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受欢迎的三角函数 >

12=3sec(θ)+5csc(θ)

解答

12 = 3 sec (θ) + 5 csc (θ)

解答

θ = π - 0.33570 \dots + 2 πn, θ = 1.07609 \dots + 2 πn, θ = 0.65383 \dots + 2 πn, θ = - 1.39422 \dots + 2 πn

+1

度数

θ = 160.76534 \dots^{\circ} + 36 0^{\circ} n, θ = 61.65590 \dots^{\circ} + 36 0^{\circ} n, θ = 37.46199 \dots^{\circ} + 36 0^{\circ} n, θ = - 79.88324 \dots^{\circ} + 36 0^{\circ} n

求解步骤

12 = 3 sec (θ) + 5 csc (θ)

两边减去

5 csc (θ)

3 sec (θ) = 12 - 5 csc (θ)

两边进行平方

(3 sec (θ))^{2} = (12 - 5 csc (θ))^{2}

两边减去

(12 - 5 csc (θ))^{2}

9 sec^{2} (θ) - 144 + 120 csc (θ) - 25 csc^{2} (θ) = 0

用 sin, cos 表示

- 144 + 120 csc (θ) - 25 csc^{2} (θ) + 9 sec^{2} (θ)

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

= - 144 + 120 \cdot \frac{1}{sin ( θ )} - 25 (\frac{1}{sin ( θ )})^{2} + 9 sec^{2} (θ)

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

= - 144 + 120 \cdot \frac{1}{sin ( θ )} - 25 (\frac{1}{sin ( θ )})^{2} + 9 (\frac{1}{cos ( θ )})^{2}

化简

- 144 + 120 \cdot \frac{1}{sin ( θ )} - 25 (\frac{1}{sin ( θ )})^{2} + 9 (\frac{1}{cos ( θ )})^{2} : \frac{- 144 cos ^{2} ( θ ) sin ^{2} ( θ ) + 120 cos ^{2} ( θ ) sin ( θ ) - 25 cos ^{2} ( θ ) + 9 sin ^{2} ( θ )}{cos ^{2} ( θ ) sin ^{2} ( θ )}

- 144 + 120 \cdot \frac{1}{sin ( θ )} - 25 (\frac{1}{sin ( θ )})^{2} + 9 (\frac{1}{cos ( θ )})^{2}

120 \cdot \frac{1}{sin ( θ )} = \frac{120}{sin ( θ )}

120 \cdot \frac{1}{sin ( θ )}

分式相乘:

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

= \frac{1 \cdot 120}{sin ( θ )}

数字相乘:

1 \cdot 120 = 120

= \frac{120}{sin ( θ )}

25 (\frac{1}{sin ( θ )})^{2} = \frac{25}{sin ^{2} ( θ )}

25 (\frac{1}{sin ( θ )})^{2}

(\frac{1}{sin ( θ )})^{2} = \frac{1}{sin ^{2} ( θ )}

(\frac{1}{sin ( θ )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{sin ^{2} ( θ )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

= 25 \cdot \frac{1}{sin ^{2} ( θ )}

分式相乘:

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

= \frac{1 \cdot 25}{sin ^{2} ( θ )}

数字相乘:

1 \cdot 25 = 25

= \frac{25}{sin ^{2} ( θ )}

9 (\frac{1}{cos ( θ )})^{2} = \frac{9}{cos ^{2} ( θ )}

9 (\frac{1}{cos ( θ )})^{2}

(\frac{1}{cos ( θ )})^{2} = \frac{1}{cos ^{2} ( θ )}

(\frac{1}{cos ( θ )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

= 9 \cdot \frac{1}{cos ^{2} ( θ )}

分式相乘:

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

= \frac{1 \cdot 9}{cos ^{2} ( θ )}

数字相乘:

1 \cdot 9 = 9

= \frac{9}{cos ^{2} ( θ )}

= - 144 + \frac{120}{sin ( θ )} - \frac{25}{sin ^{2} ( θ )} + \frac{9}{cos ^{2} ( θ )}

将项转换为分式:

144 = \frac{144 cos ^{2} ( θ )}{cos ^{2} ( θ )}

= - \frac{144 cos ^{2} ( θ )}{cos ^{2} ( θ )} + \frac{120}{sin ( θ )} - \frac{25}{sin ^{2} ( θ )} + \frac{9}{cos ^{2} ( θ )}

cos^{2} (θ), sin (θ), sin^{2} (θ), cos^{2} (θ)

的最小公倍数:

cos^{2} (θ) sin^{2} (θ)

cos^{2} (θ), sin (θ), sin^{2} (θ), cos^{2} (θ)

最小公倍数 (LCM)

计算出由至少在以下一个因式表达式中出现的因子组成的表达式

= cos^{2} (θ) sin^{2} (θ)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

cos^{2} (θ) sin^{2} (θ)

对于

\frac{144 cos ^{2} ( θ )}{cos ^{2} ( θ )} :

将分母和分子乘以

sin^{2} (θ)

\frac{144 cos ^{2} ( θ )}{cos ^{2} ( θ )} = \frac{144 cos ^{2} ( θ ) sin ^{2} ( θ )}{cos ^{2} ( θ ) sin ^{2} ( θ )}

对于

\frac{120}{sin ( θ )} :

将分母和分子乘以

cos^{2} (θ) sin (θ)

\frac{120}{sin ( θ )} = \frac{120 cos ^{2} ( θ ) sin ( θ )}{sin ( θ ) cos ^{2} ( θ ) sin ( θ )} = \frac{120 cos ^{2} ( θ ) sin ( θ )}{cos ^{2} ( θ ) sin ^{2} ( θ )}

对于

\frac{25}{sin ^{2} ( θ )} :

将分母和分子乘以

cos^{2} (θ)

\frac{25}{sin ^{2} ( θ )} = \frac{25 cos ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

对于

\frac{9}{cos ^{2} ( θ )} :

将分母和分子乘以

sin^{2} (θ)

\frac{9}{cos ^{2} ( θ )} = \frac{9 sin ^{2} ( θ )}{cos ^{2} ( θ ) sin ^{2} ( θ )}

= - \frac{144 cos ^{2} ( θ ) sin ^{2} ( θ )}{cos ^{2} ( θ ) sin ^{2} ( θ )} + \frac{120 cos ^{2} ( θ ) sin ( θ )}{cos ^{2} ( θ ) sin ^{2} ( θ )} - \frac{25 cos ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )} + \frac{9 sin ^{2} ( θ )}{cos ^{2} ( θ ) sin ^{2} ( θ )}

因为分母相等，所以合并分式:

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

= \frac{- 144 cos ^{2} ( θ ) sin ^{2} ( θ ) + 120 cos ^{2} ( θ ) sin ( θ ) - 25 cos ^{2} ( θ ) + 9 sin ^{2} ( θ )}{cos ^{2} ( θ ) sin ^{2} ( θ )}

= \frac{- 144 cos ^{2} ( θ ) sin ^{2} ( θ ) + 120 cos ^{2} ( θ ) sin ( θ ) - 25 cos ^{2} ( θ ) + 9 sin ^{2} ( θ )}{cos ^{2} ( θ ) sin ^{2} ( θ )}

\frac{- 25 cos ^{2} ( θ ) + 9 sin ^{2} ( θ ) + 120 cos ^{2} ( θ ) sin ( θ ) - 144 cos ^{2} ( θ ) sin ^{2} ( θ )}{cos ^{2} ( θ ) sin ^{2} ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- 25 cos^{2} (θ) + 9 sin^{2} (θ) + 120 cos^{2} (θ) sin (θ) - 144 cos^{2} (θ) sin^{2} (θ) = 0

使用三角恒等式改写

- 25 cos^{2} (θ) + 9 sin^{2} (θ) + 120 cos^{2} (θ) sin (θ) - 144 cos^{2} (θ) sin^{2} (θ)

使用毕达哥拉斯恒等式:

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

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

= - 25 (1 - sin^{2} (θ)) + 9 sin^{2} (θ) + 120 (1 - sin^{2} (θ)) sin (θ) - 144 (1 - sin^{2} (θ)) sin^{2} (θ)

化简

- 25 (1 - sin^{2} (θ)) + 9 sin^{2} (θ) + 120 (1 - sin^{2} (θ)) sin (θ) - 144 (1 - sin^{2} (θ)) sin^{2} (θ) : - 110 sin^{2} (θ) + 120 sin (θ) - 120 sin^{3} (θ) + 144 sin^{4} (θ) - 25

- 25 (1 - sin^{2} (θ)) + 9 sin^{2} (θ) + 120 (1 - sin^{2} (θ)) sin (θ) - 144 (1 - sin^{2} (θ)) sin^{2} (θ)

= - 25 (1 - sin^{2} (θ)) + 9 sin^{2} (θ) + 120 sin (θ) (1 - sin^{2} (θ)) - 144 sin^{2} (θ) (1 - sin^{2} (θ))

乘开

- 25 (1 - sin^{2} (θ)) : - 25 + 25 sin^{2} (θ)

- 25 (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 25, b = 1, c = sin^{2} (θ)

= - 25 \cdot 1 - (- 25) sin^{2} (θ)

使用加减运算法则

- (- a) = a

= - 25 \cdot 1 + 25 sin^{2} (θ)

数字相乘:

25 \cdot 1 = 25

= - 25 + 25 sin^{2} (θ)

= - 25 + 25 sin^{2} (θ) + 9 sin^{2} (θ) + 120 (1 - sin^{2} (θ)) sin (θ) - 144 (1 - sin^{2} (θ)) sin^{2} (θ)

乘开

120 sin (θ) (1 - sin^{2} (θ)) : 120 sin (θ) - 120 sin^{3} (θ)

120 sin (θ) (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 120 sin (θ), b = 1, c = sin^{2} (θ)

= 120 sin (θ) \cdot 1 - 120 sin (θ) sin^{2} (θ)

= 120 \cdot 1 \cdot sin (θ) - 120 sin^{2} (θ) sin (θ)

化简

120 \cdot 1 \cdot sin (θ) - 120 sin^{2} (θ) sin (θ) : 120 sin (θ) - 120 sin^{3} (θ)

120 \cdot 1 \cdot sin (θ) - 120 sin^{2} (θ) sin (θ)

120 \cdot 1 \cdot sin (θ) = 120 sin (θ)

120 \cdot 1 \cdot sin (θ)

数字相乘:

120 \cdot 1 = 120

= 120 sin (θ)

120 sin^{2} (θ) sin (θ) = 120 sin^{3} (θ)

120 sin^{2} (θ) sin (θ)

使用指数法则:

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

sin^{2} (θ) sin (θ) = sin^{2 + 1} (θ)

= 120 sin^{2 + 1} (θ)

数字相加:

2 + 1 = 3

= 120 sin^{3} (θ)

= 120 sin (θ) - 120 sin^{3} (θ)

= 120 sin (θ) - 120 sin^{3} (θ)

= - 25 + 25 sin^{2} (θ) + 9 sin^{2} (θ) + 120 sin (θ) - 120 sin^{3} (θ) - 144 (1 - sin^{2} (θ)) sin^{2} (θ)

乘开

- 144 sin^{2} (θ) (1 - sin^{2} (θ)) : - 144 sin^{2} (θ) + 144 sin^{4} (θ)

- 144 sin^{2} (θ) (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 144 sin^{2} (θ), b = 1, c = sin^{2} (θ)

= - 144 sin^{2} (θ) \cdot 1 - (- 144 sin^{2} (θ)) sin^{2} (θ)

使用加减运算法则

- (- a) = a

= - 144 \cdot 1 \cdot sin^{2} (θ) + 144 sin^{2} (θ) sin^{2} (θ)

化简

- 144 \cdot 1 \cdot sin^{2} (θ) + 144 sin^{2} (θ) sin^{2} (θ) : - 144 sin^{2} (θ) + 144 sin^{4} (θ)

- 144 \cdot 1 \cdot sin^{2} (θ) + 144 sin^{2} (θ) sin^{2} (θ)

144 \cdot 1 \cdot sin^{2} (θ) = 144 sin^{2} (θ)

144 \cdot 1 \cdot sin^{2} (θ)

数字相乘:

144 \cdot 1 = 144

= 144 sin^{2} (θ)

144 sin^{2} (θ) sin^{2} (θ) = 144 sin^{4} (θ)

144 sin^{2} (θ) sin^{2} (θ)

使用指数法则:

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

sin^{2} (θ) sin^{2} (θ) = sin^{2 + 2} (θ)

= 144 sin^{2 + 2} (θ)

数字相加:

2 + 2 = 4

= 144 sin^{4} (θ)

= - 144 sin^{2} (θ) + 144 sin^{4} (θ)

= - 144 sin^{2} (θ) + 144 sin^{4} (θ)

= - 25 + 25 sin^{2} (θ) + 9 sin^{2} (θ) + 120 sin (θ) - 120 sin^{3} (θ) - 144 sin^{2} (θ) + 144 sin^{4} (θ)

化简

- 25 + 25 sin^{2} (θ) + 9 sin^{2} (θ) + 120 sin (θ) - 120 sin^{3} (θ) - 144 sin^{2} (θ) + 144 sin^{4} (θ) : - 110 sin^{2} (θ) + 120 sin (θ) - 120 sin^{3} (θ) + 144 sin^{4} (θ) - 25

- 25 + 25 sin^{2} (θ) + 9 sin^{2} (θ) + 120 sin (θ) - 120 sin^{3} (θ) - 144 sin^{2} (θ) + 144 sin^{4} (θ)

对同类项分组

= 25 sin^{2} (θ) + 9 sin^{2} (θ) + 120 sin (θ) - 120 sin^{3} (θ) - 144 sin^{2} (θ) + 144 sin^{4} (θ) - 25

同类项相加:

25 sin^{2} (θ) + 9 sin^{2} (θ) - 144 sin^{2} (θ) = - 110 sin^{2} (θ)

= - 110 sin^{2} (θ) + 120 sin (θ) - 120 sin^{3} (θ) + 144 sin^{4} (θ) - 25

= - 110 sin^{2} (θ) + 120 sin (θ) - 120 sin^{3} (θ) + 144 sin^{4} (θ) - 25

= - 110 sin^{2} (θ) + 120 sin (θ) - 120 sin^{3} (θ) + 144 sin^{4} (θ) - 25

- 25 - 110 sin^{2} (θ) + 120 sin (θ) - 120 sin^{3} (θ) + 144 sin^{4} (θ) = 0

用替代法求解

- 25 - 110 sin^{2} (θ) + 120 sin (θ) - 120 sin^{3} (θ) + 144 sin^{4} (θ) = 0

令

: sin (θ) = u

- 25 - 110 u^{2} + 120 u - 120 u^{3} + 144 u^{4} = 0

- 25 - 110 u^{2} + 120 u - 120 u^{3} + 144 u^{4} = 0 : u \approx 0.32943 \dots, u \approx 0.88011 \dots, u \approx 0.60823 \dots, u \approx - 0.98445 \dots

- 25 - 110 u^{2} + 120 u - 120 u^{3} + 144 u^{4} = 0

改写成标准形式

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

144 u^{4} - 120 u^{3} - 110 u^{2} + 120 u - 25 = 0

使用牛顿-拉弗森方法找到

144 u^{4} - 120 u^{3} - 110 u^{2} + 120 u - 25 = 0

的一个解:

u \approx 0.32943 \dots

144 u^{4} - 120 u^{3} - 110 u^{2} + 120 u - 25 = 0

牛顿-拉弗森近似法定义

f (u) = 144 u^{4} - 120 u^{3} - 110 u^{2} + 120 u - 25

找到

f^{^{'}} (u) : 576 u^{3} - 360 u^{2} - 220 u + 120

\frac{d}{d u} (144 u^{4} - 120 u^{3} - 110 u^{2} + 120 u - 25)

使用微分加减法定则:

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

= \frac{d}{d u} (144 u^{4}) - \frac{d}{d u} (120 u^{3}) - \frac{d}{d u} (110 u^{2}) + \frac{d}{d u} (120 u) - \frac{d}{d u} (25)

\frac{d}{d u} (144 u^{4}) = 576 u^{3}

\frac{d}{d u} (144 u^{4})

将常数提出:

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

= 144 \frac{d}{d u} (u^{4})

使用幂法则:

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

= 144 \cdot 4 u^{4 - 1}

化简

= 576 u^{3}

\frac{d}{d u} (120 u^{3}) = 360 u^{2}

\frac{d}{d u} (120 u^{3})

将常数提出:

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

= 120 \frac{d}{d u} (u^{3})

使用幂法则:

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

= 120 \cdot 3 u^{3 - 1}

化简

= 360 u^{2}

\frac{d}{d u} (110 u^{2}) = 220 u

\frac{d}{d u} (110 u^{2})

将常数提出:

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

= 110 \frac{d}{d u} (u^{2})

使用幂法则:

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

= 110 \cdot 2 u^{2 - 1}

化简

= 220 u

\frac{d}{d u} (120 u) = 120

\frac{d}{d u} (120 u)

将常数提出:

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

= 120 \frac{d u}{d u}

使用常见微分定则:

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

= 120 \cdot 1

化简

= 120

\frac{d}{d u} (25) = 0

\frac{d}{d u} (25)

常数微分:

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

= 0

= 576 u^{3} - 360 u^{2} - 220 u + 120 - 0

化简

= 576 u^{3} - 360 u^{2} - 220 u + 120

令

u_{0} = 0

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.20833 \dots : Δ u_{1} = 0.20833 \dots

f (u_{0}) = 144 \cdot 0^{4} - 120 \cdot 0^{3} - 110 \cdot 0^{2} + 120 \cdot 0 - 25 = - 25

f^{^{'}} (u_{0}) = 576 \cdot 0^{3} - 360 \cdot 0^{2} - 220 \cdot 0 + 120 = 120

u_{1} = 0.20833 \dots

Δ u_{1} = ∣ 0.20833 \dots - 0 ∣ = 0.20833 \dots

Δ u_{1} = 0.20833 \dots

u_{2} = 0.29598 \dots : Δ u_{2} = 0.08765 \dots

f (u_{1}) = 144 \cdot 0.20833 \dots^{4} - 120 \cdot 0.20833 \dots^{3} - 110 \cdot 0.20833 \dots^{2} + 120 \cdot 0.20833 \dots - 25 = - 5.58810 \dots

f^{^{'}} (u_{1}) = 576 \cdot 0.20833 \dots^{3} - 360 \cdot 0.20833 \dots^{2} - 220 \cdot 0.20833 \dots + 120 = 63.75

u_{2} = 0.29598 \dots

Δ u_{2} = ∣ 0.29598 \dots - 0.20833 \dots ∣ = 0.08765 \dots

Δ u_{2} = 0.08765 \dots

u_{3} = 0.32537 \dots : Δ u_{3} = 0.02938 \dots

f (u_{2}) = 144 \cdot 0.29598 \dots^{4} - 120 \cdot 0.29598 \dots^{3} - 110 \cdot 0.29598 \dots^{2} + 120 \cdot 0.29598 \dots - 25 = - 1.12484 \dots

f^{^{'}} (u_{2}) = 576 \cdot 0.29598 \dots^{3} - 360 \cdot 0.29598 \dots^{2} - 220 \cdot 0.29598 \dots + 120 = 38.27925 \dots

u_{3} = 0.32537 \dots

Δ u_{3} = ∣ 0.32537 \dots - 0.29598 \dots ∣ = 0.02938 \dots

Δ u_{3} = 0.02938 \dots

u_{4} = 0.32936 \dots : Δ u_{4} = 0.00398 \dots

f (u_{3}) = 144 \cdot 0.32537 \dots^{4} - 120 \cdot 0.32537 \dots^{3} - 110 \cdot 0.32537 \dots^{2} + 120 \cdot 0.32537 \dots - 25 = - 0.12024 \dots

f^{^{'}} (u_{3}) = 576 \cdot 0.32537 \dots^{3} - 360 \cdot 0.32537 \dots^{2} - 220 \cdot 0.32537 \dots + 120 = 30.14620 \dots

u_{4} = 0.32936 \dots

Δ u_{4} = ∣ 0.32936 \dots - 0.32537 \dots ∣ = 0.00398 \dots

Δ u_{4} = 0.00398 \dots

u_{5} = 0.32943 \dots : Δ u_{5} = 0.00007 \dots

f (u_{4}) = 144 \cdot 0.32936 \dots^{4} - 120 \cdot 0.32936 \dots^{3} - 110 \cdot 0.32936 \dots^{2} + 120 \cdot 0.32936 \dots - 25 = - 0.00215 \dots

f^{^{'}} (u_{4}) = 576 \cdot 0.32936 \dots^{3} - 360 \cdot 0.32936 \dots^{2} - 220 \cdot 0.32936 \dots + 120 = 29.06722 \dots

u_{5} = 0.32943 \dots

Δ u_{5} = ∣ 0.32943 \dots - 0.32936 \dots ∣ = 0.00007 \dots

Δ u_{5} = 0.00007 \dots

u_{6} = 0.32943 \dots : Δ u_{6} = 2.54922 E - 8

f (u_{5}) = 144 \cdot 0.32943 \dots^{4} - 120 \cdot 0.32943 \dots^{3} - 110 \cdot 0.32943 \dots^{2} + 120 \cdot 0.32943 \dots - 25 = - 7.40478 E - 7

f^{^{'}} (u_{5}) = 576 \cdot 0.32943 \dots^{3} - 360 \cdot 0.32943 \dots^{2} - 220 \cdot 0.32943 \dots + 120 = 29.04723 \dots

u_{6} = 0.32943 \dots

Δ u_{6} = ∣ 0.32943 \dots - 0.32943 \dots ∣ = 2.54922 E - 8

Δ u_{6} = 2.54922 E - 8

u \approx 0.32943 \dots

使用长除法

Eq u a t i o n 0 : \frac{144 u ^{4} - 120 u ^{3} - 110 u ^{2} + 120 u - 25}{u - 0.32943 \dots} = 144 u^{3} - 72.56094 \dots u^{2} - 133.90432 \dots u + 75.88684 \dots

144 u^{3} - 72.56094 \dots u^{2} - 133.90432 \dots u + 75.88684 \dots \approx 0

使用牛顿-拉弗森方法找到

144 u^{3} - 72.56094 \dots u^{2} - 133.90432 \dots u + 75.88684 \dots = 0

的一个解:

u \approx 0.88011 \dots

144 u^{3} - 72.56094 \dots u^{2} - 133.90432 \dots u + 75.88684 \dots = 0

牛顿-拉弗森近似法定义

f (u) = 144 u^{3} - 72.56094 \dots u^{2} - 133.90432 \dots u + 75.88684 \dots

找到

f^{^{'}} (u) : 432 u^{2} - 145.12189 \dots u - 133.90432 \dots

\frac{d}{d u} (144 u^{3} - 72.56094 \dots u^{2} - 133.90432 \dots u + 75.88684 \dots)

使用微分加减法定则:

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

= \frac{d}{d u} (144 u^{3}) - \frac{d}{d u} (72.56094 \dots u^{2}) - \frac{d}{d u} (133.90432 \dots u) + \frac{d}{d u} (75.88684 \dots)

\frac{d}{d u} (144 u^{3}) = 432 u^{2}

\frac{d}{d u} (144 u^{3})

将常数提出:

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

= 144 \frac{d}{d u} (u^{3})

使用幂法则:

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

= 144 \cdot 3 u^{3 - 1}

化简

= 432 u^{2}

\frac{d}{d u} (72.56094 \dots u^{2}) = 145.12189 \dots u

\frac{d}{d u} (72.56094 \dots u^{2})

将常数提出:

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

= 72.56094 \dots \frac{d}{d u} (u^{2})

使用幂法则:

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

= 72.56094 \dots \cdot 2 u^{2 - 1}

化简

= 145.12189 \dots u

\frac{d}{d u} (133.90432 \dots u) = 133.90432 \dots

\frac{d}{d u} (133.90432 \dots u)

将常数提出:

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

= 133.90432 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 133.90432 \dots \cdot 1

化简

= 133.90432 \dots

\frac{d}{d u} (75.88684 \dots) = 0

\frac{d}{d u} (75.88684 \dots)

常数微分:

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

= 0

= 432 u^{2} - 145.12189 \dots u - 133.90432 \dots + 0

化简

= 432 u^{2} - 145.12189 \dots u - 133.90432 \dots

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.91226 \dots : Δ u_{1} = 0.08773 \dots

f (u_{0}) = 144 \cdot 1^{3} - 72.56094 \dots \cdot 1^{2} - 133.90432 \dots \cdot 1 + 75.88684 \dots = 13.42157 \dots

f^{^{'}} (u_{0}) = 432 \cdot 1^{2} - 145.12189 \dots \cdot 1 - 133.90432 \dots = 152.97377 \dots

u_{1} = 0.91226 \dots

Δ u_{1} = ∣ 0.91226 \dots - 1 ∣ = 0.08773 \dots

Δ u_{1} = 0.08773 \dots

u_{2} = 0.88362 \dots : Δ u_{2} = 0.02863 \dots

f (u_{1}) = 144 \cdot 0.91226 \dots^{3} - 72.56094 \dots \cdot 0.91226 \dots^{2} - 133.90432 \dots \cdot 0.91226 \dots + 75.88684 \dots = 2.66967 \dots

f^{^{'}} (u_{1}) = 432 \cdot 0.91226 \dots^{2} - 145.12189 \dots \cdot 0.91226 \dots - 133.90432 \dots = 93.22653 \dots

u_{2} = 0.88362 \dots

Δ u_{2} = ∣ 0.88362 \dots - 0.91226 \dots ∣ = 0.02863 \dots

Δ u_{2} = 0.02863 \dots

u_{3} = 0.88016 \dots : Δ u_{3} = 0.00346 \dots

f (u_{2}) = 144 \cdot 0.88362 \dots^{3} - 72.56094 \dots \cdot 0.88362 \dots^{2} - 133.90432 \dots \cdot 0.88362 \dots + 75.88684 \dots = 0.26029 \dots

f^{^{'}} (u_{2}) = 432 \cdot 0.88362 \dots^{2} - 145.12189 \dots \cdot 0.88362 \dots - 133.90432 \dots = 75.16549 \dots

u_{3} = 0.88016 \dots

Δ u_{3} = ∣ 0.88016 \dots - 0.88362 \dots ∣ = 0.00346 \dots

Δ u_{3} = 0.00346 \dots

u_{4} = 0.88011 \dots : Δ u_{4} = 0.00005 \dots

f (u_{3}) = 144 \cdot 0.88016 \dots^{3} - 72.56094 \dots \cdot 0.88016 \dots^{2} - 133.90432 \dots \cdot 0.88016 \dots + 75.88684 \dots = 0.00370 \dots

f^{^{'}} (u_{3}) = 432 \cdot 0.88016 \dots^{2} - 145.12189 \dots \cdot 0.88016 \dots - 133.90432 \dots = 73.02944 \dots

u_{4} = 0.88011 \dots

Δ u_{4} = ∣ 0.88011 \dots - 0.88016 \dots ∣ = 0.00005 \dots

Δ u_{4} = 0.00005 \dots

u_{5} = 0.88011 \dots : Δ u_{5} = 1.08272 E - 8

f (u_{4}) = 144 \cdot 0.88011 \dots^{3} - 72.56094 \dots \cdot 0.88011 \dots^{2} - 133.90432 \dots \cdot 0.88011 \dots + 75.88684 \dots = 7.90368 E - 7

f^{^{'}} (u_{4}) = 432 \cdot 0.88011 \dots^{2} - 145.12189 \dots \cdot 0.88011 \dots - 133.90432 \dots = 72.99825 \dots

u_{5} = 0.88011 \dots

Δ u_{5} = ∣ 0.88011 \dots - 0.88011 \dots ∣ = 1.08272 E - 8

Δ u_{5} = 1.08272 E - 8

u \approx 0.88011 \dots

使用长除法

Eq u a t i o n 0 : \frac{144 u ^{3} - 72.56094 \dots u ^{2} - 133.90432 \dots u + 75.88684 \dots}{u - 0.88011 \dots} = 144 u^{2} + 54.17521 \dots u - 86.22405 \dots

144 u^{2} + 54.17521 \dots u - 86.22405 \dots \approx 0

使用牛顿-拉弗森方法找到

144 u^{2} + 54.17521 \dots u - 86.22405 \dots = 0

的一个解:

u \approx 0.60823 \dots

144 u^{2} + 54.17521 \dots u - 86.22405 \dots = 0

牛顿-拉弗森近似法定义

f (u) = 144 u^{2} + 54.17521 \dots u - 86.22405 \dots

找到

f^{^{'}} (u) : 288 u + 54.17521 \dots

\frac{d}{d u} (144 u^{2} + 54.17521 \dots u - 86.22405 \dots)

使用微分加减法定则:

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

= \frac{d}{d u} (144 u^{2}) + \frac{d}{d u} (54.17521 \dots u) - \frac{d}{d u} (86.22405 \dots)

\frac{d}{d u} (144 u^{2}) = 288 u

\frac{d}{d u} (144 u^{2})

将常数提出:

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

= 144 \frac{d}{d u} (u^{2})

使用幂法则:

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

= 144 \cdot 2 u^{2 - 1}

化简

= 288 u

\frac{d}{d u} (54.17521 \dots u) = 54.17521 \dots

\frac{d}{d u} (54.17521 \dots u)

将常数提出:

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

= 54.17521 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 54.17521 \dots \cdot 1

化简

= 54.17521 \dots

\frac{d}{d u} (86.22405 \dots) = 0

\frac{d}{d u} (86.22405 \dots)

常数微分:

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

= 0

= 288 u + 54.17521 \dots - 0

化简

= 288 u + 54.17521 \dots

令

u_{0} = 2

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 1.05085 \dots : Δ u_{1} = 0.94914 \dots

f (u_{0}) = 144 \cdot 2^{2} + 54.17521 \dots \cdot 2 - 86.22405 \dots = 598.12636 \dots

f^{^{'}} (u_{0}) = 288 \cdot 2 + 54.17521 \dots = 630.17521 \dots

u_{1} = 1.05085 \dots

Δ u_{1} = ∣ 1.05085 \dots - 2 ∣ = 0.94914 \dots

Δ u_{1} = 0.94914 \dots

u_{2} = 0.68729 \dots : Δ u_{2} = 0.36355 \dots

f (u_{1}) = 144 \cdot 1.05085 \dots^{2} + 54.17521 \dots \cdot 1.05085 \dots - 86.22405 \dots = 129.72561 \dots

f^{^{'}} (u_{1}) = 288 \cdot 1.05085 \dots + 54.17521 \dots = 356.82203 \dots

u_{2} = 0.68729 \dots

Δ u_{2} = ∣ 0.68729 \dots - 1.05085 \dots ∣ = 0.36355 \dots

Δ u_{2} = 0.36355 \dots

u_{3} = 0.61180 \dots : Δ u_{3} = 0.07549 \dots

f (u_{2}) = 144 \cdot 0.68729 \dots^{2} + 54.17521 \dots \cdot 0.68729 \dots - 86.22405 \dots = 19.03314 \dots

f^{^{'}} (u_{2}) = 288 \cdot 0.68729 \dots + 54.17521 \dots = 252.11724 \dots

u_{3} = 0.61180 \dots

Δ u_{3} = ∣ 0.61180 \dots - 0.68729 \dots ∣ = 0.07549 \dots

Δ u_{3} = 0.07549 \dots

u_{4} = 0.60824 \dots : Δ u_{4} = 0.00356 \dots

f (u_{3}) = 144 \cdot 0.61180 \dots^{2} + 54.17521 \dots \cdot 0.61180 \dots - 86.22405 \dots = 0.82068 \dots

f^{^{'}} (u_{3}) = 288 \cdot 0.61180 \dots + 54.17521 \dots = 230.37518 \dots

u_{4} = 0.60824 \dots

Δ u_{4} = ∣ 0.60824 \dots - 0.61180 \dots ∣ = 0.00356 \dots

Δ u_{4} = 0.00356 \dots

u_{5} = 0.60823 \dots : Δ u_{5} = 7.96803 E - 6

f (u_{4}) = 144 \cdot 0.60824 \dots^{2} + 54.17521 \dots \cdot 0.60824 \dots - 86.22405 \dots = 0.00182 \dots

f^{^{'}} (u_{4}) = 288 \cdot 0.60824 \dots + 54.17521 \dots = 229.34921 \dots

u_{5} = 0.60823 \dots

Δ u_{5} = ∣ 0.60823 \dots - 0.60824 \dots ∣ = 7.96803 E - 6

Δ u_{5} = 7.96803 E - 6

u_{6} = 0.60823 \dots : Δ u_{6} = 3.98632 E - 11

f (u_{5}) = 144 \cdot 0.60823 \dots^{2} + 54.17521 \dots \cdot 0.60823 \dots - 86.22405 \dots = 9.1425 E - 9

f^{^{'}} (u_{5}) = 288 \cdot 0.60823 \dots + 54.17521 \dots = 229.34692 \dots

u_{6} = 0.60823 \dots

Δ u_{6} = ∣ 0.60823 \dots - 0.60823 \dots ∣ = 3.98632 E - 11

Δ u_{6} = 3.98632 E - 11

u \approx 0.60823 \dots

使用长除法

Eq u a t i o n 0 : \frac{144 u ^{2} + 54.17521 \dots u - 86.22405 \dots}{u - 0.60823 \dots} = 144 u + 141.76106 \dots

144 u + 141.76106 \dots \approx 0

u \approx - 0.98445 \dots

解为

u \approx 0.32943 \dots, u \approx 0.88011 \dots, u \approx 0.60823 \dots, u \approx - 0.98445 \dots

u = sin (θ)

代回

sin (θ) \approx 0.32943 \dots, sin (θ) \approx 0.88011 \dots, sin (θ) \approx 0.60823 \dots, sin (θ) \approx - 0.98445 \dots

sin (θ) \approx 0.32943 \dots, sin (θ) \approx 0.88011 \dots, sin (θ) \approx 0.60823 \dots, sin (θ) \approx - 0.98445 \dots

sin (θ) = 0.32943 \dots : θ = arcsin (0.32943 \dots) + 2 πn, θ = π - arcsin (0.32943 \dots) + 2 πn

sin (θ) = 0.32943 \dots

使用反三角函数性质

sin (θ) = 0.32943 \dots

sin (θ) = 0.32943 \dots

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (0.32943 \dots) + 2 πn, θ = π - arcsin (0.32943 \dots) + 2 πn

θ = arcsin (0.32943 \dots) + 2 πn, θ = π - arcsin (0.32943 \dots) + 2 πn

sin (θ) = 0.88011 \dots : θ = arcsin (0.88011 \dots) + 2 πn, θ = π - arcsin (0.88011 \dots) + 2 πn

sin (θ) = 0.88011 \dots

使用反三角函数性质

sin (θ) = 0.88011 \dots

sin (θ) = 0.88011 \dots

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (0.88011 \dots) + 2 πn, θ = π - arcsin (0.88011 \dots) + 2 πn

θ = arcsin (0.88011 \dots) + 2 πn, θ = π - arcsin (0.88011 \dots) + 2 πn

sin (θ) = 0.60823 \dots : θ = arcsin (0.60823 \dots) + 2 πn, θ = π - arcsin (0.60823 \dots) + 2 πn

sin (θ) = 0.60823 \dots

使用反三角函数性质

sin (θ) = 0.60823 \dots

sin (θ) = 0.60823 \dots

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (0.60823 \dots) + 2 πn, θ = π - arcsin (0.60823 \dots) + 2 πn

θ = arcsin (0.60823 \dots) + 2 πn, θ = π - arcsin (0.60823 \dots) + 2 πn

sin (θ) = - 0.98445 \dots : θ = arcsin (- 0.98445 \dots) + 2 πn, θ = π + arcsin (0.98445 \dots) + 2 πn

sin (θ) = - 0.98445 \dots

使用反三角函数性质

sin (θ) = - 0.98445 \dots

sin (θ) = - 0.98445 \dots

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

θ = arcsin (- 0.98445 \dots) + 2 πn, θ = π + arcsin (0.98445 \dots) + 2 πn

θ = arcsin (- 0.98445 \dots) + 2 πn, θ = π + arcsin (0.98445 \dots) + 2 πn

合并所有解

θ = arcsin (0.32943 \dots) + 2 πn, θ = π - arcsin (0.32943 \dots) + 2 πn, θ = arcsin (0.88011 \dots) + 2 πn, θ = π - arcsin (0.88011 \dots) + 2 πn, θ = arcsin (0.60823 \dots) + 2 πn, θ = π - arcsin (0.60823 \dots) + 2 πn, θ = arcsin (- 0.98445 \dots) + 2 πn, θ = π + arcsin (0.98445 \dots) + 2 πn

将解代入原方程进行验证

将它们代入

3 sec (θ) + 5 csc (θ) = 12

检验解是否符合

去除与方程不符的解。

检验

arcsin (0.32943 \dots) + 2 πn

的解:

假

arcsin (0.32943 \dots) + 2 πn

代入

n = 1

arcsin (0.32943 \dots) + 2 π 1

对于

3 sec (θ) + 5 csc (θ) = 12

代入

θ = arcsin (0.32943 \dots) + 2 π 1

3 sec (arcsin (0.32943 \dots) + 2 π 1) + 5 csc (arcsin (0.32943 \dots) + 2 π 1) = 12

整理后得

18.35473 \dots = 12

\Rightarrow 假

检验

π - arcsin (0.32943 \dots) + 2 πn

的解:

真

π - arcsin (0.32943 \dots) + 2 πn

代入

n = 1

π - arcsin (0.32943 \dots) + 2 π 1

对于

3 sec (θ) + 5 csc (θ) = 12

代入

θ = π - arcsin (0.32943 \dots) + 2 π 1

3 sec (π - arcsin (0.32943 \dots) + 2 π 1) + 5 csc (π - arcsin (0.32943 \dots) + 2 π 1) = 12

整理后得

12 = 12

\Rightarrow 真

检验

arcsin (0.88011 \dots) + 2 πn

的解:

真

arcsin (0.88011 \dots) + 2 πn

代入

n = 1

arcsin (0.88011 \dots) + 2 π 1

对于

3 sec (θ) + 5 csc (θ) = 12

代入

θ = arcsin (0.88011 \dots) + 2 π 1

3 sec (arcsin (0.88011 \dots) + 2 π 1) + 5 csc (arcsin (0.88011 \dots) + 2 π 1) = 12

整理后得

12 = 12

\Rightarrow 真

检验

π - arcsin (0.88011 \dots) + 2 πn

的解:

假

π - arcsin (0.88011 \dots) + 2 πn

代入

n = 1

π - arcsin (0.88011 \dots) + 2 π 1

对于

3 sec (θ) + 5 csc (θ) = 12

代入

θ = π - arcsin (0.88011 \dots) + 2 π 1

3 sec (π - arcsin (0.88011 \dots) + 2 π 1) + 5 csc (π - arcsin (0.88011 \dots) + 2 π 1) = 12

整理后得

- 0.63781 \dots = 12

\Rightarrow 假

检验

arcsin (0.60823 \dots) + 2 πn

的解:

真

arcsin (0.60823 \dots) + 2 πn

代入

n = 1

arcsin (0.60823 \dots) + 2 π 1

对于

3 sec (θ) + 5 csc (θ) = 12

代入

θ = arcsin (0.60823 \dots) + 2 π 1

3 sec (arcsin (0.60823 \dots) + 2 π 1) + 5 csc (arcsin (0.60823 \dots) + 2 π 1) = 12

整理后得

12 = 12

\Rightarrow 真

检验

π - arcsin (0.60823 \dots) + 2 πn

的解:

假

π - arcsin (0.60823 \dots) + 2 πn

代入

n = 1

π - arcsin (0.60823 \dots) + 2 π 1

对于

3 sec (θ) + 5 csc (θ) = 12

代入

θ = π - arcsin (0.60823 \dots) + 2 π 1

3 sec (π - arcsin (0.60823 \dots) + 2 π 1) + 5 csc (π - arcsin (0.60823 \dots) + 2 π 1) = 12

整理后得

4.44101 \dots = 12

\Rightarrow 假

检验

arcsin (- 0.98445 \dots) + 2 πn

的解:

真

arcsin (- 0.98445 \dots) + 2 πn

代入

n = 1

arcsin (- 0.98445 \dots) + 2 π 1

对于

3 sec (θ) + 5 csc (θ) = 12

代入

θ = arcsin (- 0.98445 \dots) + 2 π 1

3 sec (arcsin (- 0.98445 \dots) + 2 π 1) + 5 csc (arcsin (- 0.98445 \dots) + 2 π 1) = 12

整理后得

12 = 12

\Rightarrow 真

检验

π + arcsin (0.98445 \dots) + 2 πn

的解:

假

π + arcsin (0.98445 \dots) + 2 πn

代入

n = 1

π + arcsin (0.98445 \dots) + 2 π 1

对于

3 sec (θ) + 5 csc (θ) = 12

代入

θ = π + arcsin (0.98445 \dots) + 2 π 1

3 sec (π + arcsin (0.98445 \dots) + 2 π 1) + 5 csc (π + arcsin (0.98445 \dots) + 2 π 1) = 12

整理后得

- 22.15793 \dots = 12

\Rightarrow 假

θ = π - arcsin (0.32943 \dots) + 2 πn, θ = arcsin (0.88011 \dots) + 2 πn, θ = arcsin (0.60823 \dots) + 2 πn, θ = arcsin (- 0.98445 \dots) + 2 πn

以小数形式表示解

θ = π - 0.33570 \dots + 2 πn, θ = 1.07609 \dots + 2 πn, θ = 0.65383 \dots + 2 πn, θ = - 1.39422 \dots + 2 πn

作图

流行的例子

sin(9x+2)=cos(6x-7)

sin (9 x + 2) = cos (6 x - 7)

2cos(3x+pi/2)=-1

2 cos (3 x + \frac{π}{2}) = - 1

sin (\frac{3 θ}{2}) = 0

4 sin (\frac{π}{2} x) = 3

tan (2 t) = 0