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受欢迎的三角函数 >

tan(2t)+tan(t)=1-tan(2t)tan(t)

解答

tan (2 t) + tan (t) = 1 - tan (2 t) tan (t)

解答

t = 1.30899 \dots + πn, t = 0.26179 \dots + πn

+1

度数

t = 7 5^{\circ} + 18 0^{\circ} n, t = 1 5^{\circ} + 18 0^{\circ} n

求解步骤

tan (2 t) + tan (t) = 1 - tan (2 t) tan (t)

两边减去

1 - tan (2 t) tan (t)

tan (2 t) + tan (t) - 1 + tan (2 t) tan (t) = 0

使用三角恒等式改写

- 1 + tan (2 t) + tan (t) + tan (2 t) tan (t)

使用倍角公式:

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

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

化简

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

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

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

\frac{2 tan ( t )}{1 - tan ^{2} ( t )} tan (t)

分式相乘:

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

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

2 tan (t) tan (t) = 2 tan^{2} (t)

2 tan (t) tan (t)

使用指数法则:

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

tan (t) tan (t) = tan^{1 + 1} (t)

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

数字相加:

1 + 1 = 2

= 2 tan^{2} (t)

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

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

合并分式

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

使用法则

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

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

因式分解出通项

2 tan (t) : 2 tan (t) (tan (t) + 1)

2 tan^{2} (t) + 2 tan (t)

使用指数法则:

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

tan^{2} (t) = tan (t) tan (t)

= 2 tan (t) tan (t) + 2 tan (t)

因式分解出通项

2 tan (t)

= 2 tan (t) (tan (t) + 1)

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

分解

1 - tan^{2} (t) : - (tan (t) + 1) (tan (t) - 1)

1 - tan^{2} (t)

因式分解出通项

- 1

= - (tan^{2} (t) - 1)

分解

tan^{2} (t) - 1 : (tan (t) + 1) (tan (t) - 1)

tan^{2} (t) - 1

将

1

改写为

1^{2}

= tan^{2} (t) - 1^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

tan^{2} (t) - 1^{2} = (tan (t) + 1) (tan (t) - 1)

= (tan (t) + 1) (tan (t) - 1)

= - (tan (t) + 1) (tan (t) - 1)

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

约分:

tan (t) + 1

= - \frac{2 tan ( t )}{tan ( t ) - 1}

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

合并分式

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

- \frac{2 tan ( t )}{tan ( t ) - 1} + tan (t)

将项转换为分式:

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

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

乘开

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

- 2 tan (t) + tan (t) (tan (t) - 1)

乘开

tan (t) (tan (t) - 1) : tan^{2} (t) - tan (t)

tan (t) (tan (t) - 1)

使用分配律:

a (b - c) = ab - a c

a = tan (t), b = tan (t), c = 1

= tan (t) tan (t) - tan (t) \cdot 1

= tan (t) tan (t) - 1 \cdot tan (t)

化简

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

tan (t) tan (t) - 1 \cdot tan (t)

tan (t) tan (t) = tan^{2} (t)

tan (t) tan (t)

使用指数法则:

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

tan (t) tan (t) = tan^{1 + 1} (t)

= tan^{1 + 1} (t)

数字相加:

1 + 1 = 2

= tan^{2} (t)

1 \cdot tan (t) = tan (t)

1 \cdot tan (t)

乘以:

1 \cdot tan (t) = tan (t)

= tan (t)

= tan^{2} (t) - tan (t)

= tan^{2} (t) - tan (t)

= - 2 tan (t) + tan^{2} (t) - tan (t)

同类项相加:

- 2 tan (t) - tan (t) = - 3 tan (t)

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

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

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

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

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

用替代法求解

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

令

: tan (t) = u

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

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

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

在两边乘以

- 1 + u

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

在两边乘以

- 1 + u

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

化简

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

化简

- 1 \cdot (- 1 + u) : - (- 1 + u)

- 1 \cdot (- 1 + u)

乘以:

1 \cdot (- 1 + u) = (- 1 + u)

= - (u - 1)

化简

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

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

分式相乘:

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

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

约分:

- 1 + u

= u^{2} - 3 u

化简

0 \cdot (- 1 + u) : 0

0 \cdot (- 1 + u)

使用法则

0 \cdot a = 0

= 0

- (- 1 + u) + u^{2} - 3 u = 0

- (- 1 + u) + u^{2} - 3 u = 0

- (- 1 + u) + u^{2} - 3 u = 0

解

- (- 1 + u) + u^{2} - 3 u = 0 : u = 2 + 3, u = 2 - 3

- (- 1 + u) + u^{2} - 3 u = 0

展开

- (- 1 + u) + u^{2} - 3 u : u^{2} - 4 u + 1

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

- (- 1 + u) : 1 - u

- (- 1 + u)

打开括号

= - (- 1) - (u)

使用加减运算法则

- (- a) = a, - (a) = - a

= 1 - u

= 1 - u + u^{2} - 3 u

化简

1 - u + u^{2} - 3 u : u^{2} - 4 u + 1

1 - u + u^{2} - 3 u

对同类项分组

= u^{2} - u - 3 u + 1

同类项相加:

- u - 3 u = - 4 u

= u^{2} - 4 u + 1

= u^{2} - 4 u + 1

u^{2} - 4 u + 1 = 0

使用求根公式求解

u^{2} - 4 u + 1 = 0

二次方程求根公式:

若

a = 1, b = - 4, c = 1

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot 1 = 23

(- 4)^{2} - 4 \cdot 1 \cdot 1

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 4)^{2} = 4^{2}

= 4^{2} - 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4^{2} - 4

4^{2} = 16

= 16 - 4

数字相减:

16 - 4 = 12

= 12

12

质因数分解:

2^{2} \cdot 3

12

12

除以

212 = 6 \cdot 2

= 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

使用根式运算法则:

n ab = n a n b

= 3 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 23

u_{1, 2} = \frac{- ( - 4 ) \pm 2 3}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 4 ) + 2 3}{2 \cdot 1}, u_{2} = \frac{- ( - 4 ) - 2 3}{2 \cdot 1}

u = \frac{- ( - 4 ) + 2 3}{2 \cdot 1} : 2 + 3

\frac{- ( - 4 ) + 2 3}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{4 + 2 3}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4 + 2 3}{2}

分解

4 + 23 : 2 (2 + 3)

4 + 23

改写为

= 2 \cdot 2 + 23

因式分解出通项

2

= 2 (2 + 3)

= \frac{2 ( 2 + 3 )}{2}

数字相除:

\frac{2}{2} = 1

= 2 + 3

u = \frac{- ( - 4 ) - 2 3}{2 \cdot 1} : 2 - 3

\frac{- ( - 4 ) - 2 3}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{4 - 2 3}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4 - 2 3}{2}

分解

4 - 23 : 2 (2 - 3)

4 - 23

改写为

= 2 \cdot 2 - 23

因式分解出通项

2

= 2 (2 - 3)

= \frac{2 ( 2 - 3 )}{2}

数字相除:

\frac{2}{2} = 1

= 2 - 3

二次方程组的解是:

u = 2 + 3, u = 2 - 3

u = 2 + 3, u = 2 - 3

验证解

找到无定义的点（奇点）:

u = 1

取

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

的分母，令其等于零

解

- 1 + u = 0 : u = 1

- 1 + u = 0

将

1

到右边

- 1 + u = 0

两边加上

1

- 1 + u + 1 = 0 + 1

化简

u = 1

u = 1

以下点无定义

u = 1

将不在定义域的点与解相综合:

u = 2 + 3, u = 2 - 3

u = tan (t)

代回

tan (t) = 2 + 3, tan (t) = 2 - 3

tan (t) = 2 + 3, tan (t) = 2 - 3

tan (t) = 2 + 3 : t = arctan (2 + 3) + πn

tan (t) = 2 + 3

使用反三角函数性质

tan (t) = 2 + 3

tan (t) = 2 + 3

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

t = arctan (2 + 3) + πn

t = arctan (2 + 3) + πn

tan (t) = 2 - 3 : t = arctan (2 - 3) + πn

tan (t) = 2 - 3

使用反三角函数性质

tan (t) = 2 - 3

tan (t) = 2 - 3

的通解

tan (x) = a \Rightarrow x = arctan (a) + πn

t = arctan (2 - 3) + πn

t = arctan (2 - 3) + πn

合并所有解

t = arctan (2 + 3) + πn, t = arctan (2 - 3) + πn

以小数形式表示解

t = 1.30899 \dots + πn, t = 0.26179 \dots + πn

作图

流行的例子

2cos^3(x)-cos(x)=0

2 cos^{3} (x) - cos (x) = 0

-2cos(2x)=sqrt(3)

- 2 cos (2 x) = 3

(tan(x)-1)(2sin(x)-sqrt(3))=0

(tan (x) - 1) (2 sin (x) - 3) = 0

tan(θ)(1+tan^2(θ))=0

tan (θ) (1 + tan^{2} (θ)) = 0

tan (x) = 9