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受欢迎的三角函数 >

cot(11x+1)=tan(6x+4)

解答

cot (11 x + 1) = tan (6 x + 4)

解答

x = - \frac{5}{17} + \frac{π}{34} + \frac{2 πn}{17}, x = - \frac{5}{17} + \frac{3 π}{34} + \frac{2 πn}{17}

+1

度数

x = - 11.55758 \dots^{\circ} + 21.17647 \dots^{\circ} n, x = - 0.96934 \dots^{\circ} + 21.17647 \dots^{\circ} n

求解步骤

cot (11 x + 1) = tan (6 x + 4)

两边减去

tan (6 x + 4)

cot (11 x + 1) - tan (6 x + 4) = 0

用 sin, cos 表示

cot (1 + 11 x) - tan (4 + 6 x)

使用基本三角恒等式:

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

= \frac{cos ( 1 + 11 x )}{sin ( 1 + 11 x )} - tan (4 + 6 x)

使用基本三角恒等式:

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

= \frac{cos ( 1 + 11 x )}{sin ( 1 + 11 x )} - \frac{sin ( 4 + 6 x )}{cos ( 4 + 6 x )}

化简

\frac{cos ( 1 + 11 x )}{sin ( 1 + 11 x )} - \frac{sin ( 4 + 6 x )}{cos ( 4 + 6 x )} : \frac{cos ( 1 + 11 x ) cos ( 6 x + 4 ) - sin ( 4 + 6 x ) sin ( 11 x + 1 )}{sin ( 11 x + 1 ) cos ( 6 x + 4 )}

\frac{cos ( 1 + 11 x )}{sin ( 1 + 11 x )} - \frac{sin ( 4 + 6 x )}{cos ( 4 + 6 x )}

sin (1 + 11 x), cos (4 + 6 x)

的最小公倍数:

sin (11 x + 1) cos (6 x + 4)

sin (1 + 11 x), cos (4 + 6 x)

最小公倍数 (LCM)

计算出由出现在

sin (1 + 11 x)

或

cos (4 + 6 x)

中的因子组成的表达式

= sin (11 x + 1) cos (6 x + 4)

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 (11 x + 1) cos (6 x + 4)

对于

\frac{cos ( 1 + 11 x )}{sin ( 1 + 11 x )} :

将分母和分子乘以

cos (6 x + 4)

\frac{cos ( 1 + 11 x )}{sin ( 1 + 11 x )} = \frac{cos ( 1 + 11 x ) cos ( 6 x + 4 )}{sin ( 1 + 11 x ) cos ( 6 x + 4 )}

对于

\frac{sin ( 4 + 6 x )}{cos ( 4 + 6 x )} :

将分母和分子乘以

sin (11 x + 1)

\frac{sin ( 4 + 6 x )}{cos ( 4 + 6 x )} = \frac{sin ( 4 + 6 x ) sin ( 11 x + 1 )}{cos ( 4 + 6 x ) sin ( 11 x + 1 )}

= \frac{cos ( 1 + 11 x ) cos ( 6 x + 4 )}{sin ( 1 + 11 x ) cos ( 6 x + 4 )} - \frac{sin ( 4 + 6 x ) sin ( 11 x + 1 )}{cos ( 4 + 6 x ) sin ( 11 x + 1 )}

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 ( 1 + 11 x ) cos ( 6 x + 4 ) - sin ( 4 + 6 x ) sin ( 11 x + 1 )}{sin ( 11 x + 1 ) cos ( 6 x + 4 )}

= \frac{cos ( 1 + 11 x ) cos ( 6 x + 4 ) - sin ( 4 + 6 x ) sin ( 11 x + 1 )}{sin ( 11 x + 1 ) cos ( 6 x + 4 )}

\frac{cos ( 1 + 11 x ) cos ( 4 + 6 x ) - sin ( 1 + 11 x ) sin ( 4 + 6 x )}{cos ( 4 + 6 x ) sin ( 1 + 11 x )} = 0

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

cos (1 + 11 x) cos (4 + 6 x) - sin (1 + 11 x) sin (4 + 6 x) = 0

使用三角恒等式改写

cos (1 + 11 x) cos (4 + 6 x) - sin (1 + 11 x) sin (4 + 6 x)

使用角和恒等式:

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

= cos (1 + 11 x + 4 + 6 x)

cos (1 + 11 x + 4 + 6 x) = 0

cos (1 + 11 x + 4 + 6 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}

1 + 11 x + 4 + 6 x = \frac{π}{2} + 2 πn, 1 + 11 x + 4 + 6 x = \frac{3 π}{2} + 2 πn

1 + 11 x + 4 + 6 x = \frac{π}{2} + 2 πn, 1 + 11 x + 4 + 6 x = \frac{3 π}{2} + 2 πn

解

1 + 11 x + 4 + 6 x = \frac{π}{2} + 2 πn : x = - \frac{5}{17} + \frac{π}{34} + \frac{2 πn}{17}

1 + 11 x + 4 + 6 x = \frac{π}{2} + 2 πn

对同类项分组

11 x + 6 x + 1 + 4 = \frac{π}{2} + 2 πn

同类项相加:

11 x + 6 x = 17 x

17 x + 1 + 4 = \frac{π}{2} + 2 πn

数字相加:

1 + 4 = 5

17 x + 5 = \frac{π}{2} + 2 πn

将

5

到右边

17 x + 5 = \frac{π}{2} + 2 πn

两边减去

5

17 x + 5 - 5 = \frac{π}{2} + 2 πn - 5

化简

17 x = \frac{π}{2} + 2 πn - 5

17 x = \frac{π}{2} + 2 πn - 5

两边除以

17

17 x = \frac{π}{2} + 2 πn - 5

两边除以

17

\frac{17 x}{17} = \frac{\frac{π}{2}}{17} + \frac{2 πn}{17} - \frac{5}{17}

化简

\frac{17 x}{17} = \frac{\frac{π}{2}}{17} + \frac{2 πn}{17} - \frac{5}{17}

化简

\frac{17 x}{17} : x

\frac{17 x}{17}

数字相除:

\frac{17}{17} = 1

= x

化简

\frac{\frac{π}{2}}{17} + \frac{2 πn}{17} - \frac{5}{17} : - \frac{5}{17} + \frac{π}{34} + \frac{2 πn}{17}

\frac{\frac{π}{2}}{17} + \frac{2 πn}{17} - \frac{5}{17}

对同类项分组

= - \frac{5}{17} + \frac{2 πn}{17} + \frac{\frac{π}{2}}{17}

\frac{\frac{π}{2}}{17} = \frac{π}{34}

\frac{\frac{π}{2}}{17}

使用分式法则:

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

= \frac{π}{2 \cdot 17}

数字相乘:

2 \cdot 17 = 34

= \frac{π}{34}

= - \frac{5}{17} + \frac{2 πn}{17} + \frac{π}{34}

对同类项分组

= - \frac{5}{17} + \frac{π}{34} + \frac{2 πn}{17}

x = - \frac{5}{17} + \frac{π}{34} + \frac{2 πn}{17}

x = - \frac{5}{17} + \frac{π}{34} + \frac{2 πn}{17}

x = - \frac{5}{17} + \frac{π}{34} + \frac{2 πn}{17}

解

1 + 11 x + 4 + 6 x = \frac{3 π}{2} + 2 πn : x = - \frac{5}{17} + \frac{3 π}{34} + \frac{2 πn}{17}

1 + 11 x + 4 + 6 x = \frac{3 π}{2} + 2 πn

对同类项分组

11 x + 6 x + 1 + 4 = \frac{3 π}{2} + 2 πn

同类项相加:

11 x + 6 x = 17 x

17 x + 1 + 4 = \frac{3 π}{2} + 2 πn

数字相加:

1 + 4 = 5

17 x + 5 = \frac{3 π}{2} + 2 πn

将

5

到右边

17 x + 5 = \frac{3 π}{2} + 2 πn

两边减去

5

17 x + 5 - 5 = \frac{3 π}{2} + 2 πn - 5

化简

17 x = \frac{3 π}{2} + 2 πn - 5

17 x = \frac{3 π}{2} + 2 πn - 5

两边除以

17

17 x = \frac{3 π}{2} + 2 πn - 5

两边除以

17

\frac{17 x}{17} = \frac{\frac{3 π}{2}}{17} + \frac{2 πn}{17} - \frac{5}{17}

化简

\frac{17 x}{17} = \frac{\frac{3 π}{2}}{17} + \frac{2 πn}{17} - \frac{5}{17}

化简

\frac{17 x}{17} : x

\frac{17 x}{17}

数字相除:

\frac{17}{17} = 1

= x

化简

\frac{\frac{3 π}{2}}{17} + \frac{2 πn}{17} - \frac{5}{17} : - \frac{5}{17} + \frac{3 π}{34} + \frac{2 πn}{17}

\frac{\frac{3 π}{2}}{17} + \frac{2 πn}{17} - \frac{5}{17}

对同类项分组

= - \frac{5}{17} + \frac{2 πn}{17} + \frac{\frac{3 π}{2}}{17}

\frac{\frac{3 π}{2}}{17} = \frac{3 π}{34}

\frac{\frac{3 π}{2}}{17}

使用分式法则:

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

= \frac{3 π}{2 \cdot 17}

数字相乘:

2 \cdot 17 = 34

= \frac{3 π}{34}

= - \frac{5}{17} + \frac{2 πn}{17} + \frac{3 π}{34}

对同类项分组

= - \frac{5}{17} + \frac{3 π}{34} + \frac{2 πn}{17}

x = - \frac{5}{17} + \frac{3 π}{34} + \frac{2 πn}{17}

x = - \frac{5}{17} + \frac{3 π}{34} + \frac{2 πn}{17}

x = - \frac{5}{17} + \frac{3 π}{34} + \frac{2 πn}{17}

x = - \frac{5}{17} + \frac{π}{34} + \frac{2 πn}{17}, x = - \frac{5}{17} + \frac{3 π}{34} + \frac{2 πn}{17}

作图

流行的例子

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2 sin (2 x) = cos (2 x + 3 0^{\circ})

tan (θ) = \frac{15}{25}

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sin(x)= 1/(sqrt(2)),0<= x<= 2pi

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π

27^2=56^2+35^2-(2*56*35*cos(x))

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