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受欢迎的三角函数 >

(1+tan^2(27θ))/(1-tan^2(27θ))=sqrt(2)

解答

\frac{1 + tan ^{2} ( 27 \cdot θ )}{1 - tan ^{2} ( 27 \cdot θ )} = 2

解答

θ = \frac{0.39269 \dots}{27} + \frac{πn}{27}, θ = \frac{- 0.39269 \dots}{27} + \frac{πn}{27}

+1

度数

θ = 0.83333 \dots^{\circ} + 6.66666 \dots^{\circ} n, θ = - 0.83333 \dots^{\circ} + 6.66666 \dots^{\circ} n

求解步骤

\frac{1 + tan ^{2} ( 27 θ )}{1 - tan ^{2} ( 27 θ )} = 2

用替代法求解

\frac{1 + tan ^{2} ( 27 θ )}{1 - tan ^{2} ( 27 θ )} = 2

令

: tan (27 θ) = u

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

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

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

在两边乘以

1 - u^{2}

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

在两边乘以

1 - u^{2}

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

化简

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

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

解

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

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

将

1

到右边

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

两边减去

1

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

化简

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

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

将

2 (1 - u^{2})

para o lado esquerdo

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

两边减去

2 (1 - u^{2})

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

化简

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

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

乘开

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

- 2 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = - 2, b = 1, c = u^{2}

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

使用加减运算法则

- (- a) = a

= - 1 \cdot 2 + 2 u^{2}

乘以:

1 \cdot 2 = 2

= - 2 + 2 u^{2}

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

将

2

到右边

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

两边加上

2

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

化简

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

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

分解

u^{2} + 2 u^{2} : (1 + 2) u^{2}

u^{2} + 2 u^{2}

因式分解出通项

u^{2}

= u^{2} (1 + 2)

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

两边除以

1 + 2

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

两边除以

1 + 2

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

化简

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

化简

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

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

约分:

1 + 2

= u^{2}

化简

- \frac{1}{1 + 2} + \frac{2}{1 + 2} : 3 - 22

- \frac{1}{1 + 2} + \frac{2}{1 + 2}

使用法则

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

= \frac{- 1 + 2}{1 + 2}

乘以共轭根式

\frac{1 - 2}{1 - 2}

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

(- 1 + 2) (1 - 2) = 22 - 3

(- 1 + 2) (1 - 2)

使用 FOIL 方法:

(a + b) (c + d) = a c + a d + b c + b d

a = - 1, b = 2, c = 1, d = - 2

= (- 1) \cdot 1 + (- 1) (- 2) + 2 \cdot 1 + 2 (- 2)

使用加减运算法则

+ (- a) = - a, (- a) (- b) = ab

= - 1 \cdot 1 + 1 \cdot 2 + 1 \cdot 2 - 22

化简

- 1 \cdot 1 + 1 \cdot 2 + 1 \cdot 2 - 22 : 22 - 3

- 1 \cdot 1 + 1 \cdot 2 + 1 \cdot 2 - 22

同类项相加:

1 \cdot 2 + 1 \cdot 2 = 22

= - 1 \cdot 1 + 22 - 22

数字相乘:

1 \cdot 1 = 1

= - 1 + 22 - 22

使用根式运算法则:

a a = a

22 = 2

= - 1 + 22 - 2

数字相减:

- 1 - 2 = - 3

= 22 - 3

= 22 - 3

(1 + 2) (1 - 2) = - 1

(1 + 2) (1 - 2)

使用平方差公式:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = 2

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - (2)^{2}

(2)^{2} = 2

(2)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

使用指数法则:

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

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

= 1 - 2

数字相减:

1 - 2 = - 1

= - 1

= - 1

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

使用分式法则:

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

22 - 3 = - (3 - 22)

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

使用法则

\frac{a}{1} = a

= 3 - 22

u^{2} = 3 - 22

u^{2} = 3 - 22

u^{2} = 3 - 22

对于

x^{2} = f (a)

解为

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

u = 3 - 22, u = - 3 - 22

3 - 22 = 2 - 1

3 - 22

= 2 - 22 + 1

= (2)^{2} - 22 + (1)^{2}

1 = 1

1

使用法则

1 = 1

= 1

= (2)^{2} - 22 + 1^{2}

22 \cdot 1 = 22

22 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 22

= (2)^{2} - 22 \cdot 1 + 1^{2}

使用完全平方公式:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

(2)^{2} - 22 \cdot 1 + 1^{2} = (2 - 1)^{2}

= (2 - 1)^{2}

使用根式运算法则:

n a^{n} = a

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

= 2 - 1

- 3 - 22 = - 2 + 1

- 3 - 22

3 - 22 = 2 - 1

3 - 22

= 2 - 22 + 1

= (2)^{2} - 22 + (1)^{2}

1 = 1

1

使用法则

1 = 1

= 1

= (2)^{2} - 22 + 1^{2}

22 \cdot 1 = 22

22 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 22

= (2)^{2} - 22 \cdot 1 + 1^{2}

使用完全平方公式:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

(2)^{2} - 22 \cdot 1 + 1^{2} = (2 - 1)^{2}

= (2 - 1)^{2}

使用根式运算法则:

n a^{n} = a

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

= 2 - 1

= - (2 - 1)

打开括号

= - (2) - (- 1)

使用加减运算法则

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

= - 2 + 1

u = 2 - 1, u = - 2 + 1

u = 2 - 1, u = - 2 + 1

验证解

找到无定义的点（奇点）:

u = 1, u = - 1

取

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

u = tan (27 θ)

代回

tan (27 θ) = 2 - 1, tan (27 θ) = - 2 + 1

tan (27 θ) = 2 - 1, tan (27 θ) = - 2 + 1

tan (27 θ) = 2 - 1 : θ = \frac{arctan ( 2 - 1 )}{27} + \frac{πn}{27}

tan (27 θ) = 2 - 1

使用反三角函数性质

tan (27 θ) = 2 - 1

tan (27 θ) = 2 - 1

的通解

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

27 θ = arctan (2 - 1) + πn

27 θ = arctan (2 - 1) + πn

解

27 θ = arctan (2 - 1) + πn : θ = \frac{arctan ( 2 - 1 )}{27} + \frac{πn}{27}

27 θ = arctan (2 - 1) + πn

两边除以

27

27 θ = arctan (2 - 1) + πn

两边除以

27

\frac{27 θ}{27} = \frac{arctan ( 2 - 1 )}{27} + \frac{πn}{27}

化简

θ = \frac{arctan ( 2 - 1 )}{27} + \frac{πn}{27}

θ = \frac{arctan ( 2 - 1 )}{27} + \frac{πn}{27}

θ = \frac{arctan ( 2 - 1 )}{27} + \frac{πn}{27}

tan (27 θ) = - 2 + 1 : θ = \frac{arctan ( - 2 + 1 )}{27} + \frac{πn}{27}

tan (27 θ) = - 2 + 1

使用反三角函数性质

tan (27 θ) = - 2 + 1

tan (27 θ) = - 2 + 1

的通解

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

27 θ = arctan (- 2 + 1) + πn

27 θ = arctan (- 2 + 1) + πn

解

27 θ = arctan (- 2 + 1) + πn : θ = \frac{arctan ( - 2 + 1 )}{27} + \frac{πn}{27}

27 θ = arctan (- 2 + 1) + πn

两边除以

27

27 θ = arctan (- 2 + 1) + πn

两边除以

27

\frac{27 θ}{27} = \frac{arctan ( - 2 + 1 )}{27} + \frac{πn}{27}

化简

θ = \frac{arctan ( - 2 + 1 )}{27} + \frac{πn}{27}

θ = \frac{arctan ( - 2 + 1 )}{27} + \frac{πn}{27}

θ = \frac{arctan ( - 2 + 1 )}{27} + \frac{πn}{27}

合并所有解

θ = \frac{arctan ( 2 - 1 )}{27} + \frac{πn}{27}, θ = \frac{arctan ( - 2 + 1 )}{27} + \frac{πn}{27}

以小数形式表示解

θ = \frac{0.39269 \dots}{27} + \frac{πn}{27}, θ = \frac{- 0.39269 \dots}{27} + \frac{πn}{27}

作图

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