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受欢迎的三角函数 >

1/(tan(α))+tan(α)= 1/(sin(α))

解答

\frac{1}{tan ( α )} + tan (α) = \frac{1}{sin ( α )}

解答

α \in R 无解

求解步骤

\frac{1}{tan ( α )} + tan (α) = \frac{1}{sin ( α )}

两边减去

\frac{1}{sin ( α )}

\frac{1}{tan ( α )} + tan (α) - \frac{1}{sin ( α )} = 0

化简

\frac{1}{tan ( α )} + tan (α) - \frac{1}{sin ( α )} : \frac{sin ( α ) + tan ^{2} ( α ) sin ( α ) - tan ( α )}{tan ( α ) sin ( α )}

\frac{1}{tan ( α )} + tan (α) - \frac{1}{sin ( α )}

将项转换为分式:

tan (α) = \frac{tan ( α )}{1}

= \frac{1}{tan ( α )} + \frac{tan ( α )}{1} - \frac{1}{sin ( α )}

tan (α), 1, sin (α)

的最小公倍数:

tan (α) sin (α)

tan (α), 1, sin (α)

最小公倍数 (LCM)

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

= tan (α) sin (α)

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

tan (α) sin (α)

对于

\frac{1}{tan ( α )} :

将分母和分子乘以

sin (α)

\frac{1}{tan ( α )} = \frac{1 \cdot sin ( α )}{tan ( α ) sin ( α )} = \frac{sin ( α )}{tan ( α ) sin ( α )}

对于

\frac{tan ( α )}{1} :

将分母和分子乘以

tan (α) sin (α)

\frac{tan ( α )}{1} = \frac{tan ( α ) tan ( α ) sin ( α )}{1 \cdot tan ( α ) sin ( α )} = \frac{tan ^{2} ( α ) sin ( α )}{tan ( α ) sin ( α )}

对于

\frac{1}{sin ( α )} :

将分母和分子乘以

tan (α)

\frac{1}{sin ( α )} = \frac{1 \cdot tan ( α )}{sin ( α ) tan ( α )} = \frac{tan ( α )}{tan ( α ) sin ( α )}

= \frac{sin ( α )}{tan ( α ) sin ( α )} + \frac{tan ^{2} ( α ) sin ( α )}{tan ( α ) sin ( α )} - \frac{tan ( α )}{tan ( α ) sin ( α )}

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 ( α ) + tan ^{2} ( α ) sin ( α ) - tan ( α )}{tan ( α ) sin ( α )}

\frac{sin ( α ) + tan ^{2} ( α ) sin ( α ) - tan ( α )}{tan ( α ) sin ( α )} = 0

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

sin (α) + tan^{2} (α) sin (α) - tan (α) = 0

用 sin, cos 表示

sin (α) - tan (α) + sin (α) tan^{2} (α)

使用基本三角恒等式:

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

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

化简

sin (α) - \frac{sin ( α )}{cos ( α )} + sin (α) (\frac{sin ( α )}{cos ( α )})^{2} : \frac{cos ^{2} ( α ) sin ( α ) - sin ( α ) cos ( α ) + sin ^{3} ( α )}{cos ^{2} ( α )}

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

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

sin (α) (\frac{sin ( α )}{cos ( α )})^{2}

(\frac{sin ( α )}{cos ( α )})^{2} = \frac{sin ^{2} ( α )}{cos ^{2} ( α )}

(\frac{sin ( α )}{cos ( α )})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{sin ^{2} ( α )}{cos ^{2} ( α )}

= \frac{sin ^{2} ( α )}{cos ^{2} ( α )} sin (α)

分式相乘:

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

= \frac{sin ^{2} ( α ) sin ( α )}{cos ^{2} ( α )}

sin^{2} (α) sin (α) = sin^{3} (α)

sin^{2} (α) sin (α)

使用指数法则:

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

sin^{2} (α) sin (α) = sin^{2 + 1} (α)

= sin^{2 + 1} (α)

数字相加:

2 + 1 = 3

= sin^{3} (α)

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

= sin (α) - \frac{sin ( α )}{cos ( α )} + \frac{sin ^{3} ( α )}{cos ^{2} ( α )}

将项转换为分式:

sin (α) = \frac{sin ( α )}{1}

= \frac{sin ( α )}{1} - \frac{sin ( α )}{cos ( α )} + \frac{sin ^{3} ( α )}{cos ^{2} ( α )}

1, cos (α), cos^{2} (α)

的最小公倍数:

cos^{2} (α)

1, cos (α), cos^{2} (α)

最小公倍数 (LCM)

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

= cos^{2} (α)

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

cos^{2} (α)

对于

\frac{sin ( α )}{1} :

将分母和分子乘以

cos^{2} (α)

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

对于

\frac{sin ( α )}{cos ( α )} :

将分母和分子乘以

cos (α)

\frac{sin ( α )}{cos ( α )} = \frac{sin ( α ) cos ( α )}{cos ( α ) cos ( α )} = \frac{sin ( α ) cos ( α )}{cos ^{2} ( α )}

= \frac{sin ( α ) cos ^{2} ( α )}{cos ^{2} ( α )} - \frac{sin ( α ) cos ( α )}{cos ^{2} ( α )} + \frac{sin ^{3} ( α )}{cos ^{2} ( α )}

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 ( α ) cos ^{2} ( α ) - sin ( α ) cos ( α ) + sin ^{3} ( α )}{cos ^{2} ( α )}

= \frac{cos ^{2} ( α ) sin ( α ) - sin ( α ) cos ( α ) + sin ^{3} ( α )}{cos ^{2} ( α )}

\frac{sin ^{3} ( α ) - cos ( α ) sin ( α ) + cos ^{2} ( α ) sin ( α )}{cos ^{2} ( α )} = 0

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

sin^{3} (α) - cos (α) sin (α) + cos^{2} (α) sin (α) = 0

分解

sin^{3} (α) - cos (α) sin (α) + cos^{2} (α) sin (α) : sin (α) (sin^{2} (α) - cos (α) + cos^{2} (α))

sin^{3} (α) - cos (α) sin (α) + cos^{2} (α) sin (α)

使用指数法则:

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

sin^{3} (α) = sin (α) sin^{2} (α)

= sin (α) sin^{2} (α) - sin (α) cos (α) + sin (α) cos^{2} (α)

因式分解出通项

sin (α)

= sin (α) (sin^{2} (α) - cos (α) + cos^{2} (α))

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

= sin (α) (- cos (α) + 1)

sin (α) (- cos (α) + 1) = 0

分别求解每个部分

sin (α) = 0 or - cos (α) + 1 = 0

sin (α) = 0 : α = 2 πn, α = π + 2 πn

sin (α) = 0

sin (α) = 0

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

α = 0 + 2 πn, α = π + 2 πn

α = 0 + 2 πn, α = π + 2 πn

解

α = 0 + 2 πn : α = 2 πn

α = 0 + 2 πn

0 + 2 πn = 2 πn

α = 2 πn

α = 2 πn, α = π + 2 πn

- cos (α) + 1 = 0 : α = 2 πn

- cos (α) + 1 = 0

将

1

到右边

- cos (α) + 1 = 0

两边减去

1

- cos (α) + 1 - 1 = 0 - 1

化简

- cos (α) = - 1

- cos (α) = - 1

两边除以

- 1

- cos (α) = - 1

两边除以

- 1

\frac{- cos ( α )}{- 1} = \frac{- 1}{- 1}

化简

cos (α) = 1

cos (α) = 1

cos (α) = 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}

α = 0 + 2 πn

α = 0 + 2 πn

解

α = 0 + 2 πn : α = 2 πn

α = 0 + 2 πn

0 + 2 πn = 2 πn

α = 2 πn

α = 2 πn

合并所有解

α = 2 πn, α = π + 2 πn

因为方程对以下值无定义:

2 πn, π + 2 πn

α \in R 无解

作图

流行的例子

8 sin^{2} (x) - 1 = 5

csc (θ) = \frac{17}{8}

cos(x)= 60/61 ,cos(2x)

cos (x) = \frac{60}{61}, cos (2 x)

cos (b) = 0.47

tan^2(x)-4sec(x)=4

tan^{2} (x) - 4 sec (x) = 4