u^{4} = 8 u^{4} - 8 u^{2} + 1 : u = 1, u = - 1, u = \frac{1}{7}, u = - \frac{1}{7}

u^{4} = 8 u^{4} - 8 u^{2} + 1

交换两边

8 u^{4} - 8 u^{2} + 1 = u^{4}

将

u^{4}

para o lado esquerdo

8 u^{4} - 8 u^{2} + 1 = u^{4}

两边减去

u^{4}

8 u^{4} - 8 u^{2} + 1 - u^{4} = u^{4} - u^{4}

化简

7 u^{4} - 8 u^{2} + 1 = 0

7 u^{4} - 8 u^{2} + 1 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

7 v^{2} - 8 v + 1 = 0

解

7 v^{2} - 8 v + 1 = 0 : v = 1, v = \frac{1}{7}

7 v^{2} - 8 v + 1 = 0

使用求根公式求解

7 v^{2} - 8 v + 1 = 0

二次方程求根公式:

若

a = 7, b = - 8, c = 1

v_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 7 \cdot 1}{2 \cdot 7}

v_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 7 \cdot 1}{2 \cdot 7}

(- 8)^{2} - 4 \cdot 7 \cdot 1 = 6

(- 8)^{2} - 4 \cdot 7 \cdot 1

使用指数法则:

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

若

n

是偶数

(- 8)^{2} = 8^{2}

= 8^{2} - 4 \cdot 7 \cdot 1

数字相乘:

4 \cdot 7 \cdot 1 = 28

= 8^{2} - 28

8^{2} = 64

= 64 - 28

数字相减:

64 - 28 = 36

= 36

因式分解数字:

36 = 6^{2}

= 6^{2}

使用根式运算法则:

n a^{n} = a

6^{2} = 6

= 6

v_{1, 2} = \frac{- ( - 8 ) \pm 6}{2 \cdot 7}

将解分隔开

v_{1} = \frac{- ( - 8 ) + 6}{2 \cdot 7}, v_{2} = \frac{- ( - 8 ) - 6}{2 \cdot 7}

v = \frac{- ( - 8 ) + 6}{2 \cdot 7} : 1

\frac{- ( - 8 ) + 6}{2 \cdot 7}

使用法则

- (- a) = a

= \frac{8 + 6}{2 \cdot 7}

数字相加:

8 + 6 = 14

= \frac{14}{2 \cdot 7}

数字相乘:

2 \cdot 7 = 14

= \frac{14}{14}

使用法则

\frac{a}{a} = 1

= 1

v = \frac{- ( - 8 ) - 6}{2 \cdot 7} : \frac{1}{7}

\frac{- ( - 8 ) - 6}{2 \cdot 7}

使用法则

- (- a) = a

= \frac{8 - 6}{2 \cdot 7}

数字相减:

8 - 6 = 2

= \frac{2}{2 \cdot 7}

数字相乘:

2 \cdot 7 = 14

= \frac{2}{14}

约分:

2

= \frac{1}{7}

二次方程组的解是:

v = 1, v = \frac{1}{7}

v = 1, v = \frac{1}{7}

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} = \frac{1}{7} : u = \frac{1}{7}, u = - \frac{1}{7}

u^{2} = \frac{1}{7}

对于

x^{2} = f (a)

解为

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

u = \frac{1}{7}, u = - \frac{1}{7}

解为

u = 1, u = - 1, u = \frac{1}{7}, u = - \frac{1}{7}

u = cos (a)

代回

cos (a) = 1, cos (a) = - 1, cos (a) = \frac{1}{7}, cos (a) = - \frac{1}{7}

cos (a) = 1, cos (a) = - 1, cos (a) = \frac{1}{7}, cos (a) = - \frac{1}{7}

cos (a) = 1 : a = 2 πn

cos (a) = 1

cos (a) = 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}

a = 0 + 2 πn

a = 0 + 2 πn

解

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn

cos (a) = - 1 : a = π + 2 πn

cos (a) = - 1

cos (a) = - 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}

a = π + 2 πn

a = π + 2 πn

cos (a) = \frac{1}{7} : a = arccos (\frac{1}{7}) + 2 πn, a = 2 π - arccos (\frac{1}{7}) + 2 πn

cos (a) = \frac{1}{7}

使用反三角函数性质

cos (a) = \frac{1}{7}

cos (a) = \frac{1}{7}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

a = arccos (\frac{1}{7}) + 2 πn, a = 2 π - arccos (\frac{1}{7}) + 2 πn

a = arccos (\frac{1}{7}) + 2 πn, a = 2 π - arccos (\frac{1}{7}) + 2 πn

cos (a) = - \frac{1}{7} : a = arccos (- \frac{1}{7}) + 2 πn, a = - arccos (- \frac{1}{7}) + 2 πn

cos (a) = - \frac{1}{7}

使用反三角函数性质

cos (a) = - \frac{1}{7}

cos (a) = - \frac{1}{7}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

a = arccos (- \frac{1}{7}) + 2 πn, a = - arccos (- \frac{1}{7}) + 2 πn

a = arccos (- \frac{1}{7}) + 2 πn, a = - arccos (- \frac{1}{7}) + 2 πn

合并所有解

a = 2 πn, a = π + 2 πn, a = arccos (\frac{1}{7}) + 2 πn, a = 2 π - arccos (\frac{1}{7}) + 2 πn, a = arccos (- \frac{1}{7}) + 2 πn, a = - arccos (- \frac{1}{7}) + 2 πn

以小数形式表示解

a = 2 πn, a = π + 2 πn, a = 1.18319 \dots + 2 πn, a = 2 π - 1.18319 \dots + 2 πn, a = 1.95839 \dots + 2 πn, a = - 1.95839 \dots + 2 πn

作图

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