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受欢迎的三角函数 >

(1+4cos(θ))^2=(sqrt(3)sin(θ))

解答

(1 + 4 cos (θ))^{2} = (3 sin (θ))

解答

θ = 1.49220 \dots + 2 πn, θ = 2.15388 \dots + 2 πn

+1

度数

θ = 85.49705 \dots^{\circ} + 36 0^{\circ} n, θ = 123.40880 \dots^{\circ} + 36 0^{\circ} n

求解步骤

(1 + 4 cos (θ))^{2} = (3 sin (θ))

两边进行平方

((1 + 4 cos (θ))^{2})^{2} = (3 sin (θ))^{2}

两边减去

(3 sin (θ))^{2}

(1 + 4 cos (θ))^{4} - 3 sin^{2} (θ) = 0

使用三角恒等式改写

(1 + 4 cos (θ))^{4} - 3 sin^{2} (θ)

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 + 4 cos (θ))^{4} - 3 (1 - cos^{2} (θ))

化简

(1 + 4 cos (θ))^{4} - 3 (1 - cos^{2} (θ)) : 256 cos^{4} (θ) + 256 cos^{3} (θ) + 99 cos^{2} (θ) + 16 cos (θ) - 2

(1 + 4 cos (θ))^{4} - 3 (1 - cos^{2} (θ))

(1 + 4 cos (θ))^{4} : 1 + 16 cos (θ) + 96 cos^{2} (θ) + 256 cos^{3} (θ) + 256 cos^{4} (θ)

使用二项式定理:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = 1, b = 4 cos (θ)

= i = 0 \sum 4 (i 4) \cdot 1^{(4 - i)} (4 cos (θ))^{i}

展开求和

(i n) = \frac{n !}{i ! ( n - i ) !}

i = 0 : \frac{4 !}{0 ! ( 4 - 0 ) !} 1^{4} (4 cos (θ))^{0}

i = 1 : \frac{4 !}{1 ! ( 4 - 1 ) !} 1^{3} (4 cos (θ))^{1}

i = 2 : \frac{4 !}{2 ! ( 4 - 2 ) !} 1^{2} (4 cos (θ))^{2}

i = 3 : \frac{4 !}{3 ! ( 4 - 3 ) !} 1^{1} (4 cos (θ))^{3}

i = 4 : \frac{4 !}{4 ! ( 4 - 4 ) !} 1^{0} (4 cos (θ))^{4}

= \frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 1^{4} (4 cos (θ))^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 1^{3} (4 cos (θ))^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 1^{2} (4 cos (θ))^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 1^{1} (4 cos (θ))^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 1^{0} (4 cos (θ))^{4}

= \frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 1^{4} (4 cos (θ))^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 1^{3} (4 cos (θ))^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 1^{2} (4 cos (θ))^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 1^{1} (4 cos (θ))^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 1^{0} (4 cos (θ))^{4}

\frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 1^{4} (4 cos (θ))^{0} = 1

\frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 1^{4} (4 cos (θ))^{0}

使用法则

1^{a} = 1

1^{4} = 1

= 1 \cdot \frac{4 !}{0 ! ( 4 - 0 ) !} (4 cos (θ))^{0}

使用法则

a^{0} = 1, a \neq = 0

(4 cos (θ))^{0} = 1

= 1 \cdot 1 \cdot \frac{4 !}{0 ! ( 4 - 0 ) !}

\frac{4 !}{0 ! ( 4 - 0 ) !} = 1

\frac{4 !}{0 ! ( 4 - 0 ) !}

0! (4 - 0)! = 4!

0! (4 - 0)!

数字相减:

4 - 0 = 4

= 0! \cdot 4!

使用因式运算法则:

0! = 1

= 1 \cdot 4!

乘以:

1 \cdot 4! = 4!

= 4!

= \frac{4 !}{4 !}

使用法则

\frac{a}{a} = 1

= 1

= 1 \cdot 1 \cdot 1

数字相乘:

1 \cdot 1 \cdot 1 = 1

= 1

化简

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 1^{3} (4 cos (θ))^{1} : 16 cos (θ)

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 1^{3} (4 cos (θ))^{1}

使用法则

1^{a} = 1

1^{3} = 1

= 1 \cdot \frac{4 !}{1 ! ( 4 - 1 ) !} (4 cos (θ))^{1}

使用法则

a^{1} = a

(4 cos (θ))^{1} = 4 cos (θ)

= 1 \cdot 4 \cdot \frac{4 !}{1 ! ( 4 - 1 ) !} cos (θ)

分式相乘:

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

= 1 \cdot \frac{4 \cdot 4 ! cos ( θ )}{1 ! ( 4 - 1 ) !}

化简

\frac{4 ! \cdot 4 cos ( θ )}{1 ! ( 4 - 1 ) !} : 16 cos (θ)

\frac{4 ! \cdot 4 cos ( θ )}{1 ! ( 4 - 1 ) !}

数字相减:

4 - 1 = 3

= \frac{4 \cdot 4 ! cos ( θ )}{1 ! \cdot 3 !}

消去因子:

\frac{n !}{( n - m ) !} = n \cdot (n - 1) \dots (n - m + 1), n > m

\frac{4 !}{3 !} = 4

= \frac{4 \cdot 4 cos ( θ )}{1 !}

整理后得

= \frac{16 cos ( θ )}{1 !}

使用因式运算法则:

n! = 1 \cdot 2 \cdot 3 \cdot \dots \cdot n

1! = 1

= \frac{16 cos ( θ )}{1}

使用法则

\frac{a}{1} = a

= 16 cos (θ)

= 1 \cdot 16 cos (θ)

数字相乘:

1 \cdot 16 = 16

= 16 cos (θ)

化简

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 1^{2} (4 cos (θ))^{2} : 96 cos^{2} (θ)

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 1^{2} (4 cos (θ))^{2}

使用法则

1^{a} = 1

1^{2} = 1

= 1 \cdot \frac{4 !}{2 ! ( 4 - 2 ) !} (4 cos (θ))^{2}

分式相乘:

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

= 1 \cdot \frac{4 ! ( 4 cos ( θ ) ) ^{2}}{2 ! ( 4 - 2 ) !}

化简

\frac{4 ! ( 4 cos ( θ ) ) ^{2}}{2 ! ( 4 - 2 ) !} : 96 cos^{2} (θ)

\frac{4 ! ( 4 cos ( θ ) ) ^{2}}{2 ! ( 4 - 2 ) !}

数字相减:

4 - 2 = 2

= \frac{4 ! ( 4 cos ( θ ) ) ^{2}}{2 ! \cdot 2 !}

消去因子:

\frac{n !}{( n - m ) !} = n \cdot (n - 1) \dots (n - m + 1), n > m

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

= \frac{4 \cdot 3 ( 4 cos ( θ ) ) ^{2}}{2 !}

整理后得

= \frac{12 ( 4 cos ( θ ) ) ^{2}}{2 !}

(4 cos (θ))^{2} = 4^{2} cos^{2} (θ)

(4 cos (θ))^{2}

使用指数法则:

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

= 4^{2} cos^{2} (θ)

= \frac{4 ^{2} \cdot 12 cos ^{2} ( θ )}{2 !}

12 \cdot 4^{2} cos^{2} (θ) = 192 cos^{2} (θ)

12 \cdot 4^{2} cos^{2} (θ)

4^{2} = 16

= 12 \cdot 16 cos^{2} (θ)

数字相乘:

12 \cdot 16 = 192

= 192 cos^{2} (θ)

= \frac{192 cos ^{2} ( θ )}{2 !}

2! = 2

2!

使用因式运算法则:

n! = 1 \cdot 2 \cdot 3 \cdot \dots \cdot n

2! = 1 \cdot 2

= 1 \cdot 2

数字相乘:

1 \cdot 2 = 2

= 2

= \frac{192 cos ^{2} ( θ )}{2}

数字相除:

\frac{192}{2} = 96

= 96 cos^{2} (θ)

= 1 \cdot 96 cos^{2} (θ)

数字相乘:

1 \cdot 96 = 96

= 96 cos^{2} (θ)

化简

\frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 1^{1} (4 cos (θ))^{3} : 256 cos^{3} (θ)

\frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 1^{1} (4 cos (θ))^{3}

使用法则

1^{a} = 1

1^{1} = 1

= 1 \cdot \frac{4 !}{3 ! ( 4 - 3 ) !} (4 cos (θ))^{3}

分式相乘:

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

= 1 \cdot \frac{4 ! ( 4 cos ( θ ) ) ^{3}}{3 ! ( 4 - 3 ) !}

化简

\frac{4 ! ( 4 cos ( θ ) ) ^{3}}{3 ! ( 4 - 3 ) !} : 256 cos^{3} (θ)

\frac{4 ! ( 4 cos ( θ ) ) ^{3}}{3 ! ( 4 - 3 ) !}

数字相减:

4 - 3 = 1

= \frac{4 ! ( 4 cos ( θ ) ) ^{3}}{3 ! \cdot 1 !}

消去因子:

\frac{n !}{( n - m ) !} = n \cdot (n - 1) \dots (n - m + 1), n > m

\frac{4 !}{3 !} = 4

= \frac{4 ( 4 cos ( θ ) ) ^{3}}{1 !}

(4 cos (θ))^{3} = 4^{3} cos^{3} (θ)

(4 cos (θ))^{3}

使用指数法则:

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

= 4^{3} cos^{3} (θ)

= \frac{4 ^{3} \cdot 4 cos ^{3} ( θ )}{1 !}

4 \cdot 4^{3} cos^{3} (θ) = 4^{4} cos^{3} (θ)

4 \cdot 4^{3} cos^{3} (θ)

使用指数法则:

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

4 \cdot 4^{3} = 4^{1 + 3}

= 4^{1 + 3} cos^{3} (θ)

数字相加:

1 + 3 = 4

= 4^{4} cos^{3} (θ)

= \frac{4 ^{4} cos ^{3} ( θ )}{1 !}

4^{4} = 256

= \frac{256 cos ^{3} ( θ )}{1 !}

使用因式运算法则:

n! = 1 \cdot 2 \cdot 3 \cdot \dots \cdot n

1! = 1

= \frac{256 cos ^{3} ( θ )}{1}

使用法则

\frac{a}{1} = a

= 256 cos^{3} (θ)

= 1 \cdot 256 cos^{3} (θ)

数字相乘:

1 \cdot 256 = 256

= 256 cos^{3} (θ)

化简

\frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 1^{0} (4 cos (θ))^{4} : 256 cos^{4} (θ)

\frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 1^{0} (4 cos (θ))^{4}

使用法则

1^{a} = 1

1^{0} = 1

= 1 \cdot \frac{4 !}{4 ! ( 4 - 4 ) !} (4 cos (θ))^{4}

分式相乘:

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

= 1 \cdot \frac{4 ! ( 4 cos ( θ ) ) ^{4}}{4 ! ( 4 - 4 ) !}

约分:

4!

= 1 \cdot \frac{( 4 cos ( θ ) ) ^{4}}{( 4 - 4 ) !}

化简

\frac{( 4 cos ( θ ) ) ^{4}}{( 4 - 4 ) !} : 256 cos^{4} (θ)

\frac{( 4 cos ( θ ) ) ^{4}}{( 4 - 4 ) !}

(4 - 4)! = 1

(4 - 4)!

数字相减:

4 - 4 = 0

= 0!

使用因式运算法则:

0! = 1

= 1

= \frac{( 4 cos ( θ ) ) ^{4}}{1}

使用法则

\frac{a}{1} = a

= (4 cos (θ))^{4}

使用指数法则:

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

= 4^{4} cos^{4} (θ)

4^{4} = 256

= 256 cos^{4} (θ)

= 1 \cdot 256 cos^{4} (θ)

数字相乘:

1 \cdot 256 = 256

= 256 cos^{4} (θ)

= 1 + 16 cos (θ) + 96 cos^{2} (θ) + 256 cos^{3} (θ) + 256 cos^{4} (θ)

= 1 + 16 cos (θ) + 96 cos^{2} (θ) + 256 cos^{3} (θ) + 256 cos^{4} (θ) - 3 (1 - cos^{2} (θ))

乘开

- 3 (1 - cos^{2} (θ)) : - 3 + 3 cos^{2} (θ)

- 3 (1 - cos^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 3, b = 1, c = cos^{2} (θ)

= - 3 \cdot 1 - (- 3) cos^{2} (θ)

使用加减运算法则

- (- a) = a

= - 3 \cdot 1 + 3 cos^{2} (θ)

数字相乘:

3 \cdot 1 = 3

= - 3 + 3 cos^{2} (θ)

= 1 + 16 cos (θ) + 96 cos^{2} (θ) + 256 cos^{3} (θ) + 256 cos^{4} (θ) - 3 + 3 cos^{2} (θ)

化简

1 + 16 cos (θ) + 96 cos^{2} (θ) + 256 cos^{3} (θ) + 256 cos^{4} (θ) - 3 + 3 cos^{2} (θ) : 256 cos^{4} (θ) + 256 cos^{3} (θ) + 99 cos^{2} (θ) + 16 cos (θ) - 2

1 + 16 cos (θ) + 96 cos^{2} (θ) + 256 cos^{3} (θ) + 256 cos^{4} (θ) - 3 + 3 cos^{2} (θ)

对同类项分组

= 16 cos (θ) + 96 cos^{2} (θ) + 256 cos^{3} (θ) + 256 cos^{4} (θ) + 3 cos^{2} (θ) + 1 - 3

同类项相加:

96 cos^{2} (θ) + 3 cos^{2} (θ) = 99 cos^{2} (θ)

= 16 cos (θ) + 99 cos^{2} (θ) + 256 cos^{3} (θ) + 256 cos^{4} (θ) + 1 - 3

数字相加/相减:

1 - 3 = - 2

= 256 cos^{4} (θ) + 256 cos^{3} (θ) + 99 cos^{2} (θ) + 16 cos (θ) - 2

= 256 cos^{4} (θ) + 256 cos^{3} (θ) + 99 cos^{2} (θ) + 16 cos (θ) - 2

= 256 cos^{4} (θ) + 256 cos^{3} (θ) + 99 cos^{2} (θ) + 16 cos (θ) - 2

- 2 + 16 cos (θ) + 256 cos^{3} (θ) + 256 cos^{4} (θ) + 99 cos^{2} (θ) = 0

用替代法求解

- 2 + 16 cos (θ) + 256 cos^{3} (θ) + 256 cos^{4} (θ) + 99 cos^{2} (θ) = 0

令

: cos (θ) = u

- 2 + 16 u + 256 u^{3} + 256 u^{4} + 99 u^{2} = 0

- 2 + 16 u + 256 u^{3} + 256 u^{4} + 99 u^{2} = 0 : u \approx 0.07851 \dots, u \approx - 0.55060 \dots

- 2 + 16 u + 256 u^{3} + 256 u^{4} + 99 u^{2} = 0

改写成标准形式

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

256 u^{4} + 256 u^{3} + 99 u^{2} + 16 u - 2 = 0

使用牛顿-拉弗森方法找到

256 u^{4} + 256 u^{3} + 99 u^{2} + 16 u - 2 = 0

的一个解:

u \approx 0.07851 \dots

256 u^{4} + 256 u^{3} + 99 u^{2} + 16 u - 2 = 0

牛顿-拉弗森近似法定义

f (u) = 256 u^{4} + 256 u^{3} + 99 u^{2} + 16 u - 2

找到

f^{^{'}} (u) : 1024 u^{3} + 768 u^{2} + 198 u + 16

\frac{d}{d u} (256 u^{4} + 256 u^{3} + 99 u^{2} + 16 u - 2)

使用微分加减法定则:

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

= \frac{d}{d u} (256 u^{4}) + \frac{d}{d u} (256 u^{3}) + \frac{d}{d u} (99 u^{2}) + \frac{d}{d u} (16 u) - \frac{d}{d u} (2)

\frac{d}{d u} (256 u^{4}) = 1024 u^{3}

\frac{d}{d u} (256 u^{4})

将常数提出:

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

= 256 \frac{d}{d u} (u^{4})

使用幂法则:

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

= 256 \cdot 4 u^{4 - 1}

化简

= 1024 u^{3}

\frac{d}{d u} (256 u^{3}) = 768 u^{2}

\frac{d}{d u} (256 u^{3})

将常数提出:

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

= 256 \frac{d}{d u} (u^{3})

使用幂法则:

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

= 256 \cdot 3 u^{3 - 1}

化简

= 768 u^{2}

\frac{d}{d u} (99 u^{2}) = 198 u

\frac{d}{d u} (99 u^{2})

将常数提出:

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

= 99 \frac{d}{d u} (u^{2})

使用幂法则:

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

= 99 \cdot 2 u^{2 - 1}

化简

= 198 u

\frac{d}{d u} (16 u) = 16

\frac{d}{d u} (16 u)

将常数提出:

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

= 16 \frac{d u}{d u}

使用常见微分定则:

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

= 16 \cdot 1

化简

= 16

\frac{d}{d u} (2) = 0

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

常数微分:

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

= 0

= 1024 u^{3} + 768 u^{2} + 198 u + 16 - 0

化简

= 1024 u^{3} + 768 u^{2} + 198 u + 16

令

u_{0} = 0

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.125 : Δ u_{1} = 0.125

f (u_{0}) = 256 \cdot 0^{4} + 256 \cdot 0^{3} + 99 \cdot 0^{2} + 16 \cdot 0 - 2 = - 2

f^{^{'}} (u_{0}) = 1024 \cdot 0^{3} + 768 \cdot 0^{2} + 198 \cdot 0 + 16 = 16

u_{1} = 0.125

Δ u_{1} = ∣ 0.125 - 0 ∣ = 0.125

Δ u_{1} = 0.125

u_{2} = 0.08647 \dots : Δ u_{2} = 0.03852 \dots

f (u_{1}) = 256 \cdot 0.12 5^{4} + 256 \cdot 0.12 5^{3} + 99 \cdot 0.12 5^{2} + 16 \cdot 0.125 - 2 = 2.109375

f^{^{'}} (u_{1}) = 1024 \cdot 0.12 5^{3} + 768 \cdot 0.12 5^{2} + 198 \cdot 0.125 + 16 = 54.75

u_{2} = 0.08647 \dots

Δ u_{2} = ∣ 0.08647 \dots - 0.125 ∣ = 0.03852 \dots

Δ u_{2} = 0.03852 \dots

u_{3} = 0.07878 \dots : Δ u_{3} = 0.00768 \dots

f (u_{2}) = 256 \cdot 0.08647 \dots^{4} + 256 \cdot 0.08647 \dots^{3} + 99 \cdot 0.08647 \dots^{2} + 16 \cdot 0.08647 \dots - 2 = 0.30367 \dots

f^{^{'}} (u_{2}) = 1024 \cdot 0.08647 \dots^{3} + 768 \cdot 0.08647 \dots^{2} + 198 \cdot 0.08647 \dots + 16 = 39.52642 \dots

u_{3} = 0.07878 \dots

Δ u_{3} = ∣ 0.07878 \dots - 0.08647 \dots ∣ = 0.00768 \dots

Δ u_{3} = 0.00768 \dots

u_{4} = 0.07851 \dots : Δ u_{4} = 0.00027 \dots

f (u_{3}) = 256 \cdot 0.07878 \dots^{4} + 256 \cdot 0.07878 \dots^{3} + 99 \cdot 0.07878 \dots^{2} + 16 \cdot 0.07878 \dots - 2 = 0.01028 \dots

f^{^{'}} (u_{3}) = 1024 \cdot 0.07878 \dots^{3} + 768 \cdot 0.07878 \dots^{2} + 198 \cdot 0.07878 \dots + 16 = 36.86880 \dots

u_{4} = 0.07851 \dots

Δ u_{4} = ∣ 0.07851 \dots - 0.07878 \dots ∣ = 0.00027 \dots

Δ u_{4} = 0.00027 \dots

u_{5} = 0.07851 \dots : Δ u_{5} = 3.57619 E - 7

f (u_{4}) = 256 \cdot 0.07851 \dots^{4} + 256 \cdot 0.07851 \dots^{3} + 99 \cdot 0.07851 \dots^{2} + 16 \cdot 0.07851 \dots - 2 = 0.00001 \dots

f^{^{'}} (u_{4}) = 1024 \cdot 0.07851 \dots^{3} + 768 \cdot 0.07851 \dots^{2} + 198 \cdot 0.07851 \dots + 16 = 36.77455 \dots

u_{5} = 0.07851 \dots

Δ u_{5} = ∣ 0.07851 \dots - 0.07851 \dots ∣ = 3.57619 E - 7

Δ u_{5} = 3.57619 E - 7

u \approx 0.07851 \dots

使用长除法

Eq u a t i o n 0 : \frac{256 u ^{4} + 256 u ^{3} + 99 u ^{2} + 16 u - 2}{u - 0.07851 \dots} = 256 u^{3} + 276.09864 \dots u^{2} + 120.67659 \dots u + 25.47435 \dots

256 u^{3} + 276.09864 \dots u^{2} + 120.67659 \dots u + 25.47435 \dots \approx 0

使用牛顿-拉弗森方法找到

256 u^{3} + 276.09864 \dots u^{2} + 120.67659 \dots u + 25.47435 \dots = 0

的一个解:

u \approx - 0.55060 \dots

256 u^{3} + 276.09864 \dots u^{2} + 120.67659 \dots u + 25.47435 \dots = 0

牛顿-拉弗森近似法定义

f (u) = 256 u^{3} + 276.09864 \dots u^{2} + 120.67659 \dots u + 25.47435 \dots

找到

f^{^{'}} (u) : 768 u^{2} + 552.19728 \dots u + 120.67659 \dots

\frac{d}{d u} (256 u^{3} + 276.09864 \dots u^{2} + 120.67659 \dots u + 25.47435 \dots)

使用微分加减法定则:

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

= \frac{d}{d u} (256 u^{3}) + \frac{d}{d u} (276.09864 \dots u^{2}) + \frac{d}{d u} (120.67659 \dots u) + \frac{d}{d u} (25.47435 \dots)

\frac{d}{d u} (256 u^{3}) = 768 u^{2}

\frac{d}{d u} (256 u^{3})

将常数提出:

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

= 256 \frac{d}{d u} (u^{3})

使用幂法则:

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

= 256 \cdot 3 u^{3 - 1}

化简

= 768 u^{2}

\frac{d}{d u} (276.09864 \dots u^{2}) = 552.19728 \dots u

\frac{d}{d u} (276.09864 \dots u^{2})

将常数提出:

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

= 276.09864 \dots \frac{d}{d u} (u^{2})

使用幂法则:

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

= 276.09864 \dots \cdot 2 u^{2 - 1}

化简

= 552.19728 \dots u

\frac{d}{d u} (120.67659 \dots u) = 120.67659 \dots

\frac{d}{d u} (120.67659 \dots u)

将常数提出:

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

= 120.67659 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 120.67659 \dots \cdot 1

化简

= 120.67659 \dots

\frac{d}{d u} (25.47435 \dots) = 0

\frac{d}{d u} (25.47435 \dots)

常数微分:

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

= 0

= 768 u^{2} + 552.19728 \dots u + 120.67659 \dots + 0

化简

= 768 u^{2} + 552.19728 \dots u + 120.67659 \dots

令

u_{0} = 0

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.21109 \dots : Δ u_{1} = 0.21109 \dots

f (u_{0}) = 256 \cdot 0^{3} + 276.09864 \dots \cdot 0^{2} + 120.67659 \dots \cdot 0 + 25.47435 \dots = 25.47435 \dots

f^{^{'}} (u_{0}) = 768 \cdot 0^{2} + 552.19728 \dots \cdot 0 + 120.67659 \dots = 120.67659 \dots

u_{1} = - 0.21109 \dots

Δ u_{1} = ∣ - 0.21109 \dots - 0 ∣ = 0.21109 \dots

Δ u_{1} = 0.21109 \dots

u_{2} = - 0.46923 \dots : Δ u_{2} = 0.25813 \dots

f (u_{1}) = 256 (- 0.21109 \dots)^{3} + 276.09864 \dots (- 0.21109 \dots)^{2} + 120.67659 \dots (- 0.21109 \dots) + 25.47435 \dots = 9.89525 \dots

f^{^{'}} (u_{1}) = 768 (- 0.21109 \dots)^{2} + 552.19728 \dots (- 0.21109 \dots) + 120.67659 \dots = 38.33318 \dots

u_{2} = - 0.46923 \dots

Δ u_{2} = ∣ - 0.46923 \dots - (- 0.21109 \dots) ∣ = 0.25813 \dots

Δ u_{2} = 0.25813 \dots

u_{3} = - 0.57330 \dots : Δ u_{3} = 0.10407 \dots

f (u_{2}) = 256 (- 0.46923 \dots)^{3} + 276.09864 \dots (- 0.46923 \dots)^{2} + 120.67659 \dots (- 0.46923 \dots) + 25.47435 \dots = 3.19139 \dots

f^{^{'}} (u_{2}) = 768 (- 0.46923 \dots)^{2} + 552.19728 \dots (- 0.46923 \dots) + 120.67659 \dots = 30.66554 \dots

u_{3} = - 0.57330 \dots

Δ u_{3} = ∣ - 0.57330 \dots - (- 0.46923 \dots) ∣ = 0.10407 \dots

Δ u_{3} = 0.10407 \dots

u_{4} = - 0.55205 \dots : Δ u_{4} = 0.02125 \dots

f (u_{3}) = 256 (- 0.57330 \dots)^{3} + 276.09864 \dots (- 0.57330 \dots)^{2} + 120.67659 \dots (- 0.57330 \dots) + 25.47435 \dots = - 1.20130 \dots

f^{^{'}} (u_{3}) = 768 (- 0.57330 \dots)^{2} + 552.19728 \dots (- 0.57330 \dots) + 120.67659 \dots = 56.52438 \dots

u_{4} = - 0.55205 \dots

Δ u_{4} = ∣ - 0.55205 \dots - (- 0.57330 \dots) ∣ = 0.02125 \dots

Δ u_{4} = 0.02125 \dots

u_{5} = - 0.55061 \dots : Δ u_{5} = 0.00143 \dots

f (u_{4}) = 256 (- 0.55205 \dots)^{3} + 276.09864 \dots (- 0.55205 \dots)^{2} + 120.67659 \dots (- 0.55205 \dots) + 25.47435 \dots = - 0.07170 \dots

f^{^{'}} (u_{4}) = 768 (- 0.55205 \dots)^{2} + 552.19728 \dots (- 0.55205 \dots) + 120.67659 \dots = 49.89187 \dots

u_{5} = - 0.55061 \dots

Δ u_{5} = ∣ - 0.55061 \dots - (- 0.55205 \dots) ∣ = 0.00143 \dots

Δ u_{5} = 0.00143 \dots

u_{6} = - 0.55060 \dots : Δ u_{6} = 6.15991 E - 6

f (u_{5}) = 256 (- 0.55061 \dots)^{3} + 276.09864 \dots (- 0.55061 \dots)^{2} + 120.67659 \dots (- 0.55061 \dots) + 25.47435 \dots = - 0.00030 \dots

f^{^{'}} (u_{5}) = 768 (- 0.55061 \dots)^{2} + 552.19728 \dots (- 0.55061 \dots) + 120.67659 \dots = 49.46837 \dots

u_{6} = - 0.55060 \dots

Δ u_{6} = ∣ - 0.55060 \dots - (- 0.55061 \dots) ∣ = 6.15991 E - 6

Δ u_{6} = 6.15991 E - 6

u_{7} = - 0.55060 \dots : Δ u_{7} = 1.12585 E - 10

f (u_{6}) = 256 (- 0.55060 \dots)^{3} + 276.09864 \dots (- 0.55060 \dots)^{2} + 120.67659 \dots (- 0.55060 \dots) + 25.47435 \dots = - 5.56919 E - 9

f^{^{'}} (u_{6}) = 768 (- 0.55060 \dots)^{2} + 552.19728 \dots (- 0.55060 \dots) + 120.67659 \dots = 49.46656 \dots

u_{7} = - 0.55060 \dots

Δ u_{7} = ∣ - 0.55060 \dots - (- 0.55060 \dots) ∣ = 1.12585 E - 10

Δ u_{7} = 1.12585 E - 10

u \approx - 0.55060 \dots

使用长除法

Eq u a t i o n 0 : \frac{256 u ^{3} + 276.09864 \dots u ^{2} + 120.67659 \dots u + 25.47435 \dots}{u + 0.55060 \dots} = 256 u^{2} + 135.14273 \dots u + 46.26578 \dots

256 u^{2} + 135.14273 \dots u + 46.26578 \dots \approx 0

使用牛顿-拉弗森方法找到

256 u^{2} + 135.14273 \dots u + 46.26578 \dots = 0

的一个解:

u \in R

无解

256 u^{2} + 135.14273 \dots u + 46.26578 \dots = 0

牛顿-拉弗森近似法定义

f (u) = 256 u^{2} + 135.14273 \dots u + 46.26578 \dots

找到

f^{^{'}} (u) : 512 u + 135.14273 \dots

\frac{d}{d u} (256 u^{2} + 135.14273 \dots u + 46.26578 \dots)

使用微分加减法定则:

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

= \frac{d}{d u} (256 u^{2}) + \frac{d}{d u} (135.14273 \dots u) + \frac{d}{d u} (46.26578 \dots)

\frac{d}{d u} (256 u^{2}) = 512 u

\frac{d}{d u} (256 u^{2})

将常数提出:

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

= 256 \frac{d}{d u} (u^{2})

使用幂法则:

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

= 256 \cdot 2 u^{2 - 1}

化简

= 512 u

\frac{d}{d u} (135.14273 \dots u) = 135.14273 \dots

\frac{d}{d u} (135.14273 \dots u)

将常数提出:

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

= 135.14273 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 135.14273 \dots \cdot 1

化简

= 135.14273 \dots

\frac{d}{d u} (46.26578 \dots) = 0

\frac{d}{d u} (46.26578 \dots)

常数微分:

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

= 0

= 512 u + 135.14273 \dots + 0

化简

= 512 u + 135.14273 \dots

令

u_{0} = 0

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.34234 \dots : Δ u_{1} = 0.34234 \dots

f (u_{0}) = 256 \cdot 0^{2} + 135.14273 \dots \cdot 0 + 46.26578 \dots = 46.26578 \dots

f^{^{'}} (u_{0}) = 512 \cdot 0 + 135.14273 \dots = 135.14273 \dots

u_{1} = - 0.34234 \dots

Δ u_{1} = ∣ - 0.34234 \dots - 0 ∣ = 0.34234 \dots

Δ u_{1} = 0.34234 \dots

u_{2} = 0.40514 \dots : Δ u_{2} = 0.74749 \dots

f (u_{1}) = 256 (- 0.34234 \dots)^{2} + 135.14273 \dots (- 0.34234 \dots) + 46.26578 \dots = 30.00367 \dots

f^{^{'}} (u_{1}) = 512 (- 0.34234 \dots) + 135.14273 \dots = - 40.13921 \dots

u_{2} = 0.40514 \dots

Δ u_{2} = ∣ 0.40514 \dots - (- 0.34234 \dots) ∣ = 0.74749 \dots

Δ u_{2} = 0.74749 \dots

u_{3} = - 0.01239 \dots : Δ u_{3} = 0.41753 \dots

f (u_{2}) = 256 \cdot 0.40514 \dots^{2} + 135.14273 \dots \cdot 0.40514 \dots + 46.26578 \dots = 143.03790 \dots

f^{^{'}} (u_{2}) = 512 \cdot 0.40514 \dots + 135.14273 \dots = 342.57584 \dots

u_{3} = - 0.01239 \dots

Δ u_{3} = ∣ - 0.01239 \dots - 0.40514 \dots ∣ = 0.41753 \dots

Δ u_{3} = 0.41753 \dots

u_{4} = - 0.35890 \dots : Δ u_{4} = 0.34651 \dots

f (u_{3}) = 256 (- 0.01239 \dots)^{2} + 135.14273 \dots (- 0.01239 \dots) + 46.26578 \dots = 44.63019 \dots

f^{^{'}} (u_{3}) = 512 (- 0.01239 \dots) + 135.14273 \dots = 128.79718 \dots

u_{4} = - 0.35890 \dots

Δ u_{4} = ∣ - 0.35890 \dots - (- 0.01239 \dots) ∣ = 0.34651 \dots

Δ u_{4} = 0.34651 \dots

u_{5} = 0.27333 \dots : Δ u_{5} = 0.63223 \dots

f (u_{4}) = 256 (- 0.35890 \dots)^{2} + 135.14273 \dots (- 0.35890 \dots) + 46.26578 \dots = 30.73865 \dots

f^{^{'}} (u_{4}) = 512 (- 0.35890 \dots) + 135.14273 \dots = - 48.61865 \dots

u_{5} = 0.27333 \dots

Δ u_{5} = ∣ 0.27333 \dots - (- 0.35890 \dots) ∣ = 0.63223 \dots

Δ u_{5} = 0.63223 \dots

u_{6} = - 0.09865 \dots : Δ u_{6} = 0.37199 \dots

f (u_{5}) = 256 \cdot 0.27333 \dots^{2} + 135.14273 \dots \cdot 0.27333 \dots + 46.26578 \dots = 102.33017 \dots

f^{^{'}} (u_{5}) = 512 \cdot 0.27333 \dots + 135.14273 \dots = 275.08816 \dots

u_{6} = - 0.09865 \dots

Δ u_{6} = ∣ - 0.09865 \dots - 0.27333 \dots ∣ = 0.37199 \dots

Δ u_{6} = 0.37199 \dots

u_{7} = - 0.51724 \dots : Δ u_{7} = 0.41858 \dots

f (u_{6}) = 256 (- 0.09865 \dots)^{2} + 135.14273 \dots (- 0.09865 \dots) + 46.26578 \dots = 35.42449 \dots

f^{^{'}} (u_{6}) = 512 (- 0.09865 \dots) + 135.14273 \dots = 84.62903 \dots

u_{7} = - 0.51724 \dots

Δ u_{7} = ∣ - 0.51724 \dots - (- 0.09865 \dots) ∣ = 0.41858 \dots

Δ u_{7} = 0.41858 \dots

u_{8} = - 0.17137 \dots : Δ u_{8} = 0.34586 \dots

f (u_{7}) = 256 (- 0.51724 \dots)^{2} + 135.14273 \dots (- 0.51724 \dots) + 46.26578 \dots = 44.85474 \dots

f^{^{'}} (u_{7}) = 512 (- 0.51724 \dots) + 135.14273 \dots = - 129.68675 \dots

u_{8} = - 0.17137 \dots

Δ u_{8} = ∣ - 0.17137 \dots - (- 0.51724 \dots) ∣ = 0.34586 \dots

Δ u_{8} = 0.34586 \dots

u_{9} = - 0.81747 \dots : Δ u_{9} = 0.64609 \dots

f (u_{8}) = 256 (- 0.17137 \dots)^{2} + 135.14273 \dots (- 0.17137 \dots) + 46.26578 \dots = 30.62425 \dots

f^{^{'}} (u_{8}) = 512 (- 0.17137 \dots) + 135.14273 \dots = 47.39863 \dots

u_{9} = - 0.81747 \dots

Δ u_{9} = ∣ - 0.81747 \dots - (- 0.17137 \dots) ∣ = 0.64609 \dots

Δ u_{9} = 0.64609 \dots

无法得出解

解为

u \approx 0.07851 \dots, u \approx - 0.55060 \dots

u = cos (θ)

代回

cos (θ) \approx 0.07851 \dots, cos (θ) \approx - 0.55060 \dots

cos (θ) \approx 0.07851 \dots, cos (θ) \approx - 0.55060 \dots

cos (θ) = 0.07851 \dots : θ = arccos (0.07851 \dots) + 2 πn, θ = 2 π - arccos (0.07851 \dots) + 2 πn

cos (θ) = 0.07851 \dots

使用反三角函数性质

cos (θ) = 0.07851 \dots

cos (θ) = 0.07851 \dots

的通解

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

θ = arccos (0.07851 \dots) + 2 πn, θ = 2 π - arccos (0.07851 \dots) + 2 πn

θ = arccos (0.07851 \dots) + 2 πn, θ = 2 π - arccos (0.07851 \dots) + 2 πn

cos (θ) = - 0.55060 \dots : θ = arccos (- 0.55060 \dots) + 2 πn, θ = - arccos (- 0.55060 \dots) + 2 πn

cos (θ) = - 0.55060 \dots

使用反三角函数性质

cos (θ) = - 0.55060 \dots

cos (θ) = - 0.55060 \dots

的通解

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

θ = arccos (- 0.55060 \dots) + 2 πn, θ = - arccos (- 0.55060 \dots) + 2 πn

θ = arccos (- 0.55060 \dots) + 2 πn, θ = - arccos (- 0.55060 \dots) + 2 πn

合并所有解

θ = arccos (0.07851 \dots) + 2 πn, θ = 2 π - arccos (0.07851 \dots) + 2 πn, θ = arccos (- 0.55060 \dots) + 2 πn, θ = - arccos (- 0.55060 \dots) + 2 πn

将解代入原方程进行验证

将它们代入

(1 + 4 cos (θ))^{2} = 3 sin (θ)

检验解是否符合

去除与方程不符的解。

检验

arccos (0.07851 \dots) + 2 πn

的解:

真

arccos (0.07851 \dots) + 2 πn

代入

n = 1

arccos (0.07851 \dots) + 2 π 1

对于

(1 + 4 cos (θ))^{2} = 3 sin (θ)

代入

θ = arccos (0.07851 \dots) + 2 π 1

(1 + 4 cos (arccos (0.07851 \dots) + 2 π 1))^{2} = 3 sin (arccos (0.07851 \dots) + 2 π 1)

整理后得

1.72670 \dots = 1.72670 \dots

\Rightarrow 真

检验

2 π - arccos (0.07851 \dots) + 2 πn

的解:

假

2 π - arccos (0.07851 \dots) + 2 πn

代入

n = 1

2 π - arccos (0.07851 \dots) + 2 π 1

对于

(1 + 4 cos (θ))^{2} = 3 sin (θ)

代入

θ = 2 π - arccos (0.07851 \dots) + 2 π 1

(1 + 4 cos (2 π - arccos (0.07851 \dots) + 2 π 1))^{2} = 3 sin (2 π - arccos (0.07851 \dots) + 2 π 1)

整理后得

1.72670 \dots = - 1.72670 \dots

\Rightarrow 假

检验

arccos (- 0.55060 \dots) + 2 πn

的解:

真

arccos (- 0.55060 \dots) + 2 πn

代入

n = 1

arccos (- 0.55060 \dots) + 2 π 1

对于

(1 + 4 cos (θ))^{2} = 3 sin (θ)

代入

θ = arccos (- 0.55060 \dots) + 2 π 1

(1 + 4 cos (arccos (- 0.55060 \dots) + 2 π 1))^{2} = 3 sin (arccos (- 0.55060 \dots) + 2 π 1)

整理后得

1.44585 \dots = 1.44585 \dots

\Rightarrow 真

检验

- arccos (- 0.55060 \dots) + 2 πn

的解:

假

- arccos (- 0.55060 \dots) + 2 πn

代入

n = 1

- arccos (- 0.55060 \dots) + 2 π 1

对于

(1 + 4 cos (θ))^{2} = 3 sin (θ)

代入

θ = - arccos (- 0.55060 \dots) + 2 π 1

(1 + 4 cos (- arccos (- 0.55060 \dots) + 2 π 1))^{2} = 3 sin (- arccos (- 0.55060 \dots) + 2 π 1)

整理后得

1.44585 \dots = - 1.44585 \dots

\Rightarrow 假

θ = arccos (0.07851 \dots) + 2 πn, θ = arccos (- 0.55060 \dots) + 2 πn

以小数形式表示解

θ = 1.49220 \dots + 2 πn, θ = 2.15388 \dots + 2 πn

作图

流行的例子

sin(θ)-(cos(θ))/5 =0.6377

sin (θ) - \frac{cos ( θ )}{5} = 0.6377

sin(4x)=cos(4x)

sin (4 x) = cos (4 x)

2cos^2(x)+9cos(x)+4=0

2 cos^{2} (x) + 9 cos (x) + 4 = 0

cos(6x)=sin(x-1)

cos (6 x) = sin (x - 1)

tan (θ) = \frac{72}{65}