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受欢迎的三角函数 >

8cos(2θ)=24cos(θ)-16

解答

8 cos (2 θ) = 24 cos (θ) - 16

解答

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

+1

度数

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

求解步骤

8 cos (2 θ) = 24 cos (θ) - 16

两边减去

24 cos (θ) - 16

8 cos (2 θ) - 24 cos (θ) + 16 = 0

使用三角恒等式改写

16 - 24 cos (θ) + 8 cos (2 θ)

使用倍角公式:

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

= 16 - 24 cos (θ) + 8 (2 cos^{2} (θ) - 1)

化简

16 - 24 cos (θ) + 8 (2 cos^{2} (θ) - 1) : 16 cos^{2} (θ) - 24 cos (θ) + 8

16 - 24 cos (θ) + 8 (2 cos^{2} (θ) - 1)

乘开

8 (2 cos^{2} (θ) - 1) : 16 cos^{2} (θ) - 8

8 (2 cos^{2} (θ) - 1)

使用分配律:

a (b - c) = ab - a c

a = 8, b = 2 cos^{2} (θ), c = 1

= 8 \cdot 2 cos^{2} (θ) - 8 \cdot 1

化简

8 \cdot 2 cos^{2} (θ) - 8 \cdot 1 : 16 cos^{2} (θ) - 8

8 \cdot 2 cos^{2} (θ) - 8 \cdot 1

数字相乘:

8 \cdot 2 = 16

= 16 cos^{2} (θ) - 8 \cdot 1

数字相乘:

8 \cdot 1 = 8

= 16 cos^{2} (θ) - 8

= 16 cos^{2} (θ) - 8

= 16 - 24 cos (θ) + 16 cos^{2} (θ) - 8

化简

16 - 24 cos (θ) + 16 cos^{2} (θ) - 8 : 16 cos^{2} (θ) - 24 cos (θ) + 8

16 - 24 cos (θ) + 16 cos^{2} (θ) - 8

对同类项分组

= - 24 cos (θ) + 16 cos^{2} (θ) + 16 - 8

数字相加/相减:

16 - 8 = 8

= 16 cos^{2} (θ) - 24 cos (θ) + 8

= 16 cos^{2} (θ) - 24 cos (θ) + 8

= 16 cos^{2} (θ) - 24 cos (θ) + 8

8 + 16 cos^{2} (θ) - 24 cos (θ) = 0

用替代法求解

8 + 16 cos^{2} (θ) - 24 cos (θ) = 0

令

: cos (θ) = u

8 + 16 u^{2} - 24 u = 0

8 + 16 u^{2} - 24 u = 0 : u = 1, u = \frac{1}{2}

8 + 16 u^{2} - 24 u = 0

改写成标准形式

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

16 u^{2} - 24 u + 8 = 0

使用求根公式求解

16 u^{2} - 24 u + 8 = 0

二次方程求根公式:

若

a = 16, b = - 24, c = 8

u_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 16 \cdot 8}{2 \cdot 16}

u_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 16 \cdot 8}{2 \cdot 16}

(- 24)^{2} - 4 \cdot 16 \cdot 8 = 8

(- 24)^{2} - 4 \cdot 16 \cdot 8

使用指数法则:

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

若

n

是偶数

(- 24)^{2} = 2 4^{2}

= 2 4^{2} - 4 \cdot 16 \cdot 8

数字相乘:

4 \cdot 16 \cdot 8 = 512

= 2 4^{2} - 512

2 4^{2} = 576

= 576 - 512

数字相减:

576 - 512 = 64

= 64

因式分解数字:

64 = 8^{2}

= 8^{2}

使用根式运算法则:

n a^{n} = a

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 24 ) \pm 8}{2 \cdot 16}

将解分隔开

u_{1} = \frac{- ( - 24 ) + 8}{2 \cdot 16}, u_{2} = \frac{- ( - 24 ) - 8}{2 \cdot 16}

u = \frac{- ( - 24 ) + 8}{2 \cdot 16} : 1

\frac{- ( - 24 ) + 8}{2 \cdot 16}

使用法则

- (- a) = a

= \frac{24 + 8}{2 \cdot 16}

数字相加:

24 + 8 = 32

= \frac{32}{2 \cdot 16}

数字相乘:

2 \cdot 16 = 32

= \frac{32}{32}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 24 ) - 8}{2 \cdot 16} : \frac{1}{2}

\frac{- ( - 24 ) - 8}{2 \cdot 16}

使用法则

- (- a) = a

= \frac{24 - 8}{2 \cdot 16}

数字相减:

24 - 8 = 16

= \frac{16}{2 \cdot 16}

数字相乘:

2 \cdot 16 = 32

= \frac{16}{32}

约分:

16

= \frac{1}{2}

二次方程组的解是:

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

u = cos (θ)

代回

cos (θ) = 1, cos (θ) = \frac{1}{2}

cos (θ) = 1, cos (θ) = \frac{1}{2}

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

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

合并所有解

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

作图

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