zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

7(1-cos(θ))=sin^2(θ)

解答

7 (1 - cos (θ)) = sin^{2} (θ)

解答

θ = 2 πn

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n

求解步骤

7 (1 - cos (θ)) = sin^{2} (θ)

两边减去

sin^{2} (θ)

7 (1 - cos (θ)) - sin^{2} (θ) = 0

使用三角恒等式改写

- sin^{2} (θ) + (1 - cos (θ)) \cdot 7

使用毕达哥拉斯恒等式:

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

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

= - (1 - cos^{2} (θ)) + (1 - cos (θ)) \cdot 7

化简

- (1 - cos^{2} (θ)) + (1 - cos (θ)) \cdot 7 : cos^{2} (θ) - 7 cos (θ) + 6

- (1 - cos^{2} (θ)) + (1 - cos (θ)) \cdot 7

= - (1 - cos^{2} (θ)) + 7 (1 - cos (θ))

- (1 - cos^{2} (θ)) : - 1 + cos^{2} (θ)

- (1 - cos^{2} (θ))

打开括号

= - (1) - (- cos^{2} (θ))

使用加减运算法则

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

= - 1 + cos^{2} (θ)

= - 1 + cos^{2} (θ) + (1 - cos (θ)) \cdot 7

乘开

7 (1 - cos (θ)) : 7 - 7 cos (θ)

7 (1 - cos (θ))

使用分配律:

a (b - c) = ab - a c

a = 7, b = 1, c = cos (θ)

= 7 \cdot 1 - 7 cos (θ)

数字相乘:

7 \cdot 1 = 7

= 7 - 7 cos (θ)

= - 1 + cos^{2} (θ) + 7 - 7 cos (θ)

化简

- 1 + cos^{2} (θ) + 7 - 7 cos (θ) : cos^{2} (θ) - 7 cos (θ) + 6

- 1 + cos^{2} (θ) + 7 - 7 cos (θ)

对同类项分组

= cos^{2} (θ) - 7 cos (θ) - 1 + 7

数字相加/相减:

- 1 + 7 = 6

= cos^{2} (θ) - 7 cos (θ) + 6

= cos^{2} (θ) - 7 cos (θ) + 6

= cos^{2} (θ) - 7 cos (θ) + 6

6 + cos^{2} (θ) - 7 cos (θ) = 0

用替代法求解

6 + cos^{2} (θ) - 7 cos (θ) = 0

令

: cos (θ) = u

6 + u^{2} - 7 u = 0

6 + u^{2} - 7 u = 0 : u = 6, u = 1

6 + u^{2} - 7 u = 0

改写成标准形式

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

u^{2} - 7 u + 6 = 0

使用求根公式求解

u^{2} - 7 u + 6 = 0

二次方程求根公式:

若

a = 1, b = - 7, c = 6

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot 6 = 5

(- 7)^{2} - 4 \cdot 1 \cdot 6

使用指数法则:

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

若

n

是偶数

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 1 \cdot 6

数字相乘:

4 \cdot 1 \cdot 6 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

数字相减:

49 - 24 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 7 ) \pm 5}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 7 ) + 5}{2 \cdot 1}, u_{2} = \frac{- ( - 7 ) - 5}{2 \cdot 1}

u = \frac{- ( - 7 ) + 5}{2 \cdot 1} : 6

\frac{- ( - 7 ) + 5}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{7 + 5}{2 \cdot 1}

数字相加:

7 + 5 = 12

= \frac{12}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{12}{2}

数字相除:

\frac{12}{2} = 6

= 6

u = \frac{- ( - 7 ) - 5}{2 \cdot 1} : 1

\frac{- ( - 7 ) - 5}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{7 - 5}{2 \cdot 1}

数字相减:

7 - 5 = 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = 6, u = 1

u = cos (θ)

代回

cos (θ) = 6, cos (θ) = 1

cos (θ) = 6, cos (θ) = 1

cos (θ) = 6 :

无解

cos (θ) = 6

- 1 \leq cos (x) \leq 1

无解

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

cos (θ) = 1

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = 0 + 2 πn

θ = 0 + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

合并所有解

θ = 2 πn

作图

流行的例子

2cos(x)-sin^2(x)+2=0

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

sinh (x) = 2

cos (2 t) = \frac{1}{2}

5sqrt(2)sin(θ)+4=-1

52 sin (θ) + 4 = - 1

7 sin^{2} (x) - 1 = 0