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受欢迎的三角函数 >

4cos(2θ)+11sin(θ)-4=3

解答

4 cos (2 θ) + 11 sin (θ) - 4 = 3

解答

θ = 0.38439 \dots + 2 πn, θ = π - 0.38439 \dots + 2 πn, θ = \frac{π}{2} + 2 πn

+1

度数

θ = 22.02431 \dots^{\circ} + 36 0^{\circ} n, θ = 157.97568 \dots^{\circ} + 36 0^{\circ} n, θ = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

4 cos (2 θ) + 11 sin (θ) - 4 = 3

两边减去

3

4 cos (2 θ) + 11 sin (θ) - 7 = 0

使用三角恒等式改写

- 7 + 11 sin (θ) + 4 cos (2 θ)

使用倍角公式:

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

= - 7 + 11 sin (θ) + 4 (1 - 2 sin^{2} (θ))

化简

- 7 + 11 sin (θ) + 4 (1 - 2 sin^{2} (θ)) : 11 sin (θ) - 8 sin^{2} (θ) - 3

- 7 + 11 sin (θ) + 4 (1 - 2 sin^{2} (θ))

乘开

4 (1 - 2 sin^{2} (θ)) : 4 - 8 sin^{2} (θ)

4 (1 - 2 sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = 2 sin^{2} (θ)

= 4 \cdot 1 - 4 \cdot 2 sin^{2} (θ)

化简

4 \cdot 1 - 4 \cdot 2 sin^{2} (θ) : 4 - 8 sin^{2} (θ)

4 \cdot 1 - 4 \cdot 2 sin^{2} (θ)

数字相乘:

4 \cdot 1 = 4

= 4 - 4 \cdot 2 sin^{2} (θ)

数字相乘:

4 \cdot 2 = 8

= 4 - 8 sin^{2} (θ)

= 4 - 8 sin^{2} (θ)

= - 7 + 11 sin (θ) + 4 - 8 sin^{2} (θ)

化简

- 7 + 11 sin (θ) + 4 - 8 sin^{2} (θ) : 11 sin (θ) - 8 sin^{2} (θ) - 3

- 7 + 11 sin (θ) + 4 - 8 sin^{2} (θ)

对同类项分组

= 11 sin (θ) - 8 sin^{2} (θ) - 7 + 4

数字相加/相减:

- 7 + 4 = - 3

= 11 sin (θ) - 8 sin^{2} (θ) - 3

= 11 sin (θ) - 8 sin^{2} (θ) - 3

= 11 sin (θ) - 8 sin^{2} (θ) - 3

- 3 + 11 sin (θ) - 8 sin^{2} (θ) = 0

用替代法求解

- 3 + 11 sin (θ) - 8 sin^{2} (θ) = 0

令

: sin (θ) = u

- 3 + 11 u - 8 u^{2} = 0

- 3 + 11 u - 8 u^{2} = 0 : u = \frac{3}{8}, u = 1

- 3 + 11 u - 8 u^{2} = 0

改写成标准形式

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

- 8 u^{2} + 11 u - 3 = 0

使用求根公式求解

- 8 u^{2} + 11 u - 3 = 0

二次方程求根公式:

若

a = - 8, b = 11, c = - 3

u_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 ( - 8 ) ( - 3 )}{2 ( - 8 )}

u_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 ( - 8 ) ( - 3 )}{2 ( - 8 )}

1 1^{2} - 4 (- 8) (- 3) = 5

1 1^{2} - 4 (- 8) (- 3)

使用法则

- (- a) = a

= 1 1^{2} - 4 \cdot 8 \cdot 3

数字相乘:

4 \cdot 8 \cdot 3 = 96

= 1 1^{2} - 96

1 1^{2} = 121

= 121 - 96

数字相减:

121 - 96 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 11 \pm 5}{2 ( - 8 )}

将解分隔开

u_{1} = \frac{- 11 + 5}{2 ( - 8 )}, u_{2} = \frac{- 11 - 5}{2 ( - 8 )}

u = \frac{- 11 + 5}{2 ( - 8 )} : \frac{3}{8}

\frac{- 11 + 5}{2 ( - 8 )}

去除括号:

(- a) = - a

= \frac{- 11 + 5}{- 2 \cdot 8}

数字相加/相减:

- 11 + 5 = - 6

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

数字相乘:

2 \cdot 8 = 16

= \frac{- 6}{- 16}

使用分式法则:

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

= \frac{6}{16}

约分:

2

= \frac{3}{8}

u = \frac{- 11 - 5}{2 ( - 8 )} : 1

\frac{- 11 - 5}{2 ( - 8 )}

去除括号:

(- a) = - a

= \frac{- 11 - 5}{- 2 \cdot 8}

数字相减:

- 11 - 5 = - 16

= \frac{- 16}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 16}{- 16}

使用分式法则:

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

= \frac{16}{16}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = \frac{3}{8}, u = 1

u = sin (θ)

代回

sin (θ) = \frac{3}{8}, sin (θ) = 1

sin (θ) = \frac{3}{8}, sin (θ) = 1

sin (θ) = \frac{3}{8} : θ = arcsin (\frac{3}{8}) + 2 πn, θ = π - arcsin (\frac{3}{8}) + 2 πn

sin (θ) = \frac{3}{8}

使用反三角函数性质

sin (θ) = \frac{3}{8}

sin (θ) = \frac{3}{8}

的通解

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

θ = arcsin (\frac{3}{8}) + 2 πn, θ = π - arcsin (\frac{3}{8}) + 2 πn

θ = arcsin (\frac{3}{8}) + 2 πn, θ = π - arcsin (\frac{3}{8}) + 2 πn

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = 1

sin (θ) = 1

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

合并所有解

θ = arcsin (\frac{3}{8}) + 2 πn, θ = π - arcsin (\frac{3}{8}) + 2 πn, θ = \frac{π}{2} + 2 πn

以小数形式表示解

θ = 0.38439 \dots + 2 πn, θ = π - 0.38439 \dots + 2 πn, θ = \frac{π}{2} + 2 πn

作图

流行的例子

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cos(θ)=1.1111111

sin(5θ)-17=-18