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受欢迎的三角函数 >

sin^2(x)+cos^5(x)=2

解答

sin^{2} (x) + cos^{5} (x) = 2

解答

x \in R 无解

求解步骤

sin^{2} (x) + cos^{5} (x) = 2

两边减去

2

sin^{2} (x) + cos^{5} (x) - 2 = 0

使用三角恒等式改写

- 2 + cos^{5} (x) + sin^{2} (x)

使用毕达哥拉斯恒等式:

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

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

= - 2 + cos^{5} (x) + 1 - cos^{2} (x)

化简

- 2 + cos^{5} (x) + 1 - cos^{2} (x) : cos^{5} (x) - cos^{2} (x) - 1

- 2 + cos^{5} (x) + 1 - cos^{2} (x)

对同类项分组

= cos^{5} (x) - cos^{2} (x) - 2 + 1

数字相加/相减:

- 2 + 1 = - 1

= cos^{5} (x) - cos^{2} (x) - 1

= cos^{5} (x) - cos^{2} (x) - 1

- 1 - cos^{2} (x) + cos^{5} (x) = 0

用替代法求解

- 1 - cos^{2} (x) + cos^{5} (x) = 0

令

: cos (x) = u

- 1 - u^{2} + u^{5} = 0

- 1 - u^{2} + u^{5} = 0 : u \approx 1.19385 \dots

- 1 - u^{2} + u^{5} = 0

改写成标准形式

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

u^{5} - u^{2} - 1 = 0

使用牛顿-拉弗森方法找到

u^{5} - u^{2} - 1 = 0

的一个解:

u \approx 1.19385 \dots

u^{5} - u^{2} - 1 = 0

牛顿-拉弗森近似法定义

f (u) = u^{5} - u^{2} - 1

找到

f^{^{'}} (u) : 5 u^{4} - 2 u

\frac{d}{d u} (u^{5} - u^{2} - 1)

使用微分加减法定则:

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

= \frac{d}{d u} (u^{5}) - \frac{d}{d u} (u^{2}) - \frac{d}{d u} (1)

\frac{d}{d u} (u^{5}) = 5 u^{4}

\frac{d}{d u} (u^{5})

使用幂法则:

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

= 5 u^{5 - 1}

化简

= 5 u^{4}

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

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

使用幂法则:

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

= 2 u^{2 - 1}

化简

= 2 u

\frac{d}{d u} (1) = 0

\frac{d}{d u} (1)

常数微分:

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

= 0

= 5 u^{4} - 2 u - 0

化简

= 5 u^{4} - 2 u

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 1.33333 \dots : Δ u_{1} = 0.33333 \dots

f (u_{0}) = 1^{5} - 1^{2} - 1 = - 1

f^{^{'}} (u_{0}) = 5 \cdot 1^{4} - 2 \cdot 1 = 3

u_{1} = 1.33333 \dots

Δ u_{1} = ∣ 1.33333 \dots - 1 ∣ = 0.33333 \dots

Δ u_{1} = 0.33333 \dots

u_{2} = 1.22399 \dots : Δ u_{2} = 0.10933 \dots

f (u_{1}) = 1.33333 \dots^{5} - 1.33333 \dots^{2} - 1 = 1.43621 \dots

f^{^{'}} (u_{1}) = 5 \cdot 1.33333 \dots^{4} - 2 \cdot 1.33333 \dots = 13.13580 \dots

u_{2} = 1.22399 \dots

Δ u_{2} = ∣ 1.22399 \dots - 1.33333 \dots ∣ = 0.10933 \dots

Δ u_{2} = 0.10933 \dots

u_{3} = 1.19560 \dots : Δ u_{3} = 0.02838 \dots

f (u_{2}) = 1.22399 \dots^{5} - 1.22399 \dots^{2} - 1 = 0.24910 \dots

f^{^{'}} (u_{2}) = 5 \cdot 1.22399 \dots^{4} - 2 \cdot 1.22399 \dots = 8.77456 \dots

u_{3} = 1.19560 \dots

Δ u_{3} = ∣ 1.19560 \dots - 1.22399 \dots ∣ = 0.02838 \dots

Δ u_{3} = 0.02838 \dots

u_{4} = 1.19386 \dots : Δ u_{4} = 0.00174 \dots

f (u_{3}) = 1.19560 \dots^{5} - 1.19560 \dots^{2} - 1 = 0.01363 \dots

f^{^{'}} (u_{3}) = 5 \cdot 1.19560 \dots^{4} - 2 \cdot 1.19560 \dots = 7.82581 \dots

u_{4} = 1.19386 \dots

Δ u_{4} = ∣ 1.19386 \dots - 1.19560 \dots ∣ = 0.00174 \dots

Δ u_{4} = 0.00174 \dots

u_{5} = 1.19385 \dots : Δ u_{5} = 6.27682 E - 6

f (u_{4}) = 1.19386 \dots^{5} - 1.19386 \dots^{2} - 1 = 0.00004 \dots

f^{^{'}} (u_{4}) = 5 \cdot 1.19386 \dots^{4} - 2 \cdot 1.19386 \dots = 7.76987 \dots

u_{5} = 1.19385 \dots

Δ u_{5} = ∣ 1.19385 \dots - 1.19386 \dots ∣ = 6.27682 E - 6

Δ u_{5} = 6.27682 E - 6

u_{6} = 1.19385 \dots : Δ u_{6} = 8.12152 E - 11

f (u_{5}) = 1.19385 \dots^{5} - 1.19385 \dots^{2} - 1 = 6.31015 E - 10

f^{^{'}} (u_{5}) = 5 \cdot 1.19385 \dots^{4} - 2 \cdot 1.19385 \dots = 7.76967 \dots

u_{6} = 1.19385 \dots

Δ u_{6} = ∣ 1.19385 \dots - 1.19385 \dots ∣ = 8.12152 E - 11

Δ u_{6} = 8.12152 E - 11

u \approx 1.19385 \dots

使用长除法

Eq u a t i o n 0 : \frac{u ^{5} - u ^{2} - 1}{u - 1.19385 \dots} = u^{4} + 1.19385 \dots u^{3} + 1.42529 \dots u^{2} + 0.70160 \dots u + 0.83761 \dots

u^{4} + 1.19385 \dots u^{3} + 1.42529 \dots u^{2} + 0.70160 \dots u + 0.83761 \dots \approx 0

使用牛顿-拉弗森方法找到

u^{4} + 1.19385 \dots u^{3} + 1.42529 \dots u^{2} + 0.70160 \dots u + 0.83761 \dots = 0

的一个解:

u \in R

无解

u^{4} + 1.19385 \dots u^{3} + 1.42529 \dots u^{2} + 0.70160 \dots u + 0.83761 \dots = 0

牛顿-拉弗森近似法定义

f (u) = u^{4} + 1.19385 \dots u^{3} + 1.42529 \dots u^{2} + 0.70160 \dots u + 0.83761 \dots

找到

f^{^{'}} (u) : 4 u^{3} + 3.58157 \dots u^{2} + 2.85059 \dots u + 0.70160 \dots

\frac{d}{d u} (u^{4} + 1.19385 \dots u^{3} + 1.42529 \dots u^{2} + 0.70160 \dots u + 0.83761 \dots)

使用微分加减法定则:

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

= \frac{d}{d u} (u^{4}) + \frac{d}{d u} (1.19385 \dots u^{3}) + \frac{d}{d u} (1.42529 \dots u^{2}) + \frac{d}{d u} (0.70160 \dots u) + \frac{d}{d u} (0.83761 \dots)

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

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

使用幂法则:

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

= 4 u^{4 - 1}

化简

= 4 u^{3}

\frac{d}{d u} (1.19385 \dots u^{3}) = 3.58157 \dots u^{2}

\frac{d}{d u} (1.19385 \dots u^{3})

将常数提出:

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

= 1.19385 \dots \frac{d}{d u} (u^{3})

使用幂法则:

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

= 1.19385 \dots \cdot 3 u^{3 - 1}

化简

= 3.58157 \dots u^{2}

\frac{d}{d u} (1.42529 \dots u^{2}) = 2.85059 \dots u

\frac{d}{d u} (1.42529 \dots u^{2})

将常数提出:

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

= 1.42529 \dots \frac{d}{d u} (u^{2})

使用幂法则:

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

= 1.42529 \dots \cdot 2 u^{2 - 1}

化简

= 2.85059 \dots u

\frac{d}{d u} (0.70160 \dots u) = 0.70160 \dots

\frac{d}{d u} (0.70160 \dots u)

将常数提出:

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

= 0.70160 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 0.70160 \dots \cdot 1

化简

= 0.70160 \dots

\frac{d}{d u} (0.83761 \dots) = 0

\frac{d}{d u} (0.83761 \dots)

常数微分:

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

= 0

= 4 u^{3} + 3.58157 \dots u^{2} + 2.85059 \dots u + 0.70160 \dots + 0

化简

= 4 u^{3} + 3.58157 \dots u^{2} + 2.85059 \dots u + 0.70160 \dots

令

u_{0} = - 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.46738 \dots : Δ u_{1} = 0.53261 \dots

f (u_{0}) = (- 1)^{4} + 1.19385 \dots (- 1)^{3} + 1.42529 \dots (- 1)^{2} + 0.70160 \dots (- 1) + 0.83761 \dots = 1.36745 \dots

f^{^{'}} (u_{0}) = 4 (- 1)^{3} + 3.58157 \dots (- 1)^{2} + 2.85059 \dots (- 1) + 0.70160 \dots = - 2.56741 \dots

u_{1} = - 0.46738 \dots

Δ u_{1} = ∣ - 0.46738 \dots - (- 1) ∣ = 0.53261 \dots

Δ u_{1} = 0.53261 \dots

u_{2} = 2.44193 \dots : Δ u_{2} = 2.90931 \dots

f (u_{1}) = (- 0.46738 \dots)^{4} + 1.19385 \dots (- 0.46738 \dots)^{3} + 1.42529 \dots (- 0.46738 \dots)^{2} + 0.70160 \dots (- 0.46738 \dots) + 0.83761 \dots = 0.74688 \dots

f^{^{'}} (u_{1}) = 4 (- 0.46738 \dots)^{3} + 3.58157 \dots (- 0.46738 \dots)^{2} + 2.85059 \dots (- 0.46738 \dots) + 0.70160 \dots = - 0.25672 \dots

u_{2} = 2.44193 \dots

Δ u_{2} = ∣ 2.44193 \dots - (- 0.46738 \dots) ∣ = 2.90931 \dots

Δ u_{2} = 2.90931 \dots

u_{3} = 1.70862 \dots : Δ u_{3} = 0.73330 \dots

f (u_{2}) = 2.44193 \dots^{4} + 1.19385 \dots \cdot 2.44193 \dots^{3} + 1.42529 \dots \cdot 2.44193 \dots^{2} + 0.70160 \dots \cdot 2.44193 \dots + 0.83761 \dots = 63.99240 \dots

f^{^{'}} (u_{2}) = 4 \cdot 2.44193 \dots^{3} + 3.58157 \dots \cdot 2.44193 \dots^{2} + 2.85059 \dots \cdot 2.44193 \dots + 0.70160 \dots = 87.26542 \dots

u_{3} = 1.70862 \dots

Δ u_{3} = ∣ 1.70862 \dots - 2.44193 \dots ∣ = 0.73330 \dots

Δ u_{3} = 0.73330 \dots

u_{4} = 1.13400 \dots : Δ u_{4} = 0.57462 \dots

f (u_{3}) = 1.70862 \dots^{4} + 1.19385 \dots \cdot 1.70862 \dots^{3} + 1.42529 \dots \cdot 1.70862 \dots^{2} + 0.70160 \dots \cdot 1.70862 \dots + 0.83761 \dots = 20.67565 \dots

f^{^{'}} (u_{3}) = 4 \cdot 1.70862 \dots^{3} + 3.58157 \dots \cdot 1.70862 \dots^{2} + 2.85059 \dots \cdot 1.70862 \dots + 0.70160 \dots = 35.98115 \dots

u_{4} = 1.13400 \dots

Δ u_{4} = ∣ 1.13400 \dots - 1.70862 \dots ∣ = 0.57462 \dots

Δ u_{4} = 0.57462 \dots

u_{5} = 0.65666 \dots : Δ u_{5} = 0.47733 \dots

f (u_{4}) = 1.13400 \dots^{4} + 1.19385 \dots \cdot 1.13400 \dots^{3} + 1.42529 \dots \cdot 1.13400 \dots^{2} + 0.70160 \dots \cdot 1.13400 \dots + 0.83761 \dots = 6.86084 \dots

f^{^{'}} (u_{4}) = 4 \cdot 1.13400 \dots^{3} + 3.58157 \dots \cdot 1.13400 \dots^{2} + 2.85059 \dots \cdot 1.13400 \dots + 0.70160 \dots = 14.37317 \dots

u_{5} = 0.65666 \dots

Δ u_{5} = ∣ 0.65666 \dots - 1.13400 \dots ∣ = 0.47733 \dots

Δ u_{5} = 0.47733 \dots

u_{6} = 0.19253 \dots : Δ u_{6} = 0.46412 \dots

f (u_{5}) = 0.65666 \dots^{4} + 1.19385 \dots \cdot 0.65666 \dots^{3} + 1.42529 \dots \cdot 0.65666 \dots^{2} + 0.70160 \dots \cdot 0.65666 \dots + 0.83761 \dots = 2.43695 \dots

f^{^{'}} (u_{5}) = 4 \cdot 0.65666 \dots^{3} + 3.58157 \dots \cdot 0.65666 \dots^{2} + 2.85059 \dots \cdot 0.65666 \dots + 0.70160 \dots = 5.25058 \dots

u_{6} = 0.19253 \dots

Δ u_{6} = ∣ 0.19253 \dots - 0.65666 \dots ∣ = 0.46412 \dots

Δ u_{6} = 0.46412 \dots

u_{7} = - 0.54088 \dots : Δ u_{7} = 0.73342 \dots

f (u_{6}) = 0.19253 \dots^{4} + 1.19385 \dots \cdot 0.19253 \dots^{3} + 1.42529 \dots \cdot 0.19253 \dots^{2} + 0.70160 \dots \cdot 0.19253 \dots + 0.83761 \dots = 1.03543 \dots

f^{^{'}} (u_{6}) = 4 \cdot 0.19253 \dots^{3} + 3.58157 \dots \cdot 0.19253 \dots^{2} + 2.85059 \dots \cdot 0.19253 \dots + 0.70160 \dots = 1.41178 \dots

u_{7} = - 0.54088 \dots

Δ u_{7} = ∣ - 0.54088 \dots - 0.19253 \dots ∣ = 0.73342 \dots

Δ u_{7} = 0.73342 \dots

u_{8} = 1.27339 \dots : Δ u_{8} = 1.81428 \dots

f (u_{7}) = (- 0.54088 \dots)^{4} + 1.19385 \dots (- 0.54088 \dots)^{3} + 1.42529 \dots (- 0.54088 \dots)^{2} + 0.70160 \dots (- 0.54088 \dots) + 0.83761 \dots = 0.77178 \dots

f^{^{'}} (u_{7}) = 4 (- 0.54088 \dots)^{3} + 3.58157 \dots (- 0.54088 \dots)^{2} + 2.85059 \dots (- 0.54088 \dots) + 0.70160 \dots = - 0.42539 \dots

u_{8} = 1.27339 \dots

Δ u_{8} = ∣ 1.27339 \dots - (- 0.54088 \dots) ∣ = 1.81428 \dots

Δ u_{8} = 1.81428 \dots

u_{9} = 0.77679 \dots : Δ u_{9} = 0.49659 \dots

f (u_{8}) = 1.27339 \dots^{4} + 1.19385 \dots \cdot 1.27339 \dots^{3} + 1.42529 \dots \cdot 1.27339 \dots^{2} + 0.70160 \dots \cdot 1.27339 \dots + 0.83761 \dots = 9.13680 \dots

f^{^{'}} (u_{8}) = 4 \cdot 1.27339 \dots^{3} + 3.58157 \dots \cdot 1.27339 \dots^{2} + 2.85059 \dots \cdot 1.27339 \dots + 0.70160 \dots = 18.39872 \dots

u_{9} = 0.77679 \dots

Δ u_{9} = ∣ 0.77679 \dots - 1.27339 \dots ∣ = 0.49659 \dots

Δ u_{9} = 0.49659 \dots

u_{10} = 0.32133 \dots : Δ u_{10} = 0.45546 \dots

f (u_{9}) = 0.77679 \dots^{4} + 1.19385 \dots \cdot 0.77679 \dots^{3} + 1.42529 \dots \cdot 0.77679 \dots^{2} + 0.70160 \dots \cdot 0.77679 \dots + 0.83761 \dots = 3.16638 \dots

f^{^{'}} (u_{9}) = 4 \cdot 0.77679 \dots^{3} + 3.58157 \dots \cdot 0.77679 \dots^{2} + 2.85059 \dots \cdot 0.77679 \dots + 0.70160 \dots = 6.95206 \dots

u_{10} = 0.32133 \dots

Δ u_{10} = ∣ 0.32133 \dots - 0.77679 \dots ∣ = 0.45546 \dots

Δ u_{10} = 0.45546 \dots

u_{11} = - 0.27320 \dots : Δ u_{11} = 0.59453 \dots

f (u_{10}) = 0.32133 \dots^{4} + 1.19385 \dots \cdot 0.32133 \dots^{3} + 1.42529 \dots \cdot 0.32133 \dots^{2} + 0.70160 \dots \cdot 0.32133 \dots + 0.83761 \dots = 1.26052 \dots

f^{^{'}} (u_{10}) = 4 \cdot 0.32133 \dots^{3} + 3.58157 \dots \cdot 0.32133 \dots^{2} + 2.85059 \dots \cdot 0.32133 \dots + 0.70160 \dots = 2.12016 \dots

u_{11} = - 0.27320 \dots

Δ u_{11} = ∣ - 0.27320 \dots - 0.32133 \dots ∣ = 0.59453 \dots

Δ u_{11} = 0.59453 \dots

无法得出解

解是

u \approx 1.19385 \dots

u = cos (x)

代回

cos (x) \approx 1.19385 \dots

cos (x) \approx 1.19385 \dots

cos (x) = 1.19385 \dots :

无解

cos (x) = 1.19385 \dots

- 1 \leq cos (x) \leq 1

无解

合并所有解

x \in R 无解

作图

流行的例子

16=4+9-12cos(x)

16 = 4 + 9 - 12 cos (x)

tanh (z) + 2 = 0

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cos^{2} (a) = \frac{2}{3 sin ( a )}

5 cos (5 x) = 2

4sin(x)-sin^3(x)-1=0

4 sin (x) - sin^{3} (x) - 1 = 0