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受欢迎的三角函数 >

cot(2t+5)=tan(3t-15)

解答

cot (2 t + 5) = tan (3 t - 15)

解答

t = 2 + \frac{2 πn}{5} + \frac{π}{10}, t = 2 + \frac{2 πn}{5} + \frac{3 π}{10}

+1

度数

t = 132.59155 \dots^{\circ} + 7 2^{\circ} n, t = 168.59155 \dots^{\circ} + 7 2^{\circ} n

求解步骤

cot (2 t + 5) = tan (3 t - 15)

两边减去

tan (3 t - 15)

cot (2 t + 5) - tan (3 t - 15) = 0

用 sin, cos 表示

cot (5 + 2 t) - tan (- 15 + 3 t)

使用基本三角恒等式:

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

= \frac{cos ( 5 + 2 t )}{sin ( 5 + 2 t )} - tan (- 15 + 3 t)

使用基本三角恒等式:

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

= \frac{cos ( 5 + 2 t )}{sin ( 5 + 2 t )} - \frac{sin ( - 15 + 3 t )}{cos ( - 15 + 3 t )}

化简

\frac{cos ( 5 + 2 t )}{sin ( 5 + 2 t )} - \frac{sin ( - 15 + 3 t )}{cos ( - 15 + 3 t )} : \frac{cos ( 5 + 2 t ) cos ( 3 t - 15 ) - sin ( - 15 + 3 t ) sin ( 2 t + 5 )}{sin ( 2 t + 5 ) cos ( 3 t - 15 )}

\frac{cos ( 5 + 2 t )}{sin ( 5 + 2 t )} - \frac{sin ( - 15 + 3 t )}{cos ( - 15 + 3 t )}

sin (5 + 2 t), cos (- 15 + 3 t)

的最小公倍数:

sin (2 t + 5) cos (3 t - 15)

sin (5 + 2 t), cos (- 15 + 3 t)

最小公倍数 (LCM)

计算出由出现在

sin (5 + 2 t)

或

cos (- 15 + 3 t)

中的因子组成的表达式

= sin (2 t + 5) cos (3 t - 15)

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 (2 t + 5) cos (3 t - 15)

对于

\frac{cos ( 5 + 2 t )}{sin ( 5 + 2 t )} :

将分母和分子乘以

cos (3 t - 15)

\frac{cos ( 5 + 2 t )}{sin ( 5 + 2 t )} = \frac{cos ( 5 + 2 t ) cos ( 3 t - 15 )}{sin ( 5 + 2 t ) cos ( 3 t - 15 )}

对于

\frac{sin ( - 15 + 3 t )}{cos ( - 15 + 3 t )} :

将分母和分子乘以

sin (2 t + 5)

\frac{sin ( - 15 + 3 t )}{cos ( - 15 + 3 t )} = \frac{sin ( - 15 + 3 t ) sin ( 2 t + 5 )}{cos ( - 15 + 3 t ) sin ( 2 t + 5 )}

= \frac{cos ( 5 + 2 t ) cos ( 3 t - 15 )}{sin ( 5 + 2 t ) cos ( 3 t - 15 )} - \frac{sin ( - 15 + 3 t ) sin ( 2 t + 5 )}{cos ( - 15 + 3 t ) sin ( 2 t + 5 )}

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 ( 5 + 2 t ) cos ( 3 t - 15 ) - sin ( - 15 + 3 t ) sin ( 2 t + 5 )}{sin ( 2 t + 5 ) cos ( 3 t - 15 )}

= \frac{cos ( 5 + 2 t ) cos ( 3 t - 15 ) - sin ( - 15 + 3 t ) sin ( 2 t + 5 )}{sin ( 2 t + 5 ) cos ( 3 t - 15 )}

\frac{cos ( - 15 + 3 t ) cos ( 5 + 2 t ) - sin ( - 15 + 3 t ) sin ( 5 + 2 t )}{cos ( - 15 + 3 t ) sin ( 5 + 2 t )} = 0

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

cos (- 15 + 3 t) cos (5 + 2 t) - sin (- 15 + 3 t) sin (5 + 2 t) = 0

使用三角恒等式改写

cos (- 15 + 3 t) cos (5 + 2 t) - sin (- 15 + 3 t) sin (5 + 2 t)

使用角和恒等式:

cos (s) cos (t) - sin (s) sin (t) = cos (s + t)

= cos (- 15 + 3 t + 5 + 2 t)

cos (- 15 + 3 t + 5 + 2 t) = 0

cos (- 15 + 3 t + 5 + 2 t) = 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}

- 15 + 3 t + 5 + 2 t = \frac{π}{2} + 2 πn, - 15 + 3 t + 5 + 2 t = \frac{3 π}{2} + 2 πn

- 15 + 3 t + 5 + 2 t = \frac{π}{2} + 2 πn, - 15 + 3 t + 5 + 2 t = \frac{3 π}{2} + 2 πn

解

- 15 + 3 t + 5 + 2 t = \frac{π}{2} + 2 πn : t = 2 + \frac{2 πn}{5} + \frac{π}{10}

- 15 + 3 t + 5 + 2 t = \frac{π}{2} + 2 πn

对同类项分组

3 t + 2 t - 15 + 5 = \frac{π}{2} + 2 πn

同类项相加:

3 t + 2 t = 5 t

5 t - 15 + 5 = \frac{π}{2} + 2 πn

数字相加/相减:

- 15 + 5 = - 10

5 t - 10 = \frac{π}{2} + 2 πn

将

10

到右边

5 t - 10 = \frac{π}{2} + 2 πn

两边加上

10

5 t - 10 + 10 = \frac{π}{2} + 2 πn + 10

化简

5 t = \frac{π}{2} + 2 πn + 10

5 t = \frac{π}{2} + 2 πn + 10

两边除以

5

5 t = \frac{π}{2} + 2 πn + 10

两边除以

5

\frac{5 t}{5} = \frac{\frac{π}{2}}{5} + \frac{2 πn}{5} + \frac{10}{5}

化简

\frac{5 t}{5} = \frac{\frac{π}{2}}{5} + \frac{2 πn}{5} + \frac{10}{5}

化简

\frac{5 t}{5} : t

\frac{5 t}{5}

数字相除:

\frac{5}{5} = 1

= t

化简

\frac{\frac{π}{2}}{5} + \frac{2 πn}{5} + \frac{10}{5} : 2 + \frac{2 πn}{5} + \frac{π}{10}

\frac{\frac{π}{2}}{5} + \frac{2 πn}{5} + \frac{10}{5}

对同类项分组

= \frac{10}{5} + \frac{2 πn}{5} + \frac{\frac{π}{2}}{5}

\frac{10}{5} = 2

\frac{10}{5}

数字相除:

\frac{10}{5} = 2

= 2

\frac{\frac{π}{2}}{5} = \frac{π}{10}

\frac{\frac{π}{2}}{5}

使用分式法则:

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

= \frac{π}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{π}{10}

= 2 + \frac{2 πn}{5} + \frac{π}{10}

t = 2 + \frac{2 πn}{5} + \frac{π}{10}

t = 2 + \frac{2 πn}{5} + \frac{π}{10}

t = 2 + \frac{2 πn}{5} + \frac{π}{10}

解

- 15 + 3 t + 5 + 2 t = \frac{3 π}{2} + 2 πn : t = 2 + \frac{2 πn}{5} + \frac{3 π}{10}

- 15 + 3 t + 5 + 2 t = \frac{3 π}{2} + 2 πn

对同类项分组

3 t + 2 t - 15 + 5 = \frac{3 π}{2} + 2 πn

同类项相加:

3 t + 2 t = 5 t

5 t - 15 + 5 = \frac{3 π}{2} + 2 πn

数字相加/相减:

- 15 + 5 = - 10

5 t - 10 = \frac{3 π}{2} + 2 πn

将

10

到右边

5 t - 10 = \frac{3 π}{2} + 2 πn

两边加上

10

5 t - 10 + 10 = \frac{3 π}{2} + 2 πn + 10

化简

5 t = \frac{3 π}{2} + 2 πn + 10

5 t = \frac{3 π}{2} + 2 πn + 10

两边除以

5

5 t = \frac{3 π}{2} + 2 πn + 10

两边除以

5

\frac{5 t}{5} = \frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5} + \frac{10}{5}

化简

\frac{5 t}{5} = \frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5} + \frac{10}{5}

化简

\frac{5 t}{5} : t

\frac{5 t}{5}

数字相除:

\frac{5}{5} = 1

= t

化简

\frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5} + \frac{10}{5} : 2 + \frac{2 πn}{5} + \frac{3 π}{10}

\frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5} + \frac{10}{5}

对同类项分组

= \frac{10}{5} + \frac{2 πn}{5} + \frac{\frac{3 π}{2}}{5}

\frac{10}{5} = 2

\frac{10}{5}

数字相除:

\frac{10}{5} = 2

= 2

\frac{\frac{3 π}{2}}{5} = \frac{3 π}{10}

\frac{\frac{3 π}{2}}{5}

使用分式法则:

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

= \frac{3 π}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{3 π}{10}

= 2 + \frac{2 πn}{5} + \frac{3 π}{10}

t = 2 + \frac{2 πn}{5} + \frac{3 π}{10}

t = 2 + \frac{2 πn}{5} + \frac{3 π}{10}

t = 2 + \frac{2 πn}{5} + \frac{3 π}{10}

t = 2 + \frac{2 πn}{5} + \frac{π}{10}, t = 2 + \frac{2 πn}{5} + \frac{3 π}{10}

作图

流行的例子

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