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受欢迎的三角函数 >

2cos(x)+4cos(4x)=0

解答

2 cos (x) + 4 cos (4 x) = 0

解答

x = 2.87471 \dots + 2 πn, x = - 2.87471 \dots + 2 πn, x = 0.50586 \dots + 2 πn, x = 2 π - 0.50586 \dots + 2 πn, x = 1.92046 \dots + 2 πn, x = - 1.92046 \dots + 2 πn, x = 1.12361 \dots + 2 πn, x = 2 π - 1.12361 \dots + 2 πn

+1

度数

x = 164.70891 \dots^{\circ} + 36 0^{\circ} n, x = - 164.70891 \dots^{\circ} + 36 0^{\circ} n, x = 28.98415 \dots^{\circ} + 36 0^{\circ} n, x = 331.01584 \dots^{\circ} + 36 0^{\circ} n, x = 110.03427 \dots^{\circ} + 36 0^{\circ} n, x = - 110.03427 \dots^{\circ} + 36 0^{\circ} n, x = 64.37831 \dots^{\circ} + 36 0^{\circ} n, x = 295.62168 \dots^{\circ} + 36 0^{\circ} n

求解步骤

2 cos (x) + 4 cos (4 x) = 0

使用三角恒等式改写

2 cos (x) + 4 cos (4 x)

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

cos (4 x)

改写为

= cos (2 \cdot 2 x)

使用倍角公式:

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

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

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

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

使用倍角公式:

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

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

乘开

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

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

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

使用完全平方公式:

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

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

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

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

使用指数法则:

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

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

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

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)

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

2^{2} = 4

= 4 cos^{4} (x)

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

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

数字相乘:

2 \cdot 2 \cdot 1 = 4

= 4 cos^{2} (x)

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

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

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

乘开

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

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

打开括号

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

使用加减运算法则

+ (- a) = - a

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

化简

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

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

数字相乘:

2 \cdot 4 = 8

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

数字相乘:

2 \cdot 1 = 2

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

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

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

化简

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

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

对同类项分组

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

数字相加/相减:

- 1 + 2 = 1

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

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

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

(1 - 8 cos^{2} (x) + 8 cos^{4} (x)) \cdot 4 + 2 cos (x) = 0

用替代法求解

(1 - 8 cos^{2} (x) + 8 cos^{4} (x)) \cdot 4 + 2 cos (x) = 0

令

: cos (x) = u

(1 - 8 u^{2} + 8 u^{4}) \cdot 4 + 2 u = 0

(1 - 8 u^{2} + 8 u^{4}) \cdot 4 + 2 u = 0 : u \approx - 0.96459 \dots, u \approx 0.87475 \dots, u \approx - 0.34258 \dots, u \approx 0.43242 \dots

(1 - 8 u^{2} + 8 u^{4}) \cdot 4 + 2 u = 0

展开

(1 - 8 u^{2} + 8 u^{4}) \cdot 4 + 2 u : 4 - 32 u^{2} + 32 u^{4} + 2 u

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

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

乘开

4 (1 - 8 u^{2} + 8 u^{4}) : 4 - 32 u^{2} + 32 u^{4}

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

打开括号

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

使用加减运算法则

+ (- a) = - a

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

化简

4 \cdot 1 - 4 \cdot 8 u^{2} + 4 \cdot 8 u^{4} : 4 - 32 u^{2} + 32 u^{4}

4 \cdot 1 - 4 \cdot 8 u^{2} + 4 \cdot 8 u^{4}

数字相乘:

4 \cdot 1 = 4

= 4 - 4 \cdot 8 u^{2} + 4 \cdot 8 u^{4}

数字相乘:

4 \cdot 8 = 32

= 4 - 32 u^{2} + 32 u^{4}

= 4 - 32 u^{2} + 32 u^{4}

= 4 - 32 u^{2} + 32 u^{4} + 2 u

4 - 32 u^{2} + 32 u^{4} + 2 u = 0

改写成标准形式

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

32 u^{4} - 32 u^{2} + 2 u + 4 = 0

使用牛顿-拉弗森方法找到

32 u^{4} - 32 u^{2} + 2 u + 4 = 0

的一个解:

u \approx - 0.96459 \dots

32 u^{4} - 32 u^{2} + 2 u + 4 = 0

牛顿-拉弗森近似法定义

f (u) = 32 u^{4} - 32 u^{2} + 2 u + 4

找到

f^{^{'}} (u) : 128 u^{3} - 64 u + 2

\frac{d}{d u} (32 u^{4} - 32 u^{2} + 2 u + 4)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (32 u^{4}) - \frac{d}{d u} (32 u^{2}) + \frac{d}{d u} (2 u) + \frac{d}{d u} (4)

\frac{d}{d u} (32 u^{4}) = 128 u^{3}

\frac{d}{d u} (32 u^{4})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 32 \frac{d}{d u} (u^{4})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 32 \cdot 4 u^{4 - 1}

化简

= 128 u^{3}

\frac{d}{d u} (32 u^{2}) = 64 u

\frac{d}{d u} (32 u^{2})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 32 \frac{d}{d u} (u^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 32 \cdot 2 u^{2 - 1}

化简

= 64 u

\frac{d}{d u} (2 u) = 2

\frac{d}{d u} (2 u)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d u}{d u}

使用常见微分定则:

\frac{d u}{d u} = 1

= 2 \cdot 1

化简

= 2

\frac{d}{d u} (4) = 0

\frac{d}{d u} (4)

常数微分:

\frac{d}{d x} (a) = 0

= 0

= 128 u^{3} - 64 u + 2 + 0

化简

= 128 u^{3} - 64 u + 2

令

u_{0} = - 2

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 1.57046 \dots : Δ u_{1} = 0.42953 \dots

f (u_{0}) = 32 (- 2)^{4} - 32 (- 2)^{2} + 2 (- 2) + 4 = 384

f^{^{'}} (u_{0}) = 128 (- 2)^{3} - 64 (- 2) + 2 = - 894

u_{1} = - 1.57046 \dots

Δ u_{1} = ∣ - 1.57046 \dots - (- 2) ∣ = 0.42953 \dots

Δ u_{1} = 0.42953 \dots

u_{2} = - 1.27401 \dots : Δ u_{2} = 0.29645 \dots

f (u_{1}) = 32 (- 1.57046 \dots)^{4} - 32 (- 1.57046 \dots)^{2} + 2 (- 1.57046 \dots) + 4 = 116.59128 \dots

f^{^{'}} (u_{1}) = 128 (- 1.57046 \dots)^{3} - 64 (- 1.57046 \dots) + 2 = - 393.28104 \dots

u_{2} = - 1.27401 \dots

Δ u_{2} = ∣ - 1.27401 \dots - (- 1.57046 \dots) ∣ = 0.29645 \dots

Δ u_{2} = 0.29645 \dots

u_{3} = - 1.08733 \dots : Δ u_{3} = 0.18667 \dots

f (u_{2}) = 32 (- 1.27401 \dots)^{4} - 32 (- 1.27401 \dots)^{2} + 2 (- 1.27401 \dots) + 4 = 33.81573 \dots

f^{^{'}} (u_{2}) = 128 (- 1.27401 \dots)^{3} - 64 (- 1.27401 \dots) + 2 = - 181.14887 \dots

u_{3} = - 1.08733 \dots

Δ u_{3} = ∣ - 1.08733 \dots - (- 1.27401 \dots) ∣ = 0.18667 \dots

Δ u_{3} = 0.18667 \dots

u_{4} = - 0.99350 \dots : Δ u_{4} = 0.09382 \dots

f (u_{3}) = 32 (- 1.08733 \dots)^{4} - 32 (- 1.08733 \dots)^{2} + 2 (- 1.08733 \dots) + 4 = 8.72257 \dots

f^{^{'}} (u_{3}) = 128 (- 1.08733 \dots)^{3} - 64 (- 1.08733 \dots) + 2 = - 92.96260 \dots

u_{4} = - 0.99350 \dots

Δ u_{4} = ∣ - 0.99350 \dots - (- 1.08733 \dots) ∣ = 0.09382 \dots

Δ u_{4} = 0.09382 \dots

u_{5} = - 0.96674 \dots : Δ u_{5} = 0.02676 \dots

f (u_{4}) = 32 (- 0.99350 \dots)^{4} - 32 (- 0.99350 \dots)^{2} + 2 (- 0.99350 \dots) + 4 = 1.60428 \dots

f^{^{'}} (u_{4}) = 128 (- 0.99350 \dots)^{3} - 64 (- 0.99350 \dots) + 2 = - 59.93911 \dots

u_{5} = - 0.96674 \dots

Δ u_{5} = ∣ - 0.96674 \dots - (- 0.99350 \dots) ∣ = 0.02676 \dots

Δ u_{5} = 0.02676 \dots

u_{6} = - 0.96461 \dots : Δ u_{6} = 0.00213 \dots

f (u_{5}) = 32 (- 0.96674 \dots)^{4} - 32 (- 0.96674 \dots)^{2} + 2 (- 0.96674 \dots) + 4 = 0.11041 \dots

f^{^{'}} (u_{5}) = 128 (- 0.96674 \dots)^{3} - 64 (- 0.96674 \dots) + 2 = - 51.77809 \dots

u_{6} = - 0.96461 \dots

Δ u_{6} = ∣ - 0.96461 \dots - (- 0.96674 \dots) ∣ = 0.00213 \dots

Δ u_{6} = 0.00213 \dots

u_{7} = - 0.96459 \dots : Δ u_{7} = 0.00001 \dots

f (u_{6}) = 32 (- 0.96461 \dots)^{4} - 32 (- 0.96461 \dots)^{2} + 2 (- 0.96461 \dots) + 4 = 0.00066 \dots

f^{^{'}} (u_{6}) = 128 (- 0.96461 \dots)^{3} - 64 (- 0.96461 \dots) + 2 = - 51.15092 \dots

u_{7} = - 0.96459 \dots

Δ u_{7} = ∣ - 0.96459 \dots - (- 0.96461 \dots) ∣ = 0.00001 \dots

Δ u_{7} = 0.00001 \dots

u_{8} = - 0.96459 \dots : Δ u_{8} = 4.90935 E - 10

f (u_{7}) = 32 (- 0.96459 \dots)^{4} - 32 (- 0.96459 \dots)^{2} + 2 (- 0.96459 \dots) + 4 = 2.51099 E - 8

f^{^{'}} (u_{7}) = 128 (- 0.96459 \dots)^{3} - 64 (- 0.96459 \dots) + 2 = - 51.14709 \dots

u_{8} = - 0.96459 \dots

Δ u_{8} = ∣ - 0.96459 \dots - (- 0.96459 \dots) ∣ = 4.90935 E - 10

Δ u_{8} = 4.90935 E - 10

u \approx - 0.96459 \dots

使用长除法

Eq u a t i o n 0 : \frac{32 u ^{4} - 32 u ^{2} + 2 u + 4}{u + 0.96459 \dots} = 32 u^{3} - 30.86715 \dots u^{2} - 2.22559 \dots u + 4.14680 \dots

32 u^{3} - 30.86715 \dots u^{2} - 2.22559 \dots u + 4.14680 \dots \approx 0

使用牛顿-拉弗森方法找到

32 u^{3} - 30.86715 \dots u^{2} - 2.22559 \dots u + 4.14680 \dots = 0

的一个解:

u \approx 0.87475 \dots

32 u^{3} - 30.86715 \dots u^{2} - 2.22559 \dots u + 4.14680 \dots = 0

牛顿-拉弗森近似法定义

f (u) = 32 u^{3} - 30.86715 \dots u^{2} - 2.22559 \dots u + 4.14680 \dots

找到

f^{^{'}} (u) : 96 u^{2} - 61.73430 \dots u - 2.22559 \dots

\frac{d}{d u} (32 u^{3} - 30.86715 \dots u^{2} - 2.22559 \dots u + 4.14680 \dots)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (32 u^{3}) - \frac{d}{d u} (30.86715 \dots u^{2}) - \frac{d}{d u} (2.22559 \dots u) + \frac{d}{d u} (4.14680 \dots)

\frac{d}{d u} (32 u^{3}) = 96 u^{2}

\frac{d}{d u} (32 u^{3})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 32 \frac{d}{d u} (u^{3})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 32 \cdot 3 u^{3 - 1}

化简

= 96 u^{2}

\frac{d}{d u} (30.86715 \dots u^{2}) = 61.73430 \dots u

\frac{d}{d u} (30.86715 \dots u^{2})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 30.86715 \dots \frac{d}{d u} (u^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 30.86715 \dots \cdot 2 u^{2 - 1}

化简

= 61.73430 \dots u

\frac{d}{d u} (2.22559 \dots u) = 2.22559 \dots

\frac{d}{d u} (2.22559 \dots u)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 2.22559 \dots \frac{d u}{d u}

使用常见微分定则:

\frac{d u}{d u} = 1

= 2.22559 \dots \cdot 1

化简

= 2.22559 \dots

\frac{d}{d u} (4.14680 \dots) = 0

\frac{d}{d u} (4.14680 \dots)

常数微分:

\frac{d}{d x} (a) = 0

= 0

= 96 u^{2} - 61.73430 \dots u - 2.22559 \dots + 0

化简

= 96 u^{2} - 61.73430 \dots u - 2.22559 \dots

令

u_{0} = 2

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 1.48809 \dots : Δ u_{1} = 0.51190 \dots

f (u_{0}) = 32 \cdot 2^{3} - 30.86715 \dots \cdot 2^{2} - 2.22559 \dots \cdot 2 + 4.14680 \dots = 132.22701 \dots

f^{^{'}} (u_{0}) = 96 \cdot 2^{2} - 61.73430 \dots \cdot 2 - 2.22559 \dots = 258.30580 \dots

u_{1} = 1.48809 \dots

Δ u_{1} = ∣ 1.48809 \dots - 2 ∣ = 0.51190 \dots

Δ u_{1} = 0.51190 \dots

u_{2} = 1.16798 \dots : Δ u_{2} = 0.32011 \dots

f (u_{1}) = 32 \cdot 1.48809 \dots^{3} - 30.86715 \dots \cdot 1.48809 \dots^{2} - 2.22559 \dots \cdot 1.48809 \dots + 4.14680 \dots = 37.93120 \dots

f^{^{'}} (u_{1}) = 96 \cdot 1.48809 \dots^{2} - 61.73430 \dots \cdot 1.48809 \dots - 2.22559 \dots = 118.49374 \dots

u_{2} = 1.16798 \dots

Δ u_{2} = ∣ 1.16798 \dots - 1.48809 \dots ∣ = 0.32011 \dots

Δ u_{2} = 0.32011 \dots

u_{3} = 0.98388 \dots : Δ u_{3} = 0.18410 \dots

f (u_{2}) = 32 \cdot 1.16798 \dots^{3} - 30.86715 \dots \cdot 1.16798 \dots^{2} - 2.22559 \dots \cdot 1.16798 \dots + 4.14680 \dots = 10.42612 \dots

f^{^{'}} (u_{2}) = 96 \cdot 1.16798 \dots^{2} - 61.73430 \dots \cdot 1.16798 \dots - 2.22559 \dots = 56.63221 \dots

u_{3} = 0.98388 \dots

Δ u_{3} = ∣ 0.98388 \dots - 1.16798 \dots ∣ = 0.18410 \dots

Δ u_{3} = 0.18410 \dots

u_{4} = 0.89863 \dots : Δ u_{4} = 0.08524 \dots

f (u_{3}) = 32 \cdot 0.98388 \dots^{3} - 30.86715 \dots \cdot 0.98388 \dots^{2} - 2.22559 \dots \cdot 0.98388 \dots + 4.14680 \dots = 2.55451 \dots

f^{^{'}} (u_{3}) = 96 \cdot 0.98388 \dots^{2} - 61.73430 \dots \cdot 0.98388 \dots - 2.22559 \dots = 29.96580 \dots

u_{4} = 0.89863 \dots

Δ u_{4} = ∣ 0.89863 \dots - 0.98388 \dots ∣ = 0.08524 \dots

Δ u_{4} = 0.08524 \dots

u_{5} = 0.87632 \dots : Δ u_{5} = 0.02231 \dots

f (u_{4}) = 32 \cdot 0.89863 \dots^{3} - 30.86715 \dots \cdot 0.89863 \dots^{2} - 2.22559 \dots \cdot 0.89863 \dots + 4.14680 \dots = 0.44226 \dots

f^{^{'}} (u_{4}) = 96 \cdot 0.89863 \dots^{2} - 61.73430 \dots \cdot 0.89863 \dots - 2.22559 \dots = 19.82237 \dots

u_{5} = 0.87632 \dots

Δ u_{5} = ∣ 0.87632 \dots - 0.89863 \dots ∣ = 0.02231 \dots

Δ u_{5} = 0.02231 \dots

u_{6} = 0.87476 \dots : Δ u_{6} = 0.00156 \dots

f (u_{5}) = 32 \cdot 0.87632 \dots^{3} - 30.86715 \dots \cdot 0.87632 \dots^{2} - 2.22559 \dots \cdot 0.87632 \dots + 4.14680 \dots = 0.02722 \dots

f^{^{'}} (u_{5}) = 96 \cdot 0.87632 \dots^{2} - 61.73430 \dots \cdot 0.87632 \dots - 2.22559 \dots = 17.39797 \dots

u_{6} = 0.87476 \dots

Δ u_{6} = ∣ 0.87476 \dots - 0.87632 \dots ∣ = 0.00156 \dots

Δ u_{6} = 0.00156 \dots

u_{7} = 0.87475 \dots : Δ u_{7} = 7.56069 E - 6

f (u_{6}) = 32 \cdot 0.87476 \dots^{3} - 30.86715 \dots \cdot 0.87476 \dots^{2} - 2.22559 \dots \cdot 0.87476 \dots + 4.14680 \dots = 0.00013 \dots

f^{^{'}} (u_{6}) = 96 \cdot 0.87476 \dots^{2} - 61.73430 \dots \cdot 0.87476 \dots - 2.22559 \dots = 17.23152 \dots

u_{7} = 0.87475 \dots

Δ u_{7} = ∣ 0.87475 \dots - 0.87476 \dots ∣ = 7.56069 E - 6

Δ u_{7} = 7.56069 E - 6

u_{8} = 0.87475 \dots : Δ u_{8} = 1.76195 E - 10

f (u_{7}) = 32 \cdot 0.87475 \dots^{3} - 30.86715 \dots \cdot 0.87475 \dots^{2} - 2.22559 \dots \cdot 0.87475 \dots + 4.14680 \dots = 3.03597 E - 9

f^{^{'}} (u_{7}) = 96 \cdot 0.87475 \dots^{2} - 61.73430 \dots \cdot 0.87475 \dots - 2.22559 \dots = 17.23072 \dots

u_{8} = 0.87475 \dots

Δ u_{8} = ∣ 0.87475 \dots - 0.87475 \dots ∣ = 1.76195 E - 10

Δ u_{8} = 1.76195 E - 10

u \approx 0.87475 \dots

使用长除法

Eq u a t i o n 0 : \frac{32 u ^{3} - 30.86715 \dots u ^{2} - 2.22559 \dots u + 4.14680 \dots}{u - 0.87475 \dots} = 32 u^{2} - 2.87503 \dots u - 4.74053 \dots

32 u^{2} - 2.87503 \dots u - 4.74053 \dots \approx 0

使用牛顿-拉弗森方法找到

32 u^{2} - 2.87503 \dots u - 4.74053 \dots = 0

的一个解:

u \approx - 0.34258 \dots

32 u^{2} - 2.87503 \dots u - 4.74053 \dots = 0

牛顿-拉弗森近似法定义

f (u) = 32 u^{2} - 2.87503 \dots u - 4.74053 \dots

找到

f^{^{'}} (u) : 64 u - 2.87503 \dots

\frac{d}{d u} (32 u^{2} - 2.87503 \dots u - 4.74053 \dots)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (32 u^{2}) - \frac{d}{d u} (2.87503 \dots u) - \frac{d}{d u} (4.74053 \dots)

\frac{d}{d u} (32 u^{2}) = 64 u

\frac{d}{d u} (32 u^{2})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 32 \frac{d}{d u} (u^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 32 \cdot 2 u^{2 - 1}

化简

= 64 u

\frac{d}{d u} (2.87503 \dots u) = 2.87503 \dots

\frac{d}{d u} (2.87503 \dots u)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 2.87503 \dots \frac{d u}{d u}

使用常见微分定则:

\frac{d u}{d u} = 1

= 2.87503 \dots \cdot 1

化简

= 2.87503 \dots

\frac{d}{d u} (4.74053 \dots) = 0

\frac{d}{d u} (4.74053 \dots)

常数微分:

\frac{d}{d x} (a) = 0

= 0

= 64 u - 2.87503 \dots - 0

化简

= 64 u - 2.87503 \dots

令

u_{0} = - 2

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 1.01425 \dots : Δ u_{1} = 0.98574 \dots

f (u_{0}) = 32 (- 2)^{2} - 2.87503 \dots (- 2) - 4.74053 \dots = 129.00952 \dots

f^{^{'}} (u_{0}) = 64 (- 2) - 2.87503 \dots = - 130.87503 \dots

u_{1} = - 1.01425 \dots

Δ u_{1} = ∣ - 1.01425 \dots - (- 2) ∣ = 0.98574 \dots

Δ u_{1} = 0.98574 \dots

u_{2} = - 0.55555 \dots : Δ u_{2} = 0.45870 \dots

f (u_{1}) = 32 (- 1.01425 \dots)^{2} - 2.87503 \dots (- 1.01425 \dots) - 4.74053 \dots = 31.09423 \dots

f^{^{'}} (u_{1}) = 64 (- 1.01425 \dots) - 2.87503 \dots = - 67.78729 \dots

u_{2} = - 0.55555 \dots

Δ u_{2} = ∣ - 0.55555 \dots - (- 1.01425 \dots) ∣ = 0.45870 \dots

Δ u_{2} = 0.45870 \dots

u_{3} = - 0.38034 \dots : Δ u_{3} = 0.17520 \dots

f (u_{2}) = 32 (- 0.55555 \dots)^{2} - 2.87503 \dots (- 0.55555 \dots) - 4.74053 \dots = 6.73307 \dots

f^{^{'}} (u_{2}) = 64 (- 0.55555 \dots) - 2.87503 \dots = - 38.43029 \dots

u_{3} = - 0.38034 \dots

Δ u_{3} = ∣ - 0.38034 \dots - (- 0.55555 \dots) ∣ = 0.17520 \dots

Δ u_{3} = 0.17520 \dots

u_{4} = - 0.34425 \dots : Δ u_{4} = 0.03608 \dots

f (u_{3}) = 32 (- 0.38034 \dots)^{2} - 2.87503 \dots (- 0.38034 \dots) - 4.74053 \dots = 0.98226 \dots

f^{^{'}} (u_{3}) = 64 (- 0.38034 \dots) - 2.87503 \dots = - 27.21735 \dots

u_{4} = - 0.34425 \dots

Δ u_{4} = ∣ - 0.34425 \dots - (- 0.38034 \dots) ∣ = 0.03608 \dots

Δ u_{4} = 0.03608 \dots

u_{5} = - 0.34258 \dots : Δ u_{5} = 0.00167 \dots

f (u_{4}) = 32 (- 0.34425 \dots)^{2} - 2.87503 \dots (- 0.34425 \dots) - 4.74053 \dots = 0.04167 \dots

f^{^{'}} (u_{4}) = 64 (- 0.34425 \dots) - 2.87503 \dots = - 24.90762 \dots

u_{5} = - 0.34258 \dots

Δ u_{5} = ∣ - 0.34258 \dots - (- 0.34425 \dots) ∣ = 0.00167 \dots

Δ u_{5} = 0.00167 \dots

u_{6} = - 0.34258 \dots : Δ u_{6} = 3.6129 E - 6

f (u_{5}) = 32 (- 0.34258 \dots)^{2} - 2.87503 \dots (- 0.34258 \dots) - 4.74053 \dots = 0.00008 \dots

f^{^{'}} (u_{5}) = 64 (- 0.34258 \dots) - 2.87503 \dots = - 24.80052 \dots

u_{6} = - 0.34258 \dots

Δ u_{6} = ∣ - 0.34258 \dots - (- 0.34258 \dots) ∣ = 3.6129 E - 6

Δ u_{6} = 3.6129 E - 6

u_{7} = - 0.34258 \dots : Δ u_{7} = 1.68425 E - 11

f (u_{6}) = 32 (- 0.34258 \dots)^{2} - 2.87503 \dots (- 0.34258 \dots) - 4.74053 \dots = 4.17699 E - 10

f^{^{'}} (u_{6}) = 64 (- 0.34258 \dots) - 2.87503 \dots = - 24.80029 \dots

u_{7} = - 0.34258 \dots

Δ u_{7} = ∣ - 0.34258 \dots - (- 0.34258 \dots) ∣ = 1.68425 E - 11

Δ u_{7} = 1.68425 E - 11

u \approx - 0.34258 \dots

使用长除法

Eq u a t i o n 0 : \frac{32 u ^{2} - 2.87503 \dots u - 4.74053 \dots}{u + 0.34258 \dots} = 32 u - 13.83766 \dots

32 u - 13.83766 \dots \approx 0

u \approx 0.43242 \dots

解为

u \approx - 0.96459 \dots, u \approx 0.87475 \dots, u \approx - 0.34258 \dots, u \approx 0.43242 \dots

u = cos (x)

代回

cos (x) \approx - 0.96459 \dots, cos (x) \approx 0.87475 \dots, cos (x) \approx - 0.34258 \dots, cos (x) \approx 0.43242 \dots

cos (x) \approx - 0.96459 \dots, cos (x) \approx 0.87475 \dots, cos (x) \approx - 0.34258 \dots, cos (x) \approx 0.43242 \dots

cos (x) = - 0.96459 \dots : x = arccos (- 0.96459 \dots) + 2 πn, x = - arccos (- 0.96459 \dots) + 2 πn

cos (x) = - 0.96459 \dots

使用反三角函数性质

cos (x) = - 0.96459 \dots

cos (x) = - 0.96459 \dots

的通解

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

x = arccos (- 0.96459 \dots) + 2 πn, x = - arccos (- 0.96459 \dots) + 2 πn

x = arccos (- 0.96459 \dots) + 2 πn, x = - arccos (- 0.96459 \dots) + 2 πn

cos (x) = 0.87475 \dots : x = arccos (0.87475 \dots) + 2 πn, x = 2 π - arccos (0.87475 \dots) + 2 πn

cos (x) = 0.87475 \dots

使用反三角函数性质

cos (x) = 0.87475 \dots

cos (x) = 0.87475 \dots

的通解

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

x = arccos (0.87475 \dots) + 2 πn, x = 2 π - arccos (0.87475 \dots) + 2 πn

x = arccos (0.87475 \dots) + 2 πn, x = 2 π - arccos (0.87475 \dots) + 2 πn

cos (x) = - 0.34258 \dots : x = arccos (- 0.34258 \dots) + 2 πn, x = - arccos (- 0.34258 \dots) + 2 πn

cos (x) = - 0.34258 \dots

使用反三角函数性质

cos (x) = - 0.34258 \dots

cos (x) = - 0.34258 \dots

的通解

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

x = arccos (- 0.34258 \dots) + 2 πn, x = - arccos (- 0.34258 \dots) + 2 πn

x = arccos (- 0.34258 \dots) + 2 πn, x = - arccos (- 0.34258 \dots) + 2 πn

cos (x) = 0.43242 \dots : x = arccos (0.43242 \dots) + 2 πn, x = 2 π - arccos (0.43242 \dots) + 2 πn

cos (x) = 0.43242 \dots

使用反三角函数性质

cos (x) = 0.43242 \dots

cos (x) = 0.43242 \dots

的通解

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

x = arccos (0.43242 \dots) + 2 πn, x = 2 π - arccos (0.43242 \dots) + 2 πn

x = arccos (0.43242 \dots) + 2 πn, x = 2 π - arccos (0.43242 \dots) + 2 πn

合并所有解

x = arccos (- 0.96459 \dots) + 2 πn, x = - arccos (- 0.96459 \dots) + 2 πn, x = arccos (0.87475 \dots) + 2 πn, x = 2 π - arccos (0.87475 \dots) + 2 πn, x = arccos (- 0.34258 \dots) + 2 πn, x = - arccos (- 0.34258 \dots) + 2 πn, x = arccos (0.43242 \dots) + 2 πn, x = 2 π - arccos (0.43242 \dots) + 2 πn

以小数形式表示解

x = 2.87471 \dots + 2 πn, x = - 2.87471 \dots + 2 πn, x = 0.50586 \dots + 2 πn, x = 2 π - 0.50586 \dots + 2 πn, x = 1.92046 \dots + 2 πn, x = - 1.92046 \dots + 2 πn, x = 1.12361 \dots + 2 πn, x = 2 π - 1.12361 \dots + 2 πn

作图

流行的例子

tan (θ) = \frac{24}{7}

sin(x)=0.18,0<= x<2pi

sin (x) = 0.18, 0 \leq x < 2 π

3 cos (x) = sin (x)

证明 tan(135+x)=(tan(x)-1)/(tan(x)+1)

p ro v e tan (13 5^{\circ} + x) = \frac{tan ( x ) - 1}{tan ( x ) + 1}

sinh (x) = \frac{36}{77}