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受欢迎的三角函数 >

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

解答

\frac{cos ( x ) - cos ^{2} ( x )}{sin ( x )} = sin (x) cos (x)

解答

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

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

\frac{cos ( x ) - cos ^{2} ( x )}{sin ( x )} = sin (x) cos (x)

两边减去

sin (x) cos (x)

\frac{cos ( x ) - cos ^{2} ( x )}{sin ( x )} - sin (x) cos (x) = 0

化简

\frac{cos ( x ) - cos ^{2} ( x )}{sin ( x )} - sin (x) cos (x) : \frac{cos ( x ) - cos ^{2} ( x ) - sin ^{2} ( x ) cos ( x )}{sin ( x )}

\frac{cos ( x ) - cos ^{2} ( x )}{sin ( x )} - sin (x) cos (x)

将项转换为分式:

sin (x) cos (x) = \frac{sin ( x ) cos ( x ) sin ( x )}{sin ( x )}

= \frac{cos ( x ) - cos ^{2} ( x )}{sin ( x )} - \frac{sin ( x ) cos ( x ) sin ( x )}{sin ( x )}

因为分母相等，所以合并分式:

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

= \frac{cos ( x ) - cos ^{2} ( x ) - sin ( x ) cos ( x ) sin ( x )}{sin ( x )}

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

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

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

sin (x) cos (x) sin (x)

使用指数法则:

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

sin (x) sin (x) = sin^{1 + 1} (x)

= cos (x) sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

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

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

= \frac{cos ( x ) - cos ^{2} ( x ) - sin ^{2} ( x ) cos ( x )}{sin ( x )}

\frac{cos ( x ) - cos ^{2} ( x ) - sin ^{2} ( x ) cos ( x )}{sin ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cos (x) - cos^{2} (x) - sin^{2} (x) cos (x) = 0

分解

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

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

使用指数法则:

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

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

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

因式分解出通项

cos (x)

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

cos (x) (1 - cos (x) - sin^{2} (x)) = 0

分别求解每个部分

cos (x) = 0 or 1 - cos (x) - sin^{2} (x) = 0

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

cos (x) = 0

cos (x) = 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}

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

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

1 - cos (x) - sin^{2} (x) = 0 : x = 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

1 - cos (x) - sin^{2} (x) = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

- cos (x) + cos^{2} (x) = 0

用替代法求解

- cos (x) + cos^{2} (x) = 0

令

: cos (x) = u

- u + u^{2} = 0

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

- u + u^{2} = 0

改写成标准形式

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

u^{2} - u = 0

使用求根公式求解

u^{2} - u = 0

二次方程求根公式:

若

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

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

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

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

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 0

= 1 - 0

数字相减:

1 - 0 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

1 + 1 = 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

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

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

使用法则

- (- a) = a

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

数字相减:

1 - 1 = 0

= \frac{0}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{2}

使用法则

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

= 0

二次方程组的解是:

u = 1, u = 0

u = cos (x)

代回

cos (x) = 1, cos (x) = 0

cos (x) = 1, cos (x) = 0

cos (x) = 1 : x = 2 πn

cos (x) = 1

cos (x) = 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}

x = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

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

cos (x) = 0

cos (x) = 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}

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

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

合并所有解

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

合并所有解

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

因为方程对以下值无定义:

2 πn

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

作图

流行的例子

1 = - cot^{2} (x) + 2

60*pi/(180)=2arctan(x/(17))

60 \cdot \frac{π}{180} = 2 arctan (\frac{x}{17})

sin(a)=-1/8 ,(3pi)/2 <= a<= 2pi

sin (a) = - \frac{1}{8}, \frac{3 π}{2} \leq a \leq 2 π

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

3 cos^{2} (x) - 23 cos (x) = sin^{2} (x)

sin(x)+cos(x)= 3/2

sin (x) + cos (x) = \frac{3}{2}