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受欢迎的三角函数 >

11(1-cos(θ))=sin^2(θ)

解答

11 (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

11 (1 - cos (θ)) = sin^{2} (θ)

两边减去

sin^{2} (θ)

11 (1 - cos (θ)) - sin^{2} (θ) = 0

使用三角恒等式改写

- sin^{2} (θ) + (1 - cos (θ)) \cdot 11

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 11

化简

- (1 - cos^{2} (θ)) + (1 - cos (θ)) \cdot 11 : cos^{2} (θ) - 11 cos (θ) + 10

- (1 - cos^{2} (θ)) + (1 - cos (θ)) \cdot 11

= - (1 - cos^{2} (θ)) + 11 (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 11

乘开

11 (1 - cos (θ)) : 11 - 11 cos (θ)

11 (1 - cos (θ))

使用分配律:

a (b - c) = ab - a c

a = 11, b = 1, c = cos (θ)

= 11 \cdot 1 - 11 cos (θ)

数字相乘:

11 \cdot 1 = 11

= 11 - 11 cos (θ)

= - 1 + cos^{2} (θ) + 11 - 11 cos (θ)

化简

- 1 + cos^{2} (θ) + 11 - 11 cos (θ) : cos^{2} (θ) - 11 cos (θ) + 10

- 1 + cos^{2} (θ) + 11 - 11 cos (θ)

对同类项分组

= cos^{2} (θ) - 11 cos (θ) - 1 + 11

数字相加/相减:

- 1 + 11 = 10

= cos^{2} (θ) - 11 cos (θ) + 10

= cos^{2} (θ) - 11 cos (θ) + 10

= cos^{2} (θ) - 11 cos (θ) + 10

10 + cos^{2} (θ) - 11 cos (θ) = 0

用替代法求解

10 + cos^{2} (θ) - 11 cos (θ) = 0

令

: cos (θ) = u

10 + u^{2} - 11 u = 0

10 + u^{2} - 11 u = 0 : u = 10, u = 1

10 + u^{2} - 11 u = 0

改写成标准形式

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

u^{2} - 11 u + 10 = 0

使用求根公式求解

u^{2} - 11 u + 10 = 0

二次方程求根公式:

若

a = 1, b = - 11, c = 10

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

(- 11)^{2} - 4 \cdot 1 \cdot 10 = 9

(- 11)^{2} - 4 \cdot 1 \cdot 10

使用指数法则:

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

若

n

是偶数

(- 11)^{2} = 1 1^{2}

= 1 1^{2} - 4 \cdot 1 \cdot 10

数字相乘:

4 \cdot 1 \cdot 10 = 40

= 1 1^{2} - 40

1 1^{2} = 121

= 121 - 40

数字相减:

121 - 40 = 81

= 81

因式分解数字:

81 = 9^{2}

= 9^{2}

使用根式运算法则:

n a^{n} = a

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 11 ) \pm 9}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 11 ) + 9}{2 \cdot 1}, u_{2} = \frac{- ( - 11 ) - 9}{2 \cdot 1}

u = \frac{- ( - 11 ) + 9}{2 \cdot 1} : 10

\frac{- ( - 11 ) + 9}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{11 + 9}{2 \cdot 1}

数字相加:

11 + 9 = 20

= \frac{20}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{20}{2}

数字相除:

\frac{20}{2} = 10

= 10

u = \frac{- ( - 11 ) - 9}{2 \cdot 1} : 1

\frac{- ( - 11 ) - 9}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{11 - 9}{2 \cdot 1}

数字相减:

11 - 9 = 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 = 10, u = 1

u = cos (θ)

代回

cos (θ) = 10, cos (θ) = 1

cos (θ) = 10, cos (θ) = 1

cos (θ) = 10 :

无解

cos (θ) = 10

- 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

cos^2(x)-1/2 =0,0<= x<= 2pi

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so l v e f or x, tan (x) = tan (y)

4 cos (θ) + 3 = 0

2cos^2(x)+1=3cos(x),0<= x<= 2pi

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