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受欢迎的三角函数 >

4cos(θ/2)=3+2cos(θ)

解答

4 cos (\frac{θ}{2}) = 3 + 2 cos (θ)

解答

θ = \frac{2 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

+1

度数

θ = 12 0^{\circ} + 72 0^{\circ} n, θ = 60 0^{\circ} + 72 0^{\circ} n

求解步骤

4 cos (\frac{θ}{2}) = 3 + 2 cos (θ)

两边减去

3 + 2 cos (θ)

4 cos (\frac{θ}{2}) - 3 - 2 cos (θ) = 0

令

: u = \frac{θ}{2}

4 cos (u) - 3 - 2 cos (2 u) = 0

使用三角恒等式改写

- 3 - 2 cos (2 u) + 4 cos (u)

使用倍角公式:

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

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

化简

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

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

乘开

- 2 (2 cos^{2} (u) - 1) : - 4 cos^{2} (u) + 2

- 2 (2 cos^{2} (u) - 1)

使用分配律:

a (b - c) = ab - a c

a = - 2, b = 2 cos^{2} (u), c = 1

= - 2 \cdot 2 cos^{2} (u) - (- 2) \cdot 1

使用加减运算法则

- (- a) = a

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

化简

- 2 \cdot 2 cos^{2} (u) + 2 \cdot 1 : - 4 cos^{2} (u) + 2

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

数字相乘:

2 \cdot 2 = 4

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

数字相乘:

2 \cdot 1 = 2

= - 4 cos^{2} (u) + 2

= - 4 cos^{2} (u) + 2

= - 3 - 4 cos^{2} (u) + 2 + 4 cos (u)

化简

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

- 3 - 4 cos^{2} (u) + 2 + 4 cos (u)

对同类项分组

= - 4 cos^{2} (u) + 4 cos (u) - 3 + 2

数字相加/相减:

- 3 + 2 = - 1

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

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

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

- 1 + 4 cos (u) - 4 cos^{2} (u) = 0

用替代法求解

- 1 + 4 cos (u) - 4 cos^{2} (u) = 0

令

: cos (u) = u

- 1 + 4 u - 4 u^{2} = 0

- 1 + 4 u - 4 u^{2} = 0 : u = \frac{1}{2}

- 1 + 4 u - 4 u^{2} = 0

改写成标准形式

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

- 4 u^{2} + 4 u - 1 = 0

使用求根公式求解

- 4 u^{2} + 4 u - 1 = 0

二次方程求根公式:

若

a = - 4, b = 4, c = - 1

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

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

4^{2} - 4 (- 4) (- 1) = 0

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

数字相减:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- 4 \pm 0}{2 ( - 4 )}

u = \frac{- 4}{2 ( - 4 )}

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

\frac{- 4}{2 ( - 4 )}

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

使用分式法则:

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

= \frac{4}{8}

约分:

4

= \frac{1}{2}

u = \frac{1}{2}

二次方程组的解是:

u = \frac{1}{2}

u = cos (u)

代回

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

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

cos (u) = \frac{1}{2} : u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

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

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

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}

u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

合并所有解

u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

u = \frac{θ}{2}

代回

\frac{θ}{2} = \frac{π}{3} + 2 πn : θ = \frac{2 π}{3} + 4 πn

\frac{θ}{2} = \frac{π}{3} + 2 πn

在两边乘以

2

\frac{θ}{2} = \frac{π}{3} + 2 πn

在两边乘以

2

\frac{2 θ}{2} = 2 \cdot \frac{π}{3} + 2 \cdot 2 πn

化简

\frac{2 θ}{2} = 2 \cdot \frac{π}{3} + 2 \cdot 2 πn

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

2 \cdot \frac{π}{3} + 2 \cdot 2 πn : \frac{2 π}{3} + 4 πn

2 \cdot \frac{π}{3} + 2 \cdot 2 πn

乘

2 \cdot \frac{π}{3} : \frac{2 π}{3}

2 \cdot \frac{π}{3}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{3}

= \frac{2 π}{3} + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= \frac{2 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn

\frac{θ}{2} = \frac{5 π}{3} + 2 πn : θ = \frac{10 π}{3} + 4 πn

\frac{θ}{2} = \frac{5 π}{3} + 2 πn

在两边乘以

2

\frac{θ}{2} = \frac{5 π}{3} + 2 πn

在两边乘以

2

\frac{2 θ}{2} = 2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn

化简

\frac{2 θ}{2} = 2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn : \frac{10 π}{3} + 4 πn

2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn

2 \cdot \frac{5 π}{3} = \frac{10 π}{3}

2 \cdot \frac{5 π}{3}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{5 π 2}{3}

数字相乘:

5 \cdot 2 = 10

= \frac{10 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{10 π}{3} + 4 πn

θ = \frac{10 π}{3} + 4 πn

θ = \frac{10 π}{3} + 4 πn

θ = \frac{10 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

作图

流行的例子

tan(θ+10)=cot(2θ-10)

tan (θ + 1 0^{\circ}) = cot (2 θ - 10)

4sin^2(x)=5-2cos(x)

4 sin^{2} (x) = 5 - 2 cos (x)

sin (x + \frac{π}{6}) = - 1

sec (x) = \frac{4}{3}

4cos(x)+3cos(x)=5

4 cos (x) + 3 cos (x) = 5