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受欢迎的三角函数 >

cot(u)-csc(u)=(sin(u))/(1+cos(u))

解答

cot (u) - csc (u) = \frac{sin ( u )}{1 + cos ( u )}

解答

u \in R 无解

求解步骤

cot (u) - csc (u) = \frac{sin ( u )}{1 + cos ( u )}

两边减去

\frac{sin ( u )}{1 + cos ( u )}

cot (u) - csc (u) - \frac{sin ( u )}{1 + cos ( u )} = 0

化简

cot (u) - csc (u) - \frac{sin ( u )}{1 + cos ( u )} : \frac{cot ( u ) ( 1 + cos ( u ) ) - csc ( u ) ( 1 + cos ( u ) ) - sin ( u )}{1 + cos ( u )}

cot (u) - csc (u) - \frac{sin ( u )}{1 + cos ( u )}

将项转换为分式:

cot (u) = \frac{cot ( u ) ( 1 + cos ( u ) )}{1 + cos ( u )}, csc (u) = \frac{csc ( u ) ( 1 + cos ( u ) )}{1 + cos ( u )}

= \frac{cot ( u ) ( 1 + cos ( u ) )}{1 + cos ( u )} - \frac{csc ( u ) ( 1 + cos ( u ) )}{1 + cos ( u )} - \frac{sin ( u )}{1 + cos ( u )}

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{cot ( u ) ( 1 + cos ( u ) ) - csc ( u ) ( 1 + cos ( u ) ) - sin ( u )}{1 + cos ( u )}

\frac{cot ( u ) ( 1 + cos ( u ) ) - csc ( u ) ( 1 + cos ( u ) ) - sin ( u )}{1 + cos ( u )} = 0

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

cot (u) (1 + cos (u)) - csc (u) (1 + cos (u)) - sin (u) = 0

用 sin, cos 表示

- sin (u) + (1 + cos (u)) cot (u) - (1 + cos (u)) csc (u)

使用基本三角恒等式:

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

= - sin (u) + (1 + cos (u)) \frac{cos ( u )}{sin ( u )} - (1 + cos (u)) csc (u)

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

= - sin (u) + (1 + cos (u)) \frac{cos ( u )}{sin ( u )} - (1 + cos (u)) \frac{1}{sin ( u )}

化简

- sin (u) + (1 + cos (u)) \frac{cos ( u )}{sin ( u )} - (1 + cos (u)) \frac{1}{sin ( u )} : \frac{- sin ^{2} ( u ) + cos ^{2} ( u ) - 1}{sin ( u )}

- sin (u) + (1 + cos (u)) \frac{cos ( u )}{sin ( u )} - (1 + cos (u)) \frac{1}{sin ( u )}

(1 + cos (u)) \frac{cos ( u )}{sin ( u )} = \frac{cos ( u ) ( 1 + cos ( u ) )}{sin ( u )}

(1 + cos (u)) \frac{cos ( u )}{sin ( u )}

分式相乘:

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

= \frac{cos ( u ) ( 1 + cos ( u ) )}{sin ( u )}

(1 + cos (u)) \frac{1}{sin ( u )} = \frac{1 + cos ( u )}{sin ( u )}

(1 + cos (u)) \frac{1}{sin ( u )}

分式相乘:

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

= \frac{1 \cdot ( 1 + cos ( u ) )}{sin ( u )}

1 \cdot (1 + cos (u)) = 1 + cos (u)

1 \cdot (1 + cos (u))

乘以:

1 \cdot (1 + cos (u)) = (1 + cos (u))

= (1 + cos (u))

去除括号:

(a) = a

= 1 + cos (u)

= \frac{1 + cos ( u )}{sin ( u )}

= - sin (u) + \frac{cos ( u ) ( cos ( u ) + 1 )}{sin ( u )} - \frac{cos ( u ) + 1}{sin ( u )}

合并分式

\frac{cos ( u ) ( cos ( u ) + 1 )}{sin ( u )} - \frac{cos ( u ) + 1}{sin ( u )} : \frac{cos ( u ) ( 1 + cos ( u ) ) - ( 1 + cos ( u ) )}{sin ( u )}

使用法则

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

= \frac{cos ( u ) ( cos ( u ) + 1 ) - ( cos ( u ) + 1 )}{sin ( u )}

= - sin (u) + \frac{cos ( u ) ( cos ( u ) + 1 ) - ( cos ( u ) + 1 )}{sin ( u )}

乘开

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

cos (u) (1 + cos (u)) - (1 + cos (u))

乘开

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

cos (u) (1 + cos (u))

使用分配律:

a (b + c) = ab + a c

a = cos (u), b = 1, c = cos (u)

= cos (u) \cdot 1 + cos (u) cos (u)

= 1 \cdot cos (u) + cos (u) cos (u)

化简

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

1 \cdot cos (u) + cos (u) cos (u)

1 \cdot cos (u) = cos (u)

1 \cdot cos (u)

乘以:

1 \cdot cos (u) = cos (u)

= cos (u)

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

cos (u) cos (u)

使用指数法则:

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

cos (u) cos (u) = cos^{1 + 1} (u)

= cos^{1 + 1} (u)

数字相加:

1 + 1 = 2

= cos^{2} (u)

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

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

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

- (1 + cos (u)) : - 1 - cos (u)

- (1 + cos (u))

打开括号

= - (1) - (cos (u))

使用加减运算法则

+ (- a) = - a

= - 1 - cos (u)

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

化简

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

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

对同类项分组

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

同类项相加:

cos (u) - cos (u) = 0

= cos^{2} (u) - 1

= cos^{2} (u) - 1

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

将项转换为分式:

sin (u) = \frac{sin ( u ) sin ( u )}{sin ( u )}

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

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

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

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

sin (u) sin (u) = sin^{2} (u)

sin (u) sin (u)

使用指数法则:

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

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

= sin^{1 + 1} (u)

数字相加:

1 + 1 = 2

= sin^{2} (u)

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

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

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

\frac{- 1 + cos ^{2} ( u ) - sin ^{2} ( u )}{sin ( u )} = 0

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

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

使用三角恒等式改写

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

使用倍角公式:

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

= - 1 + cos (2 u)

- 1 + cos (2 u) = 0

将

1

到右边

- 1 + cos (2 u) = 0

两边加上

1

- 1 + cos (2 u) + 1 = 0 + 1

化简

cos (2 u) = 1

cos (2 u) = 1

cos (2 u) = 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}

2 u = 0 + 2 πn

2 u = 0 + 2 πn

解

2 u = 0 + 2 πn : u = πn

2 u = 0 + 2 πn

0 + 2 πn = 2 πn

2 u = 2 πn

两边除以

2

2 u = 2 πn

两边除以

2

\frac{2 u}{2} = \frac{2 πn}{2}

化简

u = πn

u = πn

u = πn

因为方程对以下值无定义:

πn

u \in R 无解

作图

流行的例子

sin(x)=(sqrt(5))/3

sin (x) = \frac{5}{3}

2cos(2θ)-1=0,0<= θ<= 2pi

2 cos (2 θ) - 1 = 0, 0 \leq θ \leq 2 π

6cos(x)+6sin(x)tan(x)=12,0<= x<= 2pi

6 cos (x) + 6 sin (x) tan (x) = 12, 0 \leq x \leq 2 π

6 ∣ tan (x) ∣ = 6

7 sin (2 x) = - 2