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受欢迎的三角函数 >

1+cos(a)=(2cos^2(a))/2

解答

1 + cos (a) = \frac{2 cos ^{2} ( a )}{2}

解答

a = 2.23703 \dots + 2 πn, a = - 2.23703 \dots + 2 πn

+1

度数

a = 128.17270 \dots^{\circ} + 36 0^{\circ} n, a = - 128.17270 \dots^{\circ} + 36 0^{\circ} n

求解步骤

1 + cos (a) = \frac{2 cos ^{2} ( a )}{2}

用替代法求解

1 + cos (a) = \frac{2 cos ^{2} ( a )}{2}

令

: cos (a) = u

1 + u = \frac{2 u ^{2}}{2}

1 + u = \frac{2 u ^{2}}{2} : u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

1 + u = \frac{2 u ^{2}}{2}

在两边乘以

2

1 + u = \frac{2 u ^{2}}{2}

在两边乘以

2

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

化简

2 + 2 u = 2 u^{2}

2 + 2 u = 2 u^{2}

交换两边

2 u^{2} = 2 + 2 u

将

2 u

para o lado esquerdo

2 u^{2} = 2 + 2 u

两边减去

2 u

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

化简

2 u^{2} - 2 u = 2

2 u^{2} - 2 u = 2

将

2

para o lado esquerdo

2 u^{2} - 2 u = 2

两边减去

2

2 u^{2} - 2 u - 2 = 2 - 2

化简

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

(- 2)^{2} - 4 \cdot 2 (- 2) = 25

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

使用法则

- (- a) = a

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

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 2)^{2} = 2^{2}

= 2^{2} + 4 \cdot 2 \cdot 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 2^{2} + 16

2^{2} = 4

= 4 + 16

数字相加:

4 + 16 = 20

= 20

20

质因数分解:

2^{2} \cdot 5

20

20

除以

220 = 10 \cdot 2

= 2 \cdot 10

10

除以

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

使用根式运算法则:

n ab = n a n b

= 5 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 25

u_{1, 2} = \frac{- ( - 2 ) \pm 2 5}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 2 5}{2 \cdot 2}, u_{2} = \frac{- ( - 2 ) - 2 5}{2 \cdot 2}

u = \frac{- ( - 2 ) + 2 5}{2 \cdot 2} : \frac{1 + 5}{2}

\frac{- ( - 2 ) + 2 5}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{2 + 2 5}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2 + 2 5}{4}

分解

2 + 25 : 2 (1 + 5)

2 + 25

改写为

= 2 \cdot 1 + 25

因式分解出通项

2

= 2 (1 + 5)

= \frac{2 ( 1 + 5 )}{4}

约分:

2

= \frac{1 + 5}{2}

u = \frac{- ( - 2 ) - 2 5}{2 \cdot 2} : \frac{1 - 5}{2}

\frac{- ( - 2 ) - 2 5}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{2 - 2 5}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2 - 2 5}{4}

分解

2 - 25 : 2 (1 - 5)

2 - 25

改写为

= 2 \cdot 1 - 25

因式分解出通项

2

= 2 (1 - 5)

= \frac{2 ( 1 - 5 )}{4}

约分:

2

= \frac{1 - 5}{2}

二次方程组的解是:

u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

u = cos (a)

代回

cos (a) = \frac{1 + 5}{2}, cos (a) = \frac{1 - 5}{2}

cos (a) = \frac{1 + 5}{2}, cos (a) = \frac{1 - 5}{2}

cos (a) = \frac{1 + 5}{2} :

无解

cos (a) = \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

无解

cos (a) = \frac{1 - 5}{2} : a = arccos (\frac{1 - 5}{2}) + 2 πn, a = - arccos (\frac{1 - 5}{2}) + 2 πn

cos (a) = \frac{1 - 5}{2}

使用反三角函数性质

cos (a) = \frac{1 - 5}{2}

cos (a) = \frac{1 - 5}{2}

的通解

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

a = arccos (\frac{1 - 5}{2}) + 2 πn, a = - arccos (\frac{1 - 5}{2}) + 2 πn

a = arccos (\frac{1 - 5}{2}) + 2 πn, a = - arccos (\frac{1 - 5}{2}) + 2 πn

合并所有解

a = arccos (\frac{1 - 5}{2}) + 2 πn, a = - arccos (\frac{1 - 5}{2}) + 2 πn

以小数形式表示解

a = 2.23703 \dots + 2 πn, a = - 2.23703 \dots + 2 πn

作图

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