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受欢迎的三角函数 >

(sin^2(a)+1)/(tan^2(a))=1

解答

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

解答

a = 0.90455 \dots + 2 πn, a = π - 0.90455 \dots + 2 πn, a = - 0.90455 \dots + 2 πn, a = π + 0.90455 \dots + 2 πn

+1

度数

a = 51.82729 \dots^{\circ} + 36 0^{\circ} n, a = 128.17270 \dots^{\circ} + 36 0^{\circ} n, a = - 51.82729 \dots^{\circ} + 36 0^{\circ} n, a = 231.82729 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

1

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

化简

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

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

将项转换为分式:

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

= \frac{sin ^{2} ( a ) + 1}{tan ^{2} ( a )} - \frac{1 \cdot tan ^{2} ( a )}{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{sin ^{2} ( a ) + 1 - 1 \cdot tan ^{2} ( a )}{tan ^{2} ( a )}

乘以:

1 \cdot tan^{2} (a) = tan^{2} (a)

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

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

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

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

使用三角恒等式改写

1 + sin^{2} (a) - tan^{2} (a)

使用基本三角恒等式:

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

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

使用指数法则:

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

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

使用毕达哥拉斯恒等式:

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

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

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

合并分式

- \frac{sin ^{2} ( a )}{- sin ^{2} ( a ) + 1} + sin^{2} (a) : - \frac{sin ^{4} ( a )}{- sin ^{2} ( a ) + 1}

- \frac{sin ^{2} ( a )}{- sin ^{2} ( a ) + 1} + sin^{2} (a)

将项转换为分式:

sin^{2} (a) = \frac{sin ^{2} ( a ) ( 1 - sin ^{2} ( a ) )}{1 - sin ^{2} ( a )}

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

乘开

- sin^{2} (a) + sin^{2} (a) (1 - sin^{2} (a)) : - sin^{4} (a)

- sin^{2} (a) + sin^{2} (a) (1 - sin^{2} (a))

乘开

sin^{2} (a) (1 - sin^{2} (a)) : sin^{2} (a) - sin^{4} (a)

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

使用分配律:

a (b - c) = ab - a c

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

= sin^{2} (a) \cdot 1 - sin^{2} (a) sin^{2} (a)

= 1 \cdot sin^{2} (a) - sin^{2} (a) sin^{2} (a)

化简

1 \cdot sin^{2} (a) - sin^{2} (a) sin^{2} (a) : sin^{2} (a) - sin^{4} (a)

1 \cdot sin^{2} (a) - sin^{2} (a) sin^{2} (a)

1 \cdot sin^{2} (a) = sin^{2} (a)

1 \cdot sin^{2} (a)

乘以:

1 \cdot sin^{2} (a) = sin^{2} (a)

= sin^{2} (a)

sin^{2} (a) sin^{2} (a) = sin^{4} (a)

sin^{2} (a) sin^{2} (a)

使用指数法则:

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

sin^{2} (a) sin^{2} (a) = sin^{2 + 2} (a)

= sin^{2 + 2} (a)

数字相加:

2 + 2 = 4

= sin^{4} (a)

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

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

= - sin^{2} (a) + sin^{2} (a) - sin^{4} (a)

同类项相加:

- sin^{2} (a) + sin^{2} (a) = 0

= - sin^{4} (a)

= \frac{- sin ^{4} ( a )}{1 - sin ^{2} ( a )}

使用分式法则:

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

= - \frac{sin ^{4} ( a )}{1 - sin ^{2} ( a )}

1 - \frac{sin ^{4} ( a )}{1 - sin ^{2} ( a )} = 0

1 - \frac{sin ^{4} ( a )}{1 - sin ^{2} ( a )} = 0

用替代法求解

1 - \frac{sin ^{4} ( a )}{1 - sin ^{2} ( a )} = 0

令

: sin (a) = u

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

1 - \frac{u ^{4}}{1 - u ^{2}} = 0 : u = i \frac{1 + 5}{2}, u = - i \frac{1 + 5}{2}, u = \frac{5 - 1}{2}, u = - \frac{5 - 1}{2}

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

在两边乘以

1 - u^{2}

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

在两边乘以

1 - u^{2}

1 \cdot (1 - u^{2}) - \frac{u ^{4}}{1 - u ^{2}} (1 - u^{2}) = 0 \cdot (1 - u^{2})

化简

1 \cdot (1 - u^{2}) - \frac{u ^{4}}{1 - u ^{2}} (1 - u^{2}) = 0 \cdot (1 - u^{2})

化简

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

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

乘以:

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

= (1 - u^{2})

去除括号:

(a) = a

= 1 - u^{2}

化简

- \frac{u ^{4}}{1 - u ^{2}} (1 - u^{2}) : - u^{4}

- \frac{u ^{4}}{1 - u ^{2}} (1 - u^{2})

分式相乘:

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

= - \frac{u ^{4} ( 1 - u ^{2} )}{1 - u ^{2}}

约分:

1 - u^{2}

= - u^{4}

化简

0 \cdot (1 - u^{2}) : 0

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

使用法则

0 \cdot a = 0

= 0

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

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

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

解

1 - u^{2} - u^{4} = 0 : u = i \frac{1 + 5}{2}, u = - i \frac{1 + 5}{2}, u = \frac{5 - 1}{2}, u = - \frac{5 - 1}{2}

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

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a = 0

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

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

- v^{2} - v + 1 = 0

解

- v^{2} - v + 1 = 0 : v = - \frac{1 + 5}{2}, v = \frac{5 - 1}{2}

- v^{2} - v + 1 = 0

使用求根公式求解

- v^{2} - v + 1 = 0

二次方程求根公式:

若

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

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

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

(- 1)^{2} - 4 (- 1) \cdot 1 = 5

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

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

v_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 ( - 1 )}

将解分隔开

v_{1} = \frac{- ( - 1 ) + 5}{2 ( - 1 )}, v_{2} = \frac{- ( - 1 ) - 5}{2 ( - 1 )}

v = \frac{- ( - 1 ) + 5}{2 ( - 1 )} : - \frac{1 + 5}{2}

\frac{- ( - 1 ) + 5}{2 ( - 1 )}

去除括号:

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

= \frac{1 + 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 + 5}{- 2}

使用分式法则:

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

= - \frac{1 + 5}{2}

v = \frac{- ( - 1 ) - 5}{2 ( - 1 )} : \frac{5 - 1}{2}

\frac{- ( - 1 ) - 5}{2 ( - 1 )}

去除括号:

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

= \frac{1 - 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 - 5}{- 2}

使用分式法则:

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

1 - 5 = - (5 - 1)

= \frac{5 - 1}{2}

二次方程组的解是:

v = - \frac{1 + 5}{2}, v = \frac{5 - 1}{2}

v = - \frac{1 + 5}{2}, v = \frac{5 - 1}{2}

代回

v = u^{2}

，求解

u

解

u^{2} = - \frac{1 + 5}{2} : u = i \frac{1 + 5}{2}, u = - i \frac{1 + 5}{2}

u^{2} = - \frac{1 + 5}{2}

对于

x^{2} = f (a)

解为

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

u = - \frac{1 + 5}{2}, u = - - \frac{1 + 5}{2}

化简

- \frac{1 + 5}{2} : i \frac{1 + 5}{2}

- \frac{1 + 5}{2}

使用根式运算法则:

- a = - 1 a

- \frac{1 + 5}{2} = - 1 \frac{1 + 5}{2}

= - 1 \frac{1 + 5}{2}

使用虚数运算法则:

- 1 = i

= i \frac{1 + 5}{2}

化简

- - \frac{1 + 5}{2} : - i \frac{1 + 5}{2}

- - \frac{1 + 5}{2}

化简

- \frac{1 + 5}{2} : i \frac{1 + 5}{2}

- \frac{1 + 5}{2}

使用根式运算法则:

- a = - 1 a

- \frac{1 + 5}{2} = - 1 \frac{1 + 5}{2}

= - 1 \frac{1 + 5}{2}

使用虚数运算法则:

- 1 = i

= i \frac{1 + 5}{2}

= - i \frac{1 + 5}{2}

u = i \frac{1 + 5}{2}, u = - i \frac{1 + 5}{2}

解

u^{2} = \frac{5 - 1}{2} : u = \frac{5 - 1}{2}, u = - \frac{5 - 1}{2}

u^{2} = \frac{5 - 1}{2}

对于

x^{2} = f (a)

解为

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

u = \frac{5 - 1}{2}, u = - \frac{5 - 1}{2}

解为

u = i \frac{1 + 5}{2}, u = - i \frac{1 + 5}{2}, u = \frac{5 - 1}{2}, u = - \frac{5 - 1}{2}

u = i \frac{1 + 5}{2}, u = - i \frac{1 + 5}{2}, u = \frac{5 - 1}{2}, u = - \frac{5 - 1}{2}

验证解

找到无定义的点（奇点）:

u = 1, u = - 1

取

1 - \frac{u ^{4}}{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 = i \frac{1 + 5}{2}, u = - i \frac{1 + 5}{2}, u = \frac{5 - 1}{2}, u = - \frac{5 - 1}{2}

u = sin (a)

代回

sin (a) = i \frac{1 + 5}{2}, sin (a) = - i \frac{1 + 5}{2}, sin (a) = \frac{5 - 1}{2}, sin (a) = - \frac{5 - 1}{2}

sin (a) = i \frac{1 + 5}{2}, sin (a) = - i \frac{1 + 5}{2}, sin (a) = \frac{5 - 1}{2}, sin (a) = - \frac{5 - 1}{2}

sin (a) = i \frac{1 + 5}{2} :

无解

sin (a) = i \frac{1 + 5}{2}

无解

sin (a) = - i \frac{1 + 5}{2} :

无解

sin (a) = - i \frac{1 + 5}{2}

无解

sin (a) = \frac{5 - 1}{2} : a = arcsin \frac{5 - 1}{2} + 2 πn, a = π - arcsin \frac{5 - 1}{2} + 2 πn

sin (a) = \frac{5 - 1}{2}

使用反三角函数性质

sin (a) = \frac{5 - 1}{2}

sin (a) = \frac{5 - 1}{2}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

a = arcsin \frac{5 - 1}{2} + 2 πn, a = π - arcsin \frac{5 - 1}{2} + 2 πn

a = arcsin \frac{5 - 1}{2} + 2 πn, a = π - arcsin \frac{5 - 1}{2} + 2 πn

sin (a) = - \frac{5 - 1}{2} : a = arcsin - \frac{5 - 1}{2} + 2 πn, a = π + arcsin \frac{5 - 1}{2} + 2 πn

sin (a) = - \frac{5 - 1}{2}

使用反三角函数性质

sin (a) = - \frac{5 - 1}{2}

sin (a) = - \frac{5 - 1}{2}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

a = arcsin - \frac{5 - 1}{2} + 2 πn, a = π + arcsin \frac{5 - 1}{2} + 2 πn

a = arcsin - \frac{5 - 1}{2} + 2 πn, a = π + arcsin \frac{5 - 1}{2} + 2 πn

合并所有解

a = arcsin \frac{5 - 1}{2} + 2 πn, a = π - arcsin \frac{5 - 1}{2} + 2 πn, a = arcsin - \frac{5 - 1}{2} + 2 πn, a = π + arcsin \frac{5 - 1}{2} + 2 πn

以小数形式表示解

a = 0.90455 \dots + 2 πn, a = π - 0.90455 \dots + 2 πn, a = - 0.90455 \dots + 2 πn, a = π + 0.90455 \dots + 2 πn

作图

流行的例子

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

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cos^2(x)+5cos(x)-2=0

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