zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

1+cot^2(a)=tan^2(a)

解答

1 + cot^{2} (a) = tan^{2} (a)

解答

a = 0.90455 \dots + πn, a = 2.23703 \dots + πn

+1

度数

a = 51.82729 \dots^{\circ} + 18 0^{\circ} n, a = 128.17270 \dots^{\circ} + 18 0^{\circ} n

求解步骤

1 + cot^{2} (a) = tan^{2} (a)

两边减去

tan^{2} (a)

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

使用三角恒等式改写

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

使用基本三角恒等式:

tan (x) = \frac{1}{cot ( x )}

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

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

(\frac{1}{cot ( a )})^{2}

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

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

用替代法求解

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

令

: cot (a) = u

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

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

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

在两边乘以

u^{2}

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

在两边乘以

u^{2}

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

化简

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

化简

1 \cdot u^{2} : u^{2}

1 \cdot u^{2}

乘以:

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

= u^{2}

化简

u^{2} u^{2} : u^{4}

u^{2} u^{2}

使用指数法则:

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

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

= u^{2 + 2}

数字相加:

2 + 2 = 4

= u^{4}

化简

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

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

分式相乘:

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

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

约分:

u^{2}

= - 1

化简

0 \cdot u^{2} : 0

0 \cdot u^{2}

使用法则

0 \cdot a = 0

= 0

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

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

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

解

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

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

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

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

将解分隔开

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

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

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

数字相乘:

2 \cdot 1 = 2

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

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

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

数字相乘:

2 \cdot 1 = 2

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

二次方程组的解是:

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

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

代回

v = u^{2}

，求解

u

解

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

解

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

解为

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

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

的分母，令其等于零

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

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

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

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

u = cot (a)

代回

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

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

cot (a) = \frac{- 1 + 5}{2} : a = arccot \frac{- 1 + 5}{2} + πn

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

使用反三角函数性质

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

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

的通解

cot (x) = a \Rightarrow x = arccot (a) + πn

a = arccot \frac{- 1 + 5}{2} + πn

a = arccot \frac{- 1 + 5}{2} + πn

cot (a) = - \frac{- 1 + 5}{2} : a = arccot - \frac{- 1 + 5}{2} + πn

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

使用反三角函数性质

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

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

的通解

cot (x) = - a \Rightarrow x = arccot (- a) + πn

a = arccot - \frac{- 1 + 5}{2} + πn

a = arccot - \frac{- 1 + 5}{2} + πn

cot (a) = \frac{- 1 - 5}{2} : a = arccot \frac{- 1 - 5}{2} + πn

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

使用反三角函数性质

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

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

的通解

cot (x) = a \Rightarrow x = arccot (a) + πn

a = arccot \frac{- 1 - 5}{2} + πn

a = arccot \frac{- 1 - 5}{2} + πn

cot (a) = - \frac{- 1 - 5}{2} : a = arccot - \frac{- 1 - 5}{2} + πn

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

使用反三角函数性质

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

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

的通解

cot (x) = a \Rightarrow x = arccot (a) + πn

a = arccot - \frac{- 1 - 5}{2} + πn

a = arccot - \frac{- 1 - 5}{2} + πn

合并所有解

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

因为方程对以下值无定义:

arccot \frac{- 1 - 5}{2} + πn, arccot - \frac{- 1 - 5}{2} + πn

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

以小数形式表示解

a = 0.90455 \dots + πn, a = 2.23703 \dots + πn

作图

流行的例子

tan(α)=(12)/(6sqrt(3))

tan (α) = \frac{12}{6 3}

1=cos^2(α)-2cos(α)sin(a)+sin^2(α)

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

4 sin (θ) = 1

sin(x+pi/2)=0.6

sin (x + \frac{π}{2}) = 0.6

4cot(x)+1=1+2cot(x)

4 cot (x) + 1 = 1 + 2 cot (x)