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受欢迎的三角函数 >

cos(x)-sin(x)= 1/((sin(x)))-1/((cos(x)))

解答

cos (x) - sin (x) = \frac{1}{( sin ( x ) )} - \frac{1}{( cos ( x ) )}

解答

x = \frac{π}{4} + πn

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

cos (x) - sin (x) = \frac{1}{( sin ( x ) )} - \frac{1}{( cos ( x ) )}

两边减去

\frac{1}{sin ( x )} - \frac{1}{cos ( x )}

cos (x) - sin (x) - \frac{1}{sin ( x )} + \frac{1}{cos ( x )} = 0

化简

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

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

将项转换为分式:

cos (x) = \frac{cos ( x )}{1}, sin (x) = \frac{sin ( x )}{1}

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

1, 1, sin (x), cos (x)

的最小公倍数:

sin (x) cos (x)

1, 1, sin (x), cos (x)

最小公倍数 (LCM)

1, 1

的最小公倍数:

1

1, 1

最小公倍数 (LCM)

1

质因数分解

1

质因数分解

将每个因子乘以它在

1

或

1

中出现的最多次数

= 1

数字相乘:

1 = 1

= 1

计算出由至少在以下一个因式表达式中出现的因子组成的表达式

= sin (x) cos (x)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

sin (x) cos (x)

对于

\frac{cos ( x )}{1} :

将分母和分子乘以

sin (x) cos (x)

\frac{cos ( x )}{1} = \frac{cos ( x ) sin ( x ) cos ( x )}{1 \cdot sin ( x ) cos ( x )} = \frac{cos ^{2} ( x ) sin ( x )}{sin ( x ) cos ( x )}

对于

\frac{sin ( x )}{1} :

将分母和分子乘以

sin (x) cos (x)

\frac{sin ( x )}{1} = \frac{sin ( x ) sin ( x ) cos ( x )}{1 \cdot sin ( x ) cos ( x )} = \frac{sin ^{2} ( x ) cos ( x )}{sin ( x ) cos ( x )}

对于

\frac{1}{sin ( x )} :

将分母和分子乘以

cos (x)

\frac{1}{sin ( x )} = \frac{1 \cdot cos ( x )}{sin ( x ) cos ( x )} = \frac{cos ( x )}{sin ( x ) cos ( x )}

对于

\frac{1}{cos ( x )} :

将分母和分子乘以

sin (x)

\frac{1}{cos ( x )} = \frac{1 \cdot sin ( x )}{cos ( x ) sin ( x )} = \frac{sin ( x )}{sin ( x ) cos ( x )}

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

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

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

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

分解

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

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

分解

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

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

因式分解出通项

cos (x)

= cos (x) (cos (x) sin (x) - 1)

分解

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

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

使用指数法则:

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

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

= - sin (x) sin (x) cos (x) + sin (x)

因式分解出通项

sin (x)

= sin (x) (- sin (x) cos (x) + 1)

= cos (x) (cos (x) sin (x) - 1) + sin (x) (- sin (x) cos (x) + 1)

改写为

= (- 1 + cos (x) sin (x)) cos (x) - (- 1 + cos (x) sin (x)) sin (x)

因式分解出通项

(- 1 + cos (x) sin (x))

= (- 1 + cos (x) sin (x)) (cos (x) - sin (x))

(- 1 + cos (x) sin (x)) (cos (x) - sin (x)) = 0

分别求解每个部分

- 1 + cos (x) sin (x) = 0 or cos (x) - sin (x) = 0

- 1 + cos (x) sin (x) = 0 :

无解

- 1 + cos (x) sin (x) = 0

使用三角恒等式改写

- 1 + cos (x) sin (x)

使用倍角公式:

2 sin (x) cos (x) = sin (2 x)

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

= - 1 + \frac{sin ( 2 x )}{2}

- 1 + \frac{sin ( 2 x )}{2} = 0

将

1

到右边

- 1 + \frac{sin ( 2 x )}{2} = 0

两边加上

1

- 1 + \frac{sin ( 2 x )}{2} + 1 = 0 + 1

化简

\frac{sin ( 2 x )}{2} = 1

\frac{sin ( 2 x )}{2} = 1

在两边乘以

2

\frac{sin ( 2 x )}{2} = 1

在两边乘以

2

\frac{2 sin ( 2 x )}{2} = 1 \cdot 2

化简

sin (2 x) = 2

sin (2 x) = 2

- 1 \leq sin (x) \leq 1

无解

cos (x) - sin (x) = 0 : x = \frac{π}{4} + πn

cos (x) - sin (x) = 0

使用三角恒等式改写

cos (x) - sin (x) = 0

在两边除以

cos (x), cos (x) \neq = 0

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

化简

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

使用基本三角恒等式:

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

1 - tan (x) = 0

1 - tan (x) = 0

将

1

到右边

1 - tan (x) = 0

两边减去

1

1 - tan (x) - 1 = 0 - 1

化简

- tan (x) = - 1

- tan (x) = - 1

两边除以

- 1

- tan (x) = - 1

两边除以

- 1

\frac{- tan ( x )}{- 1} = \frac{- 1}{- 1}

化简

tan (x) = 1

tan (x) = 1

tan (x) = 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

合并所有解

x = \frac{π}{4} + πn

作图

流行的例子

sin^2(x)+cos^5(x)=2

sin^{2} (x) + cos^{5} (x) = 2

16=4+9-12cos(x)

16 = 4 + 9 - 12 cos (x)

tanh (z) + 2 = 0

cos^2(a)= 2/(3sin(a))

cos^{2} (a) = \frac{2}{3 sin ( a )}

5 cos (5 x) = 2