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受欢迎的三角函数 >

sin(x)=cos^3(x)

解答

sin (x) = cos^{3} (x)

解答

x = 0.59876 \dots + 2 πn, x = - 2.54282 \dots + 2 πn

+1

度数

x = 34.30680 \dots^{\circ} + 36 0^{\circ} n, x = - 145.69319 \dots^{\circ} + 36 0^{\circ} n

求解步骤

sin (x) = cos^{3} (x)

两边进行平方

sin^{2} (x) = (cos^{3} (x))^{2}

两边减去

(cos^{3} (x))^{2}

sin^{2} (x) - cos^{6} (x) = 0

使用三角恒等式改写

- cos^{6} (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)

= - cos^{6} (x) + 1 - cos^{2} (x)

1 - cos^{2} (x) - cos^{6} (x) = 0

用替代法求解

1 - cos^{2} (x) - cos^{6} (x) = 0

令

: cos (x) = u

1 - u^{2} - u^{6} = 0

1 - u^{2} - u^{6} = 0 : u = 0.68232 \dots, u = - 0.68232 \dots

1 - u^{2} - u^{6} = 0

改写成标准形式

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

- u^{6} - u^{2} + 1 = 0

用

v = u^{2}

和

v^{3} = u^{6}

改写方程式

- v^{3} - v + 1 = 0

解

- v^{3} - v + 1 = 0 : v \approx 0.68232 \dots

- v^{3} - v + 1 = 0

使用牛顿-拉弗森方法找到

- v^{3} - v + 1 = 0

的一个解:

v \approx 0.68232 \dots

- v^{3} - v + 1 = 0

牛顿-拉弗森近似法定义

f (v) = - v^{3} - v + 1

找到

f^{^{'}} (v) : - 3 v^{2} - 1

\frac{d}{d v} (- v^{3} - v + 1)

使用微分加减法定则:

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

= - \frac{d}{d v} (v^{3}) - \frac{d v}{d v} + \frac{d}{d v} (1)

\frac{d}{d v} (v^{3}) = 3 v^{2}

\frac{d}{d v} (v^{3})

使用幂法则:

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

= 3 v^{3 - 1}

化简

= 3 v^{2}

\frac{d v}{d v} = 1

\frac{d v}{d v}

使用常见微分定则:

\frac{d v}{d v} = 1

= 1

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

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

常数微分:

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

= 0

= - 3 v^{2} - 1 + 0

化简

= - 3 v^{2} - 1

令

v_{0} = 1

计算

v_{n + 1}

至

Δ v_{n + 1} < 0.000001

v_{1} = 0.75 : Δ v_{1} = 0.25

f (v_{0}) = - 1^{3} - 1 + 1 = - 1

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

v_{1} = 0.75

Δ v_{1} = ∣ 0.75 - 1 ∣ = 0.25

Δ v_{1} = 0.25

v_{2} = 0.68604 \dots : Δ v_{2} = 0.06395 \dots

f (v_{1}) = - 0.7 5^{3} - 0.75 + 1 = - 0.171875

f^{^{'}} (v_{1}) = - 3 \cdot 0.7 5^{2} - 1 = - 2.6875

v_{2} = 0.68604 \dots

Δ v_{2} = ∣ 0.68604 \dots - 0.75 ∣ = 0.06395 \dots

Δ v_{2} = 0.06395 \dots

v_{3} = 0.68233 \dots : Δ v_{3} = 0.00370 \dots

f (v_{2}) = - 0.68604 \dots^{3} - 0.68604 \dots + 1 = - 0.00894 \dots

f^{^{'}} (v_{2}) = - 3 \cdot 0.68604 \dots^{2} - 1 = - 2.41197 \dots

v_{3} = 0.68233 \dots

Δ v_{3} = ∣ 0.68233 \dots - 0.68604 \dots ∣ = 0.00370 \dots

Δ v_{3} = 0.00370 \dots

v_{4} = 0.68232 \dots : Δ v_{4} = 0.00001 \dots

f (v_{3}) = - 0.68233 \dots^{3} - 0.68233 \dots + 1 = - 0.00002 \dots

f^{^{'}} (v_{3}) = - 3 \cdot 0.68233 \dots^{2} - 1 = - 2.39676 \dots

v_{4} = 0.68232 \dots

Δ v_{4} = ∣ 0.68232 \dots - 0.68233 \dots ∣ = 0.00001 \dots

Δ v_{4} = 0.00001 \dots

v_{5} = 0.68232 \dots : Δ v_{5} = 1.18493 E - 10

f (v_{4}) = - 0.68232 \dots^{3} - 0.68232 \dots + 1 = - 2.83995 E - 10

f^{^{'}} (v_{4}) = - 3 \cdot 0.68232 \dots^{2} - 1 = - 2.39671 \dots

v_{5} = 0.68232 \dots

Δ v_{5} = ∣ 0.68232 \dots - 0.68232 \dots ∣ = 1.18493 E - 10

Δ v_{5} = 1.18493 E - 10

v \approx 0.68232 \dots

使用长除法

Eq u a t i o n 0 : \frac{- v ^{3} - v + 1}{v - 0.68232 \dots} = - v^{2} - 0.68232 \dots v - 1.46557 \dots

- v^{2} - 0.68232 \dots v - 1.46557 \dots \approx 0

使用牛顿-拉弗森方法找到

- v^{2} - 0.68232 \dots v - 1.46557 \dots = 0

的一个解:

v \in R

无解

- v^{2} - 0.68232 \dots v - 1.46557 \dots = 0

牛顿-拉弗森近似法定义

f (v) = - v^{2} - 0.68232 \dots v - 1.46557 \dots

找到

f^{^{'}} (v) : - 2 v - 0.68232 \dots

\frac{d}{d v} (- v^{2} - 0.68232 \dots v - 1.46557 \dots)

使用微分加减法定则:

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

= - \frac{d}{d v} (v^{2}) - \frac{d}{d v} (0.68232 \dots v) - \frac{d}{d v} (1.46557 \dots)

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

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

使用幂法则:

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

= 2 v^{2 - 1}

化简

= 2 v

\frac{d}{d v} (0.68232 \dots v) = 0.68232 \dots

\frac{d}{d v} (0.68232 \dots v)

将常数提出:

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

= 0.68232 \dots \frac{d v}{d v}

使用常见微分定则:

\frac{d v}{d v} = 1

= 0.68232 \dots \cdot 1

化简

= 0.68232 \dots

\frac{d}{d v} (1.46557 \dots) = 0

\frac{d}{d v} (1.46557 \dots)

常数微分:

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

= 0

= - 2 v - 0.68232 \dots - 0

化简

= - 2 v - 0.68232 \dots

令

v_{0} = - 2

计算

v_{n + 1}

至

Δ v_{n + 1} < 0.000001

v_{1} = - 0.76391 \dots : Δ v_{1} = 1.23608 \dots

f (v_{0}) = - (- 2)^{2} - 0.68232 \dots (- 2) - 1.46557 \dots = - 4.10091 \dots

f^{^{'}} (v_{0}) = - 2 (- 2) - 0.68232 \dots = 3.31767 \dots

v_{1} = - 0.76391 \dots

Δ v_{1} = ∣ - 0.76391 \dots - (- 2) ∣ = 1.23608 \dots

Δ v_{1} = 1.23608 \dots

v_{2} = 1.04316 \dots : Δ v_{2} = 1.80707 \dots

f (v_{1}) = - (- 0.76391 \dots)^{2} - 0.68232 \dots (- 0.76391 \dots) - 1.46557 \dots = - 1.52789 \dots

f^{^{'}} (v_{1}) = - 2 (- 0.76391 \dots) - 0.68232 \dots = 0.84550 \dots

v_{2} = 1.04316 \dots

Δ v_{2} = ∣ 1.04316 \dots - (- 0.76391 \dots) ∣ = 1.80707 \dots

Δ v_{2} = 1.80707 \dots

v_{3} = - 0.13630 \dots : Δ v_{3} = 1.17946 \dots

f (v_{2}) = - 1.04316 \dots^{2} - 0.68232 \dots \cdot 1.04316 \dots - 1.46557 \dots = - 3.26553 \dots

f^{^{'}} (v_{2}) = - 2 \cdot 1.04316 \dots - 0.68232 \dots = - 2.76865 \dots

v_{3} = - 0.13630 \dots

Δ v_{3} = ∣ - 0.13630 \dots - 1.04316 \dots ∣ = 1.17946 \dots

Δ v_{3} = 1.17946 \dots

v_{4} = - 3.53171 \dots : Δ v_{4} = 3.39540 \dots

f (v_{3}) = - (- 0.13630 \dots)^{2} - 0.68232 \dots (- 0.13630 \dots) - 1.46557 \dots = - 1.39114 \dots

f^{^{'}} (v_{3}) = - 2 (- 0.13630 \dots) - 0.68232 \dots = - 0.40971 \dots

v_{4} = - 3.53171 \dots

Δ v_{4} = ∣ - 3.53171 \dots - (- 0.13630 \dots) ∣ = 3.39540 \dots

Δ v_{4} = 3.39540 \dots

v_{5} = - 1.72500 \dots : Δ v_{5} = 1.80670 \dots

f (v_{4}) = - (- 3.53171 \dots)^{2} - 0.68232 \dots (- 3.53171 \dots) - 1.46557 \dots = - 11.52876 \dots

f^{^{'}} (v_{4}) = - 2 (- 3.53171 \dots) - 0.68232 \dots = 6.38109 \dots

v_{5} = - 1.72500 \dots

Δ v_{5} = ∣ - 1.72500 \dots - (- 3.53171 \dots) ∣ = 1.80670 \dots

Δ v_{5} = 1.80670 \dots

v_{6} = - 0.54560 \dots : Δ v_{6} = 1.17939 \dots

f (v_{5}) = - (- 1.72500 \dots)^{2} - 0.68232 \dots (- 1.72500 \dots) - 1.46557 \dots = - 3.26419 \dots

f^{^{'}} (v_{5}) = - 2 (- 1.72500 \dots) - 0.68232 \dots = 2.76767 \dots

v_{6} = - 0.54560 \dots

Δ v_{6} = ∣ - 0.54560 \dots - (- 1.72500 \dots) ∣ = 1.17939 \dots

Δ v_{6} = 1.17939 \dots

v_{7} = 2.85625 \dots : Δ v_{7} = 3.40185 \dots

f (v_{6}) = - (- 0.54560 \dots)^{2} - 0.68232 \dots (- 0.54560 \dots) - 1.46557 \dots = - 1.39097 \dots

f^{^{'}} (v_{6}) = - 2 (- 0.54560 \dots) - 0.68232 \dots = 0.40888 \dots

v_{7} = 2.85625 \dots

Δ v_{7} = ∣ 2.85625 \dots - (- 0.54560 \dots) ∣ = 3.40185 \dots

Δ v_{7} = 3.40185 \dots

v_{8} = 1.04656 \dots : Δ v_{8} = 1.80968 \dots

f (v_{7}) = - 2.85625 \dots^{2} - 0.68232 \dots \cdot 2.85625 \dots - 1.46557 \dots = - 11.57264 \dots

f^{^{'}} (v_{7}) = - 2 \cdot 2.85625 \dots - 0.68232 \dots = - 6.39483 \dots

v_{8} = 1.04656 \dots

Δ v_{8} = ∣ 1.04656 \dots - 2.85625 \dots ∣ = 1.80968 \dots

Δ v_{8} = 1.80968 \dots

v_{9} = - 0.13340 \dots : Δ v_{9} = 1.17997 \dots

f (v_{8}) = - 1.04656 \dots^{2} - 0.68232 \dots \cdot 1.04656 \dots - 1.46557 \dots = - 3.27496 \dots

f^{^{'}} (v_{8}) = - 2 \cdot 1.04656 \dots - 0.68232 \dots = - 2.77545 \dots

v_{9} = - 0.13340 \dots

Δ v_{9} = ∣ - 0.13340 \dots - 1.04656 \dots ∣ = 1.17997 \dots

Δ v_{9} = 1.17997 \dots

v_{10} = - 3.48434 \dots : Δ v_{10} = 3.35093 \dots

f (v_{9}) = - (- 0.13340 \dots)^{2} - 0.68232 \dots (- 0.13340 \dots) - 1.46557 \dots = - 1.39234 \dots

f^{^{'}} (v_{9}) = - 2 (- 0.13340 \dots) - 0.68232 \dots = - 0.41550 \dots

v_{10} = - 3.48434 \dots

Δ v_{10} = ∣ - 3.48434 \dots - (- 0.13340 \dots) ∣ = 3.35093 \dots

Δ v_{10} = 3.35093 \dots

无法得出解

解是

v \approx 0.68232 \dots

v \approx 0.68232 \dots

代回

v = u^{2}

，求解

u

解

u^{2} = 0.68232 \dots : u = 0.68232 \dots, u = - 0.68232 \dots

u^{2} = 0.68232 \dots

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 0.68232 \dots, u = - 0.68232 \dots

解为

u = 0.68232 \dots, u = - 0.68232 \dots

u = cos (x)

代回

cos (x) = 0.68232 \dots, cos (x) = - 0.68232 \dots

cos (x) = 0.68232 \dots, cos (x) = - 0.68232 \dots

cos (x) = 0.68232 \dots : x = arccos (0.68232 \dots) + 2 πn, x = 2 π - arccos (0.68232 \dots) + 2 πn

cos (x) = 0.68232 \dots

使用反三角函数性质

cos (x) = 0.68232 \dots

cos (x) = 0.68232 \dots

的通解

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

x = arccos (0.68232 \dots) + 2 πn, x = 2 π - arccos (0.68232 \dots) + 2 πn

x = arccos (0.68232 \dots) + 2 πn, x = 2 π - arccos (0.68232 \dots) + 2 πn

cos (x) = - 0.68232 \dots : x = arccos (- 0.68232 \dots) + 2 πn, x = - arccos (- 0.68232 \dots) + 2 πn

cos (x) = - 0.68232 \dots

使用反三角函数性质

cos (x) = - 0.68232 \dots

cos (x) = - 0.68232 \dots

的通解

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

x = arccos (- 0.68232 \dots) + 2 πn, x = - arccos (- 0.68232 \dots) + 2 πn

x = arccos (- 0.68232 \dots) + 2 πn, x = - arccos (- 0.68232 \dots) + 2 πn

合并所有解

x = arccos (0.68232 \dots) + 2 πn, x = 2 π - arccos (0.68232 \dots) + 2 πn, x = arccos (- 0.68232 \dots) + 2 πn, x = - arccos (- 0.68232 \dots) + 2 πn

将解代入原方程进行验证

将它们代入

sin (x) = cos^{3} (x)

检验解是否符合

去除与方程不符的解。

检验

arccos (0.68232 \dots) + 2 πn

的解:

真

arccos (0.68232 \dots) + 2 πn

代入

n = 1

arccos (0.68232 \dots) + 2 π 1

对于

sin (x) = cos^{3} (x)

代入

x = arccos (0.68232 \dots) + 2 π 1

sin (arccos (0.68232 \dots) + 2 π 1) = cos^{3} (arccos (0.68232 \dots) + 2 π 1)

整理后得

0.56362 \dots = 0.56362 \dots

\Rightarrow 真

检验

2 π - arccos (0.68232 \dots) + 2 πn

的解:

假

2 π - arccos (0.68232 \dots) + 2 πn

代入

n = 1

2 π - arccos (0.68232 \dots) + 2 π 1

对于

sin (x) = cos^{3} (x)

代入

x = 2 π - arccos (0.68232 \dots) + 2 π 1

sin (2 π - arccos (0.68232 \dots) + 2 π 1) = cos^{3} (2 π - arccos (0.68232 \dots) + 2 π 1)

整理后得

- 0.56362 \dots = 0.56362 \dots

\Rightarrow 假

检验

arccos (- 0.68232 \dots) + 2 πn

的解:

假

arccos (- 0.68232 \dots) + 2 πn

代入

n = 1

arccos (- 0.68232 \dots) + 2 π 1

对于

sin (x) = cos^{3} (x)

代入

x = arccos (- 0.68232 \dots) + 2 π 1

sin (arccos (- 0.68232 \dots) + 2 π 1) = cos^{3} (arccos (- 0.68232 \dots) + 2 π 1)

整理后得

0.56362 \dots = - 0.56362 \dots

\Rightarrow 假

检验

- arccos (- 0.68232 \dots) + 2 πn

的解:

真

- arccos (- 0.68232 \dots) + 2 πn

代入

n = 1

- arccos (- 0.68232 \dots) + 2 π 1

对于

sin (x) = cos^{3} (x)

代入

x = - arccos (- 0.68232 \dots) + 2 π 1

sin (- arccos (- 0.68232 \dots) + 2 π 1) = cos^{3} (- arccos (- 0.68232 \dots) + 2 π 1)

整理后得

- 0.56362 \dots = - 0.56362 \dots

\Rightarrow 真

x = arccos (0.68232 \dots) + 2 πn, x = - arccos (- 0.68232 \dots) + 2 πn

以小数形式表示解

x = 0.59876 \dots + 2 πn, x = - 2.54282 \dots + 2 πn

作图

流行的例子

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