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受欢迎的三角函数 >

tan(2θ)+tan(2θ)=tan(4θ)

解答

tan (2 θ) + tan (2 θ) = tan (4 θ)

解答

θ = \frac{πn}{2}

+1

度数

θ = 0^{\circ} + 9 0^{\circ} n

求解步骤

tan (2 θ) + tan (2 θ) = tan (4 θ)

两边减去

tan (4 θ)

2 tan (2 θ) - tan (4 θ) = 0

令

: u = 2 θ

2 tan (u) - tan (2 u) = 0

使用三角恒等式改写

- tan (2 u) + 2 tan (u)

使用倍角公式:

tan (2 x) = \frac{2 tan ( x )}{1 - tan ^{2} ( x )}

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

化简

- \frac{2 tan ( u )}{1 - tan ^{2} ( u )} + 2 tan (u) : - \frac{2 tan ^{3} ( u )}{1 - tan ^{2} ( u )}

- \frac{2 tan ( u )}{1 - tan ^{2} ( u )} + 2 tan (u)

将项转换为分式:

2 tan (u) = \frac{2 tan ( u ) ( 1 - tan ^{2} ( u ) )}{1 - tan ^{2} ( u )}

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

乘开

- 2 tan (u) + 2 tan (u) (1 - tan^{2} (u)) : - 2 tan^{3} (u)

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

乘开

2 tan (u) (1 - tan^{2} (u)) : 2 tan (u) - 2 tan^{3} (u)

2 tan (u) (1 - tan^{2} (u))

使用分配律:

a (b - c) = ab - a c

a = 2 tan (u), b = 1, c = tan^{2} (u)

= 2 tan (u) \cdot 1 - 2 tan (u) tan^{2} (u)

= 2 \cdot 1 \cdot tan (u) - 2 tan^{2} (u) tan (u)

化简

2 \cdot 1 \cdot tan (u) - 2 tan^{2} (u) tan (u) : 2 tan (u) - 2 tan^{3} (u)

2 \cdot 1 \cdot tan (u) - 2 tan^{2} (u) tan (u)

2 \cdot 1 \cdot tan (u) = 2 tan (u)

2 \cdot 1 \cdot tan (u)

数字相乘:

2 \cdot 1 = 2

= 2 tan (u)

2 tan^{2} (u) tan (u) = 2 tan^{3} (u)

2 tan^{2} (u) tan (u)

使用指数法则:

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

tan^{2} (u) tan (u) = tan^{2 + 1} (u)

= 2 tan^{2 + 1} (u)

数字相加:

2 + 1 = 3

= 2 tan^{3} (u)

= 2 tan (u) - 2 tan^{3} (u)

= 2 tan (u) - 2 tan^{3} (u)

= - 2 tan (u) + 2 tan (u) - 2 tan^{3} (u)

同类项相加:

- 2 tan (u) + 2 tan (u) = 0

= - 2 tan^{3} (u)

= \frac{- 2 tan ^{3} ( u )}{1 - tan ^{2} ( u )}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2 tan ^{3} ( u )}{1 - tan ^{2} ( u )}

= - \frac{2 tan ^{3} ( u )}{1 - tan ^{2} ( u )}

- \frac{2 tan ^{3} ( u )}{1 - tan ^{2} ( u )} = 0

用替代法求解

- \frac{2 tan ^{3} ( u )}{1 - tan ^{2} ( u )} = 0

令

: tan (u) = u

- \frac{2 u ^{3}}{1 - u ^{2}} = 0

- \frac{2 u ^{3}}{1 - u ^{2}} = 0 : u = 0

- \frac{2 u ^{3}}{1 - u ^{2}} = 0

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

2 u^{3} = 0

解

2 u^{3} = 0 : u = 0

2 u^{3} = 0

两边除以

2

2 u^{3} = 0

两边除以

2

2 u^{3} = 0

两边除以

2

\frac{2 u ^{3}}{2} = \frac{0}{2}

化简

u^{3} = 0

u^{3} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

u = 0

u = 0

验证解

找到无定义的点（奇点）:

u = 1, u = - 1

取

- \frac{2 u ^{3}}{1 - u ^{2}}

的分母，令其等于零

解

1 - u^{2} = 0 : u = 1, u = - 1

1 - u^{2} = 0

将

1

到右边

1 - u^{2} = 0

两边减去

1

1 - u^{2} - 1 = 0 - 1

化简

- u^{2} = - 1

- u^{2} = - 1

两边除以

- 1

- u^{2} = - 1

两边除以

- 1

\frac{- u ^{2}}{- 1} = \frac{- 1}{- 1}

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

使用根式运算法则:

1 = 1

= 1

- 1 = - 1

- 1

使用根式运算法则:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

以下点无定义

u = 1, u = - 1

将不在定义域的点与解相综合:

u = 0

u = tan (u)

代回

tan (u) = 0

tan (u) = 0

tan (u) = 0 : u = πn

tan (u) = 0

tan (u) = 0

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}

u = 0 + πn

u = 0 + πn

解

u = 0 + πn : u = πn

u = 0 + πn

0 + πn = πn

u = πn

u = πn

合并所有解

u = πn

u = 2 θ

代回

2 θ = πn : θ = \frac{πn}{2}

2 θ = πn

两边除以

2

2 θ = πn

两边除以

2

\frac{2 θ}{2} = \frac{πn}{2}

化简

θ = \frac{πn}{2}

θ = \frac{πn}{2}

θ = \frac{πn}{2}

作图

流行的例子

1 - 7 cos (θ) = 8

tanh (x) = \frac{12}{13}

sin (θ) = \frac{5}{7}

8sin^2(x)+2sin(x)-1=0

8 sin^{2} (x) + 2 sin (x) - 1 = 0

2sin^2(x)=5sin(x)-3

2 sin^{2} (x) = 5 sin (x) - 3