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受欢迎的三角函数 >

sin(x)cos(2x)= 1/2 (1+sin(x))

解答

sin (x) cos (2 x) = \frac{1}{2} (1 + sin (x))

解答

x = - 0.86437 \dots + 2 πn, x = π + 0.86437 \dots + 2 πn

+1

度数

x = - 49.52505 \dots^{\circ} + 36 0^{\circ} n, x = 229.52505 \dots^{\circ} + 36 0^{\circ} n

求解步骤

sin (x) cos (2 x) = \frac{1}{2} (1 + sin (x))

两边减去

\frac{1}{2} (1 + sin (x))

sin (x) cos (2 x) - \frac{1}{2} (1 + sin (x)) = 0

化简

sin (x) cos (2 x) - \frac{1}{2} (1 + sin (x)) : \frac{2 sin ( x ) cos ( 2 x ) - 1 - sin ( x )}{2}

sin (x) cos (2 x) - \frac{1}{2} (1 + sin (x))

\frac{1}{2} (1 + sin (x)) = \frac{1 + sin ( x )}{2}

\frac{1}{2} (1 + sin (x))

分式相乘:

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

= \frac{1 \cdot ( 1 + sin ( x ) )}{2}

1 \cdot (1 + sin (x)) = 1 + sin (x)

1 \cdot (1 + sin (x))

乘以:

1 \cdot (1 + sin (x)) = (1 + sin (x))

= (1 + sin (x))

去除括号:

(a) = a

= 1 + sin (x)

= \frac{1 + sin ( x )}{2}

= sin (x) cos (2 x) - \frac{sin ( x ) + 1}{2}

将项转换为分式:

sin (x) cos (2 x) = \frac{sin ( x ) cos ( 2 x ) 2}{2}

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

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

乘开

sin (x) cos (2 x) \cdot 2 - (1 + sin (x)) : sin (x) cos (2 x) \cdot 2 - 1 - sin (x)

sin (x) cos (2 x) \cdot 2 - (1 + sin (x))

= 2 sin (x) cos (2 x) - (1 + sin (x))

- (1 + sin (x)) : - 1 - sin (x)

- (1 + sin (x))

打开括号

= - (1) - (sin (x))

使用加减运算法则

+ (- a) = - a

= - 1 - sin (x)

= sin (x) cos (2 x) \cdot 2 - 1 - sin (x)

= \frac{2 sin ( x ) cos ( 2 x ) - 1 - sin ( x )}{2}

\frac{2 sin ( x ) cos ( 2 x ) - 1 - sin ( x )}{2} = 0

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

2 sin (x) cos (2 x) - 1 - sin (x) = 0

使用三角恒等式改写

- 1 - sin (x) + 2 cos (2 x) sin (x)

使用倍角公式:

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

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

化简

- 1 - sin (x) + 2 (1 - 2 sin^{2} (x)) sin (x) : - 1 + sin (x) - 4 sin^{3} (x)

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

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

乘开

2 sin (x) (1 - 2 sin^{2} (x)) : 2 sin (x) - 4 sin^{3} (x)

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

使用分配律:

a (b - c) = ab - a c

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

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

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

化简

2 \cdot 1 \cdot sin (x) - 2 \cdot 2 sin^{2} (x) sin (x) : 2 sin (x) - 4 sin^{3} (x)

2 \cdot 1 \cdot sin (x) - 2 \cdot 2 sin^{2} (x) sin (x)

2 \cdot 1 \cdot sin (x) = 2 sin (x)

2 \cdot 1 \cdot sin (x)

数字相乘:

2 \cdot 1 = 2

= 2 sin (x)

2 \cdot 2 sin^{2} (x) sin (x) = 4 sin^{3} (x)

2 \cdot 2 sin^{2} (x) sin (x)

数字相乘:

2 \cdot 2 = 4

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

使用指数法则:

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

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

= 4 sin^{2 + 1} (x)

数字相加:

2 + 1 = 3

= 4 sin^{3} (x)

= 2 sin (x) - 4 sin^{3} (x)

= 2 sin (x) - 4 sin^{3} (x)

= - 1 - sin (x) + 2 sin (x) - 4 sin^{3} (x)

同类项相加:

- sin (x) + 2 sin (x) = sin (x)

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

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

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

用替代法求解

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

令

: sin (x) = u

- 1 + u - 4 u^{3} = 0

- 1 + u - 4 u^{3} = 0 : u \approx - 0.76068 \dots

- 1 + u - 4 u^{3} = 0

改写成标准形式

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

- 4 u^{3} + u - 1 = 0

使用牛顿-拉弗森方法找到

- 4 u^{3} + u - 1 = 0

的一个解:

u \approx - 0.76068 \dots

- 4 u^{3} + u - 1 = 0

牛顿-拉弗森近似法定义

f (u) = - 4 u^{3} + u - 1

找到

f^{^{'}} (u) : - 12 u^{2} + 1

\frac{d}{d u} (- 4 u^{3} + u - 1)

使用微分加减法定则:

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

= - \frac{d}{d u} (4 u^{3}) + \frac{d u}{d u} - \frac{d}{d u} (1)

\frac{d}{d u} (4 u^{3}) = 12 u^{2}

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

将常数提出:

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

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

使用幂法则:

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

= 4 \cdot 3 u^{3 - 1}

化简

= 12 u^{2}

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

\frac{d u}{d u}

使用常见微分定则:

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

= 1

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

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

常数微分:

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

= 0

= - 12 u^{2} + 1 - 0

化简

= - 12 u^{2} + 1

令

u_{0} = - 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.81818 \dots : Δ u_{1} = 0.18181 \dots

f (u_{0}) = - 4 (- 1)^{3} + (- 1) - 1 = 2

f^{^{'}} (u_{0}) = - 12 (- 1)^{2} + 1 = - 11

u_{1} = - 0.81818 \dots

Δ u_{1} = ∣ - 0.81818 \dots - (- 1) ∣ = 0.18181 \dots

Δ u_{1} = 0.18181 \dots

u_{2} = - 0.76519 \dots : Δ u_{2} = 0.05298 \dots

f (u_{1}) = - 4 (- 0.81818 \dots)^{3} + (- 0.81818 \dots) - 1 = 0.37265 \dots

f^{^{'}} (u_{1}) = - 12 (- 0.81818 \dots)^{2} + 1 = - 7.03305 \dots

u_{2} = - 0.76519 \dots

Δ u_{2} = ∣ - 0.76519 \dots - (- 0.81818 \dots) ∣ = 0.05298 \dots

Δ u_{2} = 0.05298 \dots

u_{3} = - 0.76072 \dots : Δ u_{3} = 0.00447 \dots

f (u_{2}) = - 4 (- 0.76519 \dots)^{3} + (- 0.76519 \dots) - 1 = 0.02696 \dots

f^{^{'}} (u_{2}) = - 12 (- 0.76519 \dots)^{2} + 1 = - 6.02629 \dots

u_{3} = - 0.76072 \dots

Δ u_{3} = ∣ - 0.76072 \dots - (- 0.76519 \dots) ∣ = 0.00447 \dots

Δ u_{3} = 0.00447 \dots

u_{4} = - 0.76068 \dots : Δ u_{4} = 0.00003 \dots

f (u_{3}) = - 4 (- 0.76072 \dots)^{3} + (- 0.76072 \dots) - 1 = 0.00018 \dots

f^{^{'}} (u_{3}) = - 12 (- 0.76072 \dots)^{2} + 1 = - 5.94435 \dots

u_{4} = - 0.76068 \dots

Δ u_{4} = ∣ - 0.76068 \dots - (- 0.76072 \dots) ∣ = 0.00003 \dots

Δ u_{4} = 0.00003 \dots

u_{5} = - 0.76068 \dots : Δ u_{5} = 1.46429 E - 9

f (u_{4}) = - 4 (- 0.76068 \dots)^{3} + (- 0.76068 \dots) - 1 = 8.70343 E - 9

f^{^{'}} (u_{4}) = - 12 (- 0.76068 \dots)^{2} + 1 = - 5.94378 \dots

u_{5} = - 0.76068 \dots

Δ u_{5} = ∣ - 0.76068 \dots - (- 0.76068 \dots) ∣ = 1.46429 E - 9

Δ u_{5} = 1.46429 E - 9

u \approx - 0.76068 \dots

使用长除法

Eq u a t i o n 0 : \frac{- 4 u ^{3} + u - 1}{u + 0.76068 \dots} = - 4 u^{2} + 3.04275 \dots u - 1.31459 \dots

- 4 u^{2} + 3.04275 \dots u - 1.31459 \dots \approx 0

使用牛顿-拉弗森方法找到

- 4 u^{2} + 3.04275 \dots u - 1.31459 \dots = 0

的一个解:

u \in R

无解

- 4 u^{2} + 3.04275 \dots u - 1.31459 \dots = 0

牛顿-拉弗森近似法定义

f (u) = - 4 u^{2} + 3.04275 \dots u - 1.31459 \dots

找到

f^{^{'}} (u) : - 8 u + 3.04275 \dots

\frac{d}{d u} (- 4 u^{2} + 3.04275 \dots u - 1.31459 \dots)

使用微分加减法定则:

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

= - \frac{d}{d u} (4 u^{2}) + \frac{d}{d u} (3.04275 \dots u) - \frac{d}{d u} (1.31459 \dots)

\frac{d}{d u} (4 u^{2}) = 8 u

\frac{d}{d u} (4 u^{2})

将常数提出:

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

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

使用幂法则:

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

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

化简

= 8 u

\frac{d}{d u} (3.04275 \dots u) = 3.04275 \dots

\frac{d}{d u} (3.04275 \dots u)

将常数提出:

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

= 3.04275 \dots \frac{d u}{d u}

使用常见微分定则:

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

= 3.04275 \dots \cdot 1

化简

= 3.04275 \dots

\frac{d}{d u} (1.31459 \dots) = 0

\frac{d}{d u} (1.31459 \dots)

常数微分:

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

= 0

= - 8 u + 3.04275 \dots - 0

化简

= - 8 u + 3.04275 \dots

令

u_{0} = 0

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.43204 \dots : Δ u_{1} = 0.43204 \dots

f (u_{0}) = - 4 \cdot 0^{2} + 3.04275 \dots \cdot 0 - 1.31459 \dots = - 1.31459 \dots

f^{^{'}} (u_{0}) = - 8 \cdot 0 + 3.04275 \dots = 3.04275 \dots

u_{1} = 0.43204 \dots

Δ u_{1} = ∣ 0.43204 \dots - 0 ∣ = 0.43204 \dots

Δ u_{1} = 0.43204 \dots

u_{2} = - 1.37331 \dots : Δ u_{2} = 1.80535 \dots

f (u_{1}) = - 4 \cdot 0.43204 \dots^{2} + 3.04275 \dots \cdot 0.43204 \dots - 1.31459 \dots = - 0.74663 \dots

f^{^{'}} (u_{1}) = - 8 \cdot 0.43204 \dots + 3.04275 \dots = - 0.41356 \dots

u_{2} = - 1.37331 \dots

Δ u_{2} = ∣ - 1.37331 \dots - 0.43204 \dots ∣ = 1.80535 \dots

Δ u_{2} = 1.80535 \dots

u_{3} = - 0.44402 \dots : Δ u_{3} = 0.92928 \dots

f (u_{2}) = - 4 (- 1.37331 \dots)^{2} + 3.04275 \dots (- 1.37331 \dots) - 1.31459 \dots = - 13.03728 \dots

f^{^{'}} (u_{2}) = - 8 (- 1.37331 \dots) + 3.04275 \dots = 14.02930 \dots

u_{3} = - 0.44402 \dots

Δ u_{3} = ∣ - 0.44402 \dots - (- 1.37331 \dots) ∣ = 0.92928 \dots

Δ u_{3} = 0.92928 \dots

u_{4} = 0.07974 \dots : Δ u_{4} = 0.52377 \dots

f (u_{3}) = - 4 (- 0.44402 \dots)^{2} + 3.04275 \dots (- 0.44402 \dots) - 1.31459 \dots = - 3.45431 \dots

f^{^{'}} (u_{3}) = - 8 (- 0.44402 \dots) + 3.04275 \dots = 6.59499 \dots

u_{4} = 0.07974 \dots

Δ u_{4} = ∣ 0.07974 \dots - (- 0.44402 \dots) ∣ = 0.52377 \dots

Δ u_{4} = 0.52377 \dots

u_{5} = 0.53608 \dots : Δ u_{5} = 0.45633 \dots

f (u_{4}) = - 4 \cdot 0.07974 \dots^{2} + 3.04275 \dots \cdot 0.07974 \dots - 1.31459 \dots = - 1.09737 \dots

f^{^{'}} (u_{4}) = - 8 \cdot 0.07974 \dots + 3.04275 \dots = 2.40476 \dots

u_{5} = 0.53608 \dots

Δ u_{5} = ∣ 0.53608 \dots - 0.07974 \dots ∣ = 0.45633 \dots

Δ u_{5} = 0.45633 \dots

u_{6} = - 0.13247 \dots : Δ u_{6} = 0.66855 \dots

f (u_{5}) = - 4 \cdot 0.53608 \dots^{2} + 3.04275 \dots \cdot 0.53608 \dots - 1.31459 \dots = - 0.83296 \dots

f^{^{'}} (u_{5}) = - 8 \cdot 0.53608 \dots + 3.04275 \dots = - 1.24592 \dots

u_{6} = - 0.13247 \dots

Δ u_{6} = ∣ - 0.13247 \dots - 0.53608 \dots ∣ = 0.66855 \dots

Δ u_{6} = 0.66855 \dots

u_{7} = 0.30332 \dots : Δ u_{7} = 0.43579 \dots

f (u_{6}) = - 4 (- 0.13247 \dots)^{2} + 3.04275 \dots (- 0.13247 \dots) - 1.31459 \dots = - 1.78786 \dots

f^{^{'}} (u_{6}) = - 8 (- 0.13247 \dots) + 3.04275 \dots = 4.10252 \dots

u_{7} = 0.30332 \dots

Δ u_{7} = ∣ 0.30332 \dots - (- 0.13247 \dots) ∣ = 0.43579 \dots

Δ u_{7} = 0.43579 \dots

u_{8} = 1.53626 \dots : Δ u_{8} = 1.23293 \dots

f (u_{7}) = - 4 \cdot 0.30332 \dots^{2} + 3.04275 \dots \cdot 0.30332 \dots - 1.31459 \dots = - 0.75967 \dots

f^{^{'}} (u_{7}) = - 8 \cdot 0.30332 \dots + 3.04275 \dots = 0.61615 \dots

u_{8} = 1.53626 \dots

Δ u_{8} = ∣ 1.53626 \dots - 0.30332 \dots ∣ = 1.23293 \dots

Δ u_{8} = 1.23293 \dots

u_{9} = 0.87871 \dots : Δ u_{9} = 0.65754 \dots

f (u_{8}) = - 4 \cdot 1.53626 \dots^{2} + 3.04275 \dots \cdot 1.53626 \dots - 1.31459 \dots = - 6.08050 \dots

f^{^{'}} (u_{8}) = - 8 \cdot 1.53626 \dots + 3.04275 \dots = - 9.24732 \dots

u_{9} = 0.87871 \dots

Δ u_{9} = ∣ 0.87871 \dots - 1.53626 \dots ∣ = 0.65754 \dots

Δ u_{9} = 0.65754 \dots

u_{10} = 0.44494 \dots : Δ u_{10} = 0.43377 \dots

f (u_{9}) = - 4 \cdot 0.87871 \dots^{2} + 3.04275 \dots \cdot 0.87871 \dots - 1.31459 \dots = - 1.72944 \dots

f^{^{'}} (u_{9}) = - 8 \cdot 0.87871 \dots + 3.04275 \dots = - 3.98698 \dots

u_{10} = 0.44494 \dots

Δ u_{10} = ∣ 0.44494 \dots - 0.87871 \dots ∣ = 0.43377 \dots

Δ u_{10} = 0.43377 \dots

u_{11} = - 1.01142 \dots : Δ u_{11} = 1.45637 \dots

f (u_{10}) = - 4 \cdot 0.44494 \dots^{2} + 3.04275 \dots \cdot 0.44494 \dots - 1.31459 \dots = - 0.75263 \dots

f^{^{'}} (u_{10}) = - 8 \cdot 0.44494 \dots + 3.04275 \dots = - 0.51679 \dots

u_{11} = - 1.01142 \dots

Δ u_{11} = ∣ - 1.01142 \dots - 0.44494 \dots ∣ = 1.45637 \dots

Δ u_{11} = 1.45637 \dots

无法得出解

解是

u \approx - 0.76068 \dots

u = sin (x)

代回

sin (x) \approx - 0.76068 \dots

sin (x) \approx - 0.76068 \dots

sin (x) = - 0.76068 \dots : x = arcsin (- 0.76068 \dots) + 2 πn, x = π + arcsin (0.76068 \dots) + 2 πn

sin (x) = - 0.76068 \dots

使用反三角函数性质

sin (x) = - 0.76068 \dots

sin (x) = - 0.76068 \dots

的通解

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

x = arcsin (- 0.76068 \dots) + 2 πn, x = π + arcsin (0.76068 \dots) + 2 πn

x = arcsin (- 0.76068 \dots) + 2 πn, x = π + arcsin (0.76068 \dots) + 2 πn

合并所有解

x = arcsin (- 0.76068 \dots) + 2 πn, x = π + arcsin (0.76068 \dots) + 2 πn

以小数形式表示解

x = - 0.86437 \dots + 2 πn, x = π + 0.86437 \dots + 2 πn

作图

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