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受欢迎的三角函数 >

tan(2A)=((2tan(A)))/([1+(tan(A))^2])

解答

tan (2 A) = \frac{( 2 tan ( A ) )}{[ 1 + ( tan ( A ) ) ^{2} ]}

解答

A = πn

+1

度数

A = 0^{\circ} + 18 0^{\circ} n

求解步骤

tan (2 A) = \frac{( 2 tan ( A ) )}{[ 1 + ( tan ( A ) ) ^{2} ]}

两边减去

\frac{2 tan ( A )}{[ 1 + ( tan ( A ) ) ^{2} ]}

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

化简

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

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

将项转换为分式:

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

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

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

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

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

tan (2 A) (1 + tan^{2} (A)) - 2 tan (A) = 0

使用三角恒等式改写

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

使用倍角公式:

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

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

化简

- 2 tan (A) + \frac{2 tan ( A )}{1 - tan ^{2} ( A )} (1 + tan^{2} (A)) : \frac{4 tan ^{3} ( A )}{1 - tan ^{2} ( A )}

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

乘

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

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

分式相乘:

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

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

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

将项转换为分式:

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

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

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

乘开

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

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

乘开

2 tan (A) (1 + tan^{2} (A)) : 2 tan (A) + 2 tan^{3} (A)

2 tan (A) (1 + tan^{2} (A))

使用分配律:

a (b + c) = ab + a c

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

= 2 tan (A) \cdot 1 + 2 tan (A) tan^{2} (A)

= 2 \cdot 1 \cdot tan (A) + 2 tan^{2} (A) tan (A)

化简

2 \cdot 1 \cdot tan (A) + 2 tan^{2} (A) tan (A) : 2 tan (A) + 2 tan^{3} (A)

2 \cdot 1 \cdot tan (A) + 2 tan^{2} (A) tan (A)

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

2 \cdot 1 \cdot tan (A)

数字相乘:

2 \cdot 1 = 2

= 2 tan (A)

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

2 tan^{2} (A) tan (A)

使用指数法则:

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

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

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

数字相加:

2 + 1 = 3

= 2 tan^{3} (A)

= 2 tan (A) + 2 tan^{3} (A)

= 2 tan (A) + 2 tan^{3} (A)

= 2 tan (A) + 2 tan^{3} (A) - 2 tan (A) (1 - tan^{2} (A))

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

使用加减运算法则

- (- a) = a

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

化简

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

- 2 \cdot 1 \cdot tan (A) + 2 tan^{2} (A) tan (A)

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

2 \cdot 1 \cdot tan (A)

数字相乘:

2 \cdot 1 = 2

= 2 tan (A)

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

2 tan^{2} (A) tan (A)

使用指数法则:

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

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

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

数字相加:

2 + 1 = 3

= 2 tan^{3} (A)

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

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

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

化简

2 tan (A) + 2 tan^{3} (A) - 2 tan (A) + 2 tan^{3} (A) : 4 tan^{3} (A)

2 tan (A) + 2 tan^{3} (A) - 2 tan (A) + 2 tan^{3} (A)

同类项相加:

2 tan^{3} (A) + 2 tan^{3} (A) = 4 tan^{3} (A)

= 2 tan (A) + 4 tan^{3} (A) - 2 tan (A)

同类项相加:

2 tan (A) - 2 tan (A) = 0

= 4 tan^{3} (A)

= 4 tan^{3} (A)

= \frac{4 tan ^{3} ( A )}{1 - tan ^{2} ( A )}

= \frac{4 tan ^{3} ( A )}{1 - tan ^{2} ( A )}

\frac{4 tan ^{3} ( A )}{1 - tan ^{2} ( A )} = 0

用替代法求解

\frac{4 tan ^{3} ( A )}{1 - tan ^{2} ( A )} = 0

令

: tan (A) = u

\frac{4 u ^{3}}{1 - u ^{2}} = 0

\frac{4 u ^{3}}{1 - u ^{2}} = 0 : u = 0

\frac{4 u ^{3}}{1 - u ^{2}} = 0

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

4 u^{3} = 0

解

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

4 u^{3} = 0

两边除以

4

4 u^{3} = 0

两边除以

4

4 u^{3} = 0

两边除以

4

\frac{4 u ^{3}}{4} = \frac{0}{4}

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{4 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 (A)

代回

tan (A) = 0

tan (A) = 0

tan (A) = 0 : A = πn

tan (A) = 0

tan (A) = 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}

A = 0 + πn

A = 0 + πn

解

A = 0 + πn : A = πn

A = 0 + πn

0 + πn = πn

A = πn

A = πn

合并所有解

A = πn

作图

流行的例子

cos(4x)+1=3sin(2x)

cos (4 x) + 1 = 3 sin (2 x)

tan(2t+15)=cot(6t-5)

tan (2 t + 15) = cot (6 t - 5)

2cos^3(x)-cos^2(x)+2cos(x)-1=0

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

sin (x) = \frac{2 π}{3}

cos(3x)+sin(2x)+cos(x)=0

cos (3 x) + sin (2 x) + cos (x) = 0