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受欢迎的三角函数 >

cos(θ)-1=cos(2θ)

解答

cos (θ) - 1 = cos (2 θ)

解答

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

+1

度数

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

求解步骤

cos (θ) - 1 = cos (2 θ)

两边减去

cos (2 θ)

cos (θ) - 1 - cos (2 θ) = 0

使用三角恒等式改写

- 1 - cos (2 θ) + cos (θ)

使用倍角公式:

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

= - 1 - (2 cos^{2} (θ) - 1) + cos (θ)

化简

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

- 1 - (2 cos^{2} (θ) - 1) + cos (θ)

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

- (2 cos^{2} (θ) - 1)

打开括号

= - (2 cos^{2} (θ)) - (- 1)

使用加减运算法则

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

= - 2 cos^{2} (θ) + 1

= - 1 - 2 cos^{2} (θ) + 1 + cos (θ)

化简

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

- 1 - 2 cos^{2} (θ) + 1 + cos (θ)

对同类项分组

= - 2 cos^{2} (θ) + cos (θ) - 1 + 1

- 1 + 1 = 0

= cos (θ) - 2 cos^{2} (θ)

= cos (θ) - 2 cos^{2} (θ)

= cos (θ) - 2 cos^{2} (θ)

cos (θ) - 2 cos^{2} (θ) = 0

用替代法求解

cos (θ) - 2 cos^{2} (θ) = 0

令

: cos (θ) = u

u - 2 u^{2} = 0

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

u - 2 u^{2} = 0

改写成标准形式

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

- 2 u^{2} + u = 0

使用求根公式求解

- 2 u^{2} + u = 0

二次方程求根公式:

若

a = - 2, b = 1, c = 0

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

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

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

1^{2} - 4 (- 2) \cdot 0

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 0

使用法则

- (- a) = a

= 1 + 4 \cdot 2 \cdot 0

使用法则

0 \cdot a = 0

= 1 + 0

数字相加:

1 + 0 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

u = \frac{- 1 + 1}{2 ( - 2 )} : 0

\frac{- 1 + 1}{2 ( - 2 )}

去除括号:

(- a) = - a

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

数字相加/相减:

- 1 + 1 = 0

= \frac{0}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{0}{- 4}

使用分式法则:

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

= - \frac{0}{4}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

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

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

去除括号:

(- a) = - a

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

数字相减:

- 1 - 1 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

使用分式法则:

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

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

u = 0, u = \frac{1}{2}

u = cos (θ)

代回

cos (θ) = 0, cos (θ) = \frac{1}{2}

cos (θ) = 0, cos (θ) = \frac{1}{2}

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

cos (θ) = 0

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

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

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

cos (θ) = \frac{1}{2}

cos (θ) = \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}

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

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

合并所有解

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

作图

流行的例子

5tan(x)=9sin(x)

5 tan (x) = 9 sin (x)

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8tan(θ/2)+8cos(θ)tan(θ/2)=1

8 tan (\frac{θ}{2}) + 8 cos (θ) tan (\frac{θ}{2}) = 1

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