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受欢迎的三角函数 >

sqrt(4cos(θ))=1-cos(θ)

解答

4 cos (θ) = 1 - cos (θ)

解答

θ = 1.39837 \dots + 2 πn, θ = 2 π - 1.39837 \dots + 2 πn

+1

度数

θ = 80.12071 \dots^{\circ} + 36 0^{\circ} n, θ = 279.87928 \dots^{\circ} + 36 0^{\circ} n

求解步骤

4 cos (θ) = 1 - cos (θ)

用替代法求解

4 cos (θ) = 1 - cos (θ)

令

: cos (θ) = u

4 u = 1 - u

4 u = 1 - u : u = 3 - 22

4 u = 1 - u

两边进行平方:

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

4 u = 1 - u

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

展开

(4 u)^{2} : 4 u

(4 u)^{2}

使用根式运算法则:

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

= ((4 u)^{\frac{1}{2}})^{2}

使用指数法则:

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

= (4 u)^{\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

= 4 u

展开

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

(1 - u)^{2}

使用完全平方公式:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 1, b = u

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

化简

1^{2} - 2 \cdot 1 \cdot u + u^{2} : 1 - 2 u + u^{2}

1^{2} - 2 \cdot 1 \cdot u + u^{2}

使用法则

1^{a} = 1

1^{2} = 1

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

数字相乘:

2 \cdot 1 = 2

= 1 - 2 u + u^{2}

= 1 - 2 u + u^{2}

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

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

解

4 u = 1 - 2 u + u^{2} : u = 3 + 22, u = 3 - 22

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

交换两边

1 - 2 u + u^{2} = 4 u

将

4 u

para o lado esquerdo

1 - 2 u + u^{2} = 4 u

两边减去

4 u

1 - 2 u + u^{2} - 4 u = 4 u - 4 u

化简

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 1, b = - 6, c = 1

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 1 = 42

(- 6)^{2} - 4 \cdot 1 \cdot 1

使用指数法则:

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

若

n

是偶数

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 6^{2} - 4

6^{2} = 36

= 36 - 4

数字相减:

36 - 4 = 32

= 32

32

质因数分解:

2^{5}

32

32

除以

232 = 16 \cdot 2

= 2 \cdot 16

16

除以

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 2^{5}

使用指数法则:

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

= 2^{4} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{4}

使用根式运算法则:

n a^{m} = a^{\frac{m}{n}}

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 2

整理后得

= 42

u_{1, 2} = \frac{- ( - 6 ) \pm 4 2}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 6 ) + 4 2}{2 \cdot 1}, u_{2} = \frac{- ( - 6 ) - 4 2}{2 \cdot 1}

u = \frac{- ( - 6 ) + 4 2}{2 \cdot 1} : 3 + 22

\frac{- ( - 6 ) + 4 2}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{6 + 4 2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{6 + 4 2}{2}

分解

6 + 42 : 2 (3 + 22)

6 + 42

改写为

= 2 \cdot 3 + 2 \cdot 22

因式分解出通项

2

= 2 (3 + 22)

= \frac{2 ( 3 + 2 2 )}{2}

数字相除:

\frac{2}{2} = 1

= 3 + 22

u = \frac{- ( - 6 ) - 4 2}{2 \cdot 1} : 3 - 22

\frac{- ( - 6 ) - 4 2}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{6 - 4 2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{6 - 4 2}{2}

分解

6 - 42 : 2 (3 - 22)

6 - 42

改写为

= 2 \cdot 3 - 2 \cdot 22

因式分解出通项

2

= 2 (3 - 22)

= \frac{2 ( 3 - 2 2 )}{2}

数字相除:

\frac{2}{2} = 1

= 3 - 22

二次方程组的解是:

u = 3 + 22, u = 3 - 22

u = 3 + 22, u = 3 - 22

验证解:

u = 3 + 22

假

, u = 3 - 22

真

将它们代入

4 u = 1 - u

检验解是否符合

去除与方程不符的解。

代入

u = 3 + 22 :

假

4 (3 + 22) = 1 - (3 + 22)

1 - (3 + 22) = - 2 - 22

1 - (3 + 22)

- (3 + 22) : - 3 - 22

- (3 + 22)

打开括号

= - (3) - (22)

使用加减运算法则

+ (- a) = - a

= - 3 - 22

= 1 - 3 - 22

数字相减:

1 - 3 = - 2

= - 2 - 22

4 (3 + 22) = - 2 - 22

假

代入

u = 3 - 22 :

真

4 (3 - 22) = 1 - (3 - 22)

1 - (3 - 22) = 22 - 2

1 - (3 - 22)

- (3 - 22) : - 3 + 22

- (3 - 22)

打开括号

= - (3) - (- 22)

使用加减运算法则

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

= - 3 + 22

= 1 - 3 + 22

数字相减:

1 - 3 = - 2

= 22 - 2

4 (3 - 22) = 22 - 2

真

解是

u = 3 - 22

u = cos (θ)

代回

cos (θ) = 3 - 22

cos (θ) = 3 - 22

cos (θ) = 3 - 22 : θ = arccos (3 - 22) + 2 πn, θ = 2 π - arccos (3 - 22) + 2 πn

cos (θ) = 3 - 22

使用反三角函数性质

cos (θ) = 3 - 22

cos (θ) = 3 - 22

的通解

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

θ = arccos (3 - 22) + 2 πn, θ = 2 π - arccos (3 - 22) + 2 πn

θ = arccos (3 - 22) + 2 πn, θ = 2 π - arccos (3 - 22) + 2 πn

合并所有解

θ = arccos (3 - 22) + 2 πn, θ = 2 π - arccos (3 - 22) + 2 πn

以小数形式表示解

θ = 1.39837 \dots + 2 πn, θ = 2 π - 1.39837 \dots + 2 πn

作图

流行的例子

sin (\frac{x}{3}) = 0

cos (x) = 0.95

2cos(2θ)+2cos(θ)+2=3cos(θ)

2 cos (2 θ) + 2 cos (θ) + 2 = 3 cos (θ)

1+csc(x)=cot^2(x)

1 + csc (x) = cot^{2} (x)

cos(1/(4θ))=(-sqrt(2))/2

cos (\frac{1}{4 θ}) = \frac{- 2}{2}