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 >

(1+4cos(θ))^2=(sqrt(3)sin(θ))^2

解答

(1 + 4 cos (θ))^{2} = (3 sin (θ))^{2}

解答

θ = 1.39363 \dots + 2 πn, θ = 2 π - 1.39363 \dots + 2 πn, θ = 2.21091 \dots + 2 πn, θ = - 2.21091 \dots + 2 πn

+1

度数

θ = 79.84945 \dots^{\circ} + 36 0^{\circ} n, θ = 280.15054 \dots^{\circ} + 36 0^{\circ} n, θ = 126.67590 \dots^{\circ} + 36 0^{\circ} n, θ = - 126.67590 \dots^{\circ} + 36 0^{\circ} n

求解步骤

(1 + 4 cos (θ))^{2} = (3 sin (θ))^{2}

两边减去

(3 sin (θ))^{2}

(1 + 4 cos (θ))^{2} - 3 sin^{2} (θ) = 0

使用三角恒等式改写

(1 + 4 cos (θ))^{2} - 3 sin^{2} (θ)

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 + 4 cos (θ))^{2} - 3 (1 - cos^{2} (θ))

化简

(1 + 4 cos (θ))^{2} - 3 (1 - cos^{2} (θ)) : 19 cos^{2} (θ) + 8 cos (θ) - 2

(1 + 4 cos (θ))^{2} - 3 (1 - cos^{2} (θ))

(1 + 4 cos (θ))^{2} : 1 + 8 cos (θ) + 16 cos^{2} (θ)

使用完全平方公式:

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

a = 1, b = 4 cos (θ)

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

化简

1^{2} + 2 \cdot 1 \cdot 4 cos (θ) + (4 cos (θ))^{2} : 1 + 8 cos (θ) + 16 cos^{2} (θ)

1^{2} + 2 \cdot 1 \cdot 4 cos (θ) + (4 cos (θ))^{2}

使用法则

1^{a} = 1

1^{2} = 1

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

2 \cdot 1 \cdot 4 cos (θ) = 8 cos (θ)

2 \cdot 1 \cdot 4 cos (θ)

数字相乘:

2 \cdot 1 \cdot 4 = 8

= 8 cos (θ)

(4 cos (θ))^{2} = 16 cos^{2} (θ)

(4 cos (θ))^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 4^{2} cos^{2} (θ)

4^{2} = 16

= 16 cos^{2} (θ)

= 1 + 8 cos (θ) + 16 cos^{2} (θ)

= 1 + 8 cos (θ) + 16 cos^{2} (θ)

= 1 + 8 cos (θ) + 16 cos^{2} (θ) - 3 (1 - cos^{2} (θ))

乘开

- 3 (1 - cos^{2} (θ)) : - 3 + 3 cos^{2} (θ)

- 3 (1 - cos^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 3, b = 1, c = cos^{2} (θ)

= - 3 \cdot 1 - (- 3) cos^{2} (θ)

使用加减运算法则

- (- a) = a

= - 3 \cdot 1 + 3 cos^{2} (θ)

数字相乘:

3 \cdot 1 = 3

= - 3 + 3 cos^{2} (θ)

= 1 + 8 cos (θ) + 16 cos^{2} (θ) - 3 + 3 cos^{2} (θ)

化简

1 + 8 cos (θ) + 16 cos^{2} (θ) - 3 + 3 cos^{2} (θ) : 19 cos^{2} (θ) + 8 cos (θ) - 2

1 + 8 cos (θ) + 16 cos^{2} (θ) - 3 + 3 cos^{2} (θ)

对同类项分组

= 8 cos (θ) + 16 cos^{2} (θ) + 3 cos^{2} (θ) + 1 - 3

同类项相加:

16 cos^{2} (θ) + 3 cos^{2} (θ) = 19 cos^{2} (θ)

= 8 cos (θ) + 19 cos^{2} (θ) + 1 - 3

数字相加/相减:

1 - 3 = - 2

= 19 cos^{2} (θ) + 8 cos (θ) - 2

= 19 cos^{2} (θ) + 8 cos (θ) - 2

= 19 cos^{2} (θ) + 8 cos (θ) - 2

- 2 + 19 cos^{2} (θ) + 8 cos (θ) = 0

用替代法求解

- 2 + 19 cos^{2} (θ) + 8 cos (θ) = 0

令

: cos (θ) = u

- 2 + 19 u^{2} + 8 u = 0

- 2 + 19 u^{2} + 8 u = 0 : u = \frac{- 4 + 3 6}{19}, u = - \frac{4 + 3 6}{19}

- 2 + 19 u^{2} + 8 u = 0

改写成标准形式

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

19 u^{2} + 8 u - 2 = 0

使用求根公式求解

19 u^{2} + 8 u - 2 = 0

二次方程求根公式:

若

a = 19, b = 8, c = - 2

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 19 ( - 2 )}{2 \cdot 19}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 19 ( - 2 )}{2 \cdot 19}

8^{2} - 4 \cdot 19 (- 2) = 66

8^{2} - 4 \cdot 19 (- 2)

使用法则

- (- a) = a

= 8^{2} + 4 \cdot 19 \cdot 2

数字相乘:

4 \cdot 19 \cdot 2 = 152

= 8^{2} + 152

8^{2} = 64

= 64 + 152

数字相加:

64 + 152 = 216

= 216

216

质因数分解:

2^{3} \cdot 3^{3}

216

216

除以

2216 = 108 \cdot 2

= 2 \cdot 108

108

除以

2108 = 54 \cdot 2

= 2 \cdot 2 \cdot 54

54

除以

254 = 27 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 27

27

除以

327 = 9 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 9

9

除以

39 = 3 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3

= 2^{3} \cdot 3^{3}

= 2^{3} \cdot 3^{3}

使用指数法则:

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

= 2^{2} \cdot 3^{2} \cdot 2 \cdot 3

使用根式运算法则:

n ab = n a n b

= 2^{2} 3^{2} 2 \cdot 3

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 3^{2} 2 \cdot 3

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 2 \cdot 3 2 \cdot 3

整理后得

= 66

u_{1, 2} = \frac{- 8 \pm 6 6}{2 \cdot 19}

将解分隔开

u_{1} = \frac{- 8 + 6 6}{2 \cdot 19}, u_{2} = \frac{- 8 - 6 6}{2 \cdot 19}

u = \frac{- 8 + 6 6}{2 \cdot 19} : \frac{- 4 + 3 6}{19}

\frac{- 8 + 6 6}{2 \cdot 19}

数字相乘:

2 \cdot 19 = 38

= \frac{- 8 + 6 6}{38}

分解

- 8 + 66 : 2 (- 4 + 36)

- 8 + 66

改写为

= - 2 \cdot 4 + 2 \cdot 36

因式分解出通项

2

= 2 (- 4 + 36)

= \frac{2 ( - 4 + 3 6 )}{38}

约分:

2

= \frac{- 4 + 3 6}{19}

u = \frac{- 8 - 6 6}{2 \cdot 19} : - \frac{4 + 3 6}{19}

\frac{- 8 - 6 6}{2 \cdot 19}

数字相乘:

2 \cdot 19 = 38

= \frac{- 8 - 6 6}{38}

分解

- 8 - 66 : - 2 (4 + 36)

- 8 - 66

改写为

= - 2 \cdot 4 - 2 \cdot 36

因式分解出通项

2

= - 2 (4 + 36)

= - \frac{2 ( 4 + 3 6 )}{38}

约分:

2

= - \frac{4 + 3 6}{19}

二次方程组的解是:

u = \frac{- 4 + 3 6}{19}, u = - \frac{4 + 3 6}{19}

u = cos (θ)

代回

cos (θ) = \frac{- 4 + 3 6}{19}, cos (θ) = - \frac{4 + 3 6}{19}

cos (θ) = \frac{- 4 + 3 6}{19}, cos (θ) = - \frac{4 + 3 6}{19}

cos (θ) = \frac{- 4 + 3 6}{19} : θ = arccos (\frac{- 4 + 3 6}{19}) + 2 πn, θ = 2 π - arccos (\frac{- 4 + 3 6}{19}) + 2 πn

cos (θ) = \frac{- 4 + 3 6}{19}

使用反三角函数性质

cos (θ) = \frac{- 4 + 3 6}{19}

cos (θ) = \frac{- 4 + 3 6}{19}

的通解

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

θ = arccos (\frac{- 4 + 3 6}{19}) + 2 πn, θ = 2 π - arccos (\frac{- 4 + 3 6}{19}) + 2 πn

θ = arccos (\frac{- 4 + 3 6}{19}) + 2 πn, θ = 2 π - arccos (\frac{- 4 + 3 6}{19}) + 2 πn

cos (θ) = - \frac{4 + 3 6}{19} : θ = arccos (- \frac{4 + 3 6}{19}) + 2 πn, θ = - arccos (- \frac{4 + 3 6}{19}) + 2 πn

cos (θ) = - \frac{4 + 3 6}{19}

使用反三角函数性质

cos (θ) = - \frac{4 + 3 6}{19}

cos (θ) = - \frac{4 + 3 6}{19}

的通解

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

θ = arccos (- \frac{4 + 3 6}{19}) + 2 πn, θ = - arccos (- \frac{4 + 3 6}{19}) + 2 πn

θ = arccos (- \frac{4 + 3 6}{19}) + 2 πn, θ = - arccos (- \frac{4 + 3 6}{19}) + 2 πn

合并所有解

θ = arccos (\frac{- 4 + 3 6}{19}) + 2 πn, θ = 2 π - arccos (\frac{- 4 + 3 6}{19}) + 2 πn, θ = arccos (- \frac{4 + 3 6}{19}) + 2 πn, θ = - arccos (- \frac{4 + 3 6}{19}) + 2 πn

以小数形式表示解

θ = 1.39363 \dots + 2 πn, θ = 2 π - 1.39363 \dots + 2 πn, θ = 2.21091 \dots + 2 πn, θ = - 2.21091 \dots + 2 πn

作图

流行的例子

\frac{1}{tan ( x )} = 4

sin^2(x)+2-3sin(x)=cos^2(x)

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

4sin^2(θ)-4cos(θ)=1

4 sin^{2} (θ) - 4 cos (θ) = 1

(33)/(sin(x))=(29)/(sin(100))

\frac{33}{sin ( x )} = \frac{29}{sin ( 10 0 ^{\circ} )}

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

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