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受欢迎的三角函数 >

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

解答

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

解答

x = - 2 \cdot 0.45227 \dots + 4 πn, x = 2 π + 2 \cdot 0.45227 \dots + 4 πn, x = 2 \cdot 0.45227 \dots + 4 πn, x = 2 π - 2 \cdot 0.45227 \dots + 4 πn

+1

度数

x = - 51.82729 \dots^{\circ} + 72 0^{\circ} n, x = 411.82729 \dots^{\circ} + 72 0^{\circ} n, x = 51.82729 \dots^{\circ} + 72 0^{\circ} n, x = 308.17270 \dots^{\circ} + 72 0^{\circ} n

求解步骤

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

两边减去

1

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

令

: u = \frac{x}{2}

sin^{2} (2 u) + 2 sin^{2} (u) - 1 = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

= 2 sin^{2} (u) - cos^{2} (2 u)

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

分解

- cos^{2} (2 u) + 2 sin^{2} (u) : (2 sin (u) + cos (2 u)) (2 sin (u) - cos (2 u))

- cos^{2} (2 u) + 2 sin^{2} (u)

将

2 sin^{2} (u) - cos^{2} (2 u)

改写为

(2 sin (u))^{2} - cos^{2} (2 u)

2 sin^{2} (u) - cos^{2} (2 u)

使用根式运算法则:

a = (a)^{2}

2 = (2)^{2}

= (2)^{2} sin^{2} (u) - cos^{2} (2 u)

使用指数法则:

a^{m} b^{m} = (ab)^{m}

(2)^{2} sin^{2} (u) = (2 sin (u))^{2}

= (2 sin (u))^{2} - cos^{2} (2 u)

= (2 sin (u))^{2} - cos^{2} (2 u)

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(2 sin (u))^{2} - cos^{2} (2 u) = (2 sin (u) + cos (2 u)) (2 sin (u) - cos (2 u))

= (2 sin (u) + cos (2 u)) (2 sin (u) - cos (2 u))

(2 sin (u) + cos (2 u)) (2 sin (u) - cos (2 u)) = 0

分别求解每个部分

2 sin (u) + cos (2 u) = 0 or 2 sin (u) - cos (2 u) = 0

2 sin (u) + cos (2 u) = 0 : u = arcsin (- \frac{- 2 + 10}{4}) + 2 πn, u = π + arcsin (\frac{- 2 + 10}{4}) + 2 πn

2 sin (u) + cos (2 u) = 0

使用三角恒等式改写

cos (2 u) + sin (u) 2

使用倍角公式:

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

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

1 - 2 sin^{2} (u) + sin (u) 2 = 0

用替代法求解

1 - 2 sin^{2} (u) + sin (u) 2 = 0

令

: sin (u) = u

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

1 - 2 u^{2} + u 2 = 0 : u = - \frac{- 2 + 10}{4}, u = \frac{2 + 10}{4}

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

改写成标准形式

a x^{2} + b x + c = 0

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 2, b = 2, c = 1

u_{1, 2} = \frac{- 2 \pm ( 2 ) ^{2} - 4 ( - 2 ) \cdot 1}{2 ( - 2 )}

u_{1, 2} = \frac{- 2 \pm ( 2 ) ^{2} - 4 ( - 2 ) \cdot 1}{2 ( - 2 )}

(2)^{2} - 4 (- 2) \cdot 1 = 10

(2)^{2} - 4 (- 2) \cdot 1

使用法则

- (- a) = a

= (2)^{2} + 4 \cdot 2 \cdot 1

(2)^{2} = 2

(2)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

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

使用指数法则:

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

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 8

= 2 + 8

数字相加:

2 + 8 = 10

= 10

u_{1, 2} = \frac{- 2 \pm 10}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 2 + 10}{2 ( - 2 )}, u_{2} = \frac{- 2 - 10}{2 ( - 2 )}

u = \frac{- 2 + 10}{2 ( - 2 )} : - \frac{- 2 + 10}{4}

\frac{- 2 + 10}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 2 + 10}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 + 10}{- 4}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{- 2 + 10}{4}

u = \frac{- 2 - 10}{2 ( - 2 )} : \frac{2 + 10}{4}

\frac{- 2 - 10}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 2 - 10}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 - 10}{- 4}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

- 2 - 10 = - (2 + 10)

= \frac{2 + 10}{4}

二次方程组的解是:

u = - \frac{- 2 + 10}{4}, u = \frac{2 + 10}{4}

u = sin (u)

代回

sin (u) = - \frac{- 2 + 10}{4}, sin (u) = \frac{2 + 10}{4}

sin (u) = - \frac{- 2 + 10}{4}, sin (u) = \frac{2 + 10}{4}

sin (u) = - \frac{- 2 + 10}{4} : u = arcsin (- \frac{- 2 + 10}{4}) + 2 πn, u = π + arcsin (\frac{- 2 + 10}{4}) + 2 πn

sin (u) = - \frac{- 2 + 10}{4}

使用反三角函数性质

sin (u) = - \frac{- 2 + 10}{4}

sin (u) = - \frac{- 2 + 10}{4}

的通解

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

u = arcsin (- \frac{- 2 + 10}{4}) + 2 πn, u = π + arcsin (\frac{- 2 + 10}{4}) + 2 πn

u = arcsin (- \frac{- 2 + 10}{4}) + 2 πn, u = π + arcsin (\frac{- 2 + 10}{4}) + 2 πn

sin (u) = \frac{2 + 10}{4} :

无解

sin (u) = \frac{2 + 10}{4}

- 1 \leq sin (x) \leq 1

无解

合并所有解

u = arcsin (- \frac{- 2 + 10}{4}) + 2 πn, u = π + arcsin (\frac{- 2 + 10}{4}) + 2 πn

2 sin (u) - cos (2 u) = 0 : u = arcsin (\frac{- 2 + 10}{4}) + 2 πn, u = π - arcsin (\frac{- 2 + 10}{4}) + 2 πn

2 sin (u) - cos (2 u) = 0

使用三角恒等式改写

- cos (2 u) + sin (u) 2

使用倍角公式:

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

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

- (1 - 2 sin^{2} (u)) : - 1 + 2 sin^{2} (u)

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

打开括号

= - (1) - (- 2 sin^{2} (u))

使用加减运算法则

- (- a) = a, - (a) = - a

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

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

- 1 + 2 sin^{2} (u) + sin (u) 2 = 0

用替代法求解

- 1 + 2 sin^{2} (u) + sin (u) 2 = 0

令

: sin (u) = u

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

- 1 + 2 u^{2} + u 2 = 0 : u = \frac{- 2 + 10}{4}, u = \frac{- 2 - 10}{4}

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

改写成标准形式

a x^{2} + b x + c = 0

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

使用求根公式求解

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

二次方程求根公式:

若

a = 2, b = 2, c = - 1

u_{1, 2} = \frac{- 2 \pm ( 2 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

u_{1, 2} = \frac{- 2 \pm ( 2 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

(2)^{2} - 4 \cdot 2 (- 1) = 10

(2)^{2} - 4 \cdot 2 (- 1)

使用法则

- (- a) = a

= (2)^{2} + 4 \cdot 2 \cdot 1

(2)^{2} = 2

(2)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

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

使用指数法则:

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

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 8

= 2 + 8

数字相加:

2 + 8 = 10

= 10

u_{1, 2} = \frac{- 2 \pm 10}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- 2 + 10}{2 \cdot 2}, u_{2} = \frac{- 2 - 10}{2 \cdot 2}

u = \frac{- 2 + 10}{2 \cdot 2} : \frac{- 2 + 10}{4}

\frac{- 2 + 10}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 + 10}{4}

u = \frac{- 2 - 10}{2 \cdot 2} : \frac{- 2 - 10}{4}

\frac{- 2 - 10}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 - 10}{4}

二次方程组的解是:

u = \frac{- 2 + 10}{4}, u = \frac{- 2 - 10}{4}

u = sin (u)

代回

sin (u) = \frac{- 2 + 10}{4}, sin (u) = \frac{- 2 - 10}{4}

sin (u) = \frac{- 2 + 10}{4}, sin (u) = \frac{- 2 - 10}{4}

sin (u) = \frac{- 2 + 10}{4} : u = arcsin (\frac{- 2 + 10}{4}) + 2 πn, u = π - arcsin (\frac{- 2 + 10}{4}) + 2 πn

sin (u) = \frac{- 2 + 10}{4}

使用反三角函数性质

sin (u) = \frac{- 2 + 10}{4}

sin (u) = \frac{- 2 + 10}{4}

的通解

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

u = arcsin (\frac{- 2 + 10}{4}) + 2 πn, u = π - arcsin (\frac{- 2 + 10}{4}) + 2 πn

u = arcsin (\frac{- 2 + 10}{4}) + 2 πn, u = π - arcsin (\frac{- 2 + 10}{4}) + 2 πn

sin (u) = \frac{- 2 - 10}{4} :

无解

sin (u) = \frac{- 2 - 10}{4}

- 1 \leq sin (x) \leq 1

无解

合并所有解

u = arcsin (\frac{- 2 + 10}{4}) + 2 πn, u = π - arcsin (\frac{- 2 + 10}{4}) + 2 πn

合并所有解

u = arcsin (- \frac{- 2 + 10}{4}) + 2 πn, u = π + arcsin (\frac{- 2 + 10}{4}) + 2 πn, u = arcsin (\frac{- 2 + 10}{4}) + 2 πn, u = π - arcsin (\frac{- 2 + 10}{4}) + 2 πn

u = \frac{x}{2}

代回

\frac{x}{2} = arcsin (- \frac{- 2 + 10}{4}) + 2 πn : x = - 2 arcsin (\frac{10 - 2}{4}) + 4 πn

\frac{x}{2} = arcsin (- \frac{- 2 + 10}{4}) + 2 πn

化简

arcsin (- \frac{- 2 + 10}{4}) + 2 πn : - arcsin (\frac{10 - 2}{4}) + 2 πn

arcsin (- \frac{- 2 + 10}{4}) + 2 πn

利用以下特性:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{10 - 2}{4}) = - arcsin (\frac{10 - 2}{4})

= - arcsin (\frac{10 - 2}{4}) + 2 πn

\frac{x}{2} = - arcsin (\frac{10 - 2}{4}) + 2 πn

在两边乘以

2

\frac{x}{2} = - arcsin (\frac{10 - 2}{4}) + 2 πn

在两边乘以

2

\frac{2 x}{2} = - 2 arcsin (\frac{10 - 2}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} = - 2 arcsin (\frac{10 - 2}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

- 2 arcsin (\frac{10 - 2}{4}) + 2 \cdot 2 πn : - 2 arcsin (\frac{10 - 2}{4}) + 4 πn

- 2 arcsin (\frac{10 - 2}{4}) + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= - 2 arcsin (\frac{10 - 2}{4}) + 4 πn

x = - 2 arcsin (\frac{10 - 2}{4}) + 4 πn

x = - 2 arcsin (\frac{10 - 2}{4}) + 4 πn

x = - 2 arcsin (\frac{10 - 2}{4}) + 4 πn

\frac{x}{2} = π + arcsin (\frac{- 2 + 10}{4}) + 2 πn : x = 2 π + 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn

\frac{x}{2} = π + arcsin (\frac{- 2 + 10}{4}) + 2 πn

在两边乘以

2

\frac{x}{2} = π + arcsin (\frac{- 2 + 10}{4}) + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 π + 2 arcsin (\frac{- 2 + 10}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} = 2 π + 2 arcsin (\frac{- 2 + 10}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 π + 2 arcsin (\frac{- 2 + 10}{4}) + 2 \cdot 2 πn : 2 π + 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn

2 π + 2 arcsin (\frac{- 2 + 10}{4}) + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 2 π + 2 arcsin (\frac{10 - 2}{4}) + 4 πn

x = 2 π + 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn

x = 2 π + 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn

x = 2 π + 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn

\frac{x}{2} = arcsin (\frac{- 2 + 10}{4}) + 2 πn : x = 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn

\frac{x}{2} = arcsin (\frac{- 2 + 10}{4}) + 2 πn

在两边乘以

2

\frac{x}{2} = arcsin (\frac{- 2 + 10}{4}) + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 arcsin (\frac{- 2 + 10}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} = 2 arcsin (\frac{- 2 + 10}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 arcsin (\frac{- 2 + 10}{4}) + 2 \cdot 2 πn : 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn

2 arcsin (\frac{- 2 + 10}{4}) + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 2 arcsin (\frac{10 - 2}{4}) + 4 πn

x = 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn

x = 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn

x = 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn

\frac{x}{2} = π - arcsin (\frac{- 2 + 10}{4}) + 2 πn : x = 2 π - 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn

\frac{x}{2} = π - arcsin (\frac{- 2 + 10}{4}) + 2 πn

在两边乘以

2

\frac{x}{2} = π - arcsin (\frac{- 2 + 10}{4}) + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 π - 2 arcsin (\frac{- 2 + 10}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} = 2 π - 2 arcsin (\frac{- 2 + 10}{4}) + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 π - 2 arcsin (\frac{- 2 + 10}{4}) + 2 \cdot 2 πn : 2 π - 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn

2 π - 2 arcsin (\frac{- 2 + 10}{4}) + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 2 π - 2 arcsin (\frac{10 - 2}{4}) + 4 πn

x = 2 π - 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn

x = 2 π - 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn

x = 2 π - 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn

x = - 2 arcsin (\frac{10 - 2}{4}) + 4 πn, x = 2 π + 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn, x = 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn, x = 2 π - 2 arcsin (\frac{- 2 + 10}{4}) + 4 πn

以小数形式表示解

x = - 2 \cdot 0.45227 \dots + 4 πn, x = 2 π + 2 \cdot 0.45227 \dots + 4 πn, x = 2 \cdot 0.45227 \dots + 4 πn, x = 2 π - 2 \cdot 0.45227 \dots + 4 πn

作图

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