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 >

sec(a)=1+tan(a)

解答

sec (a) = 1 + tan (a)

解答

a = 2 πn

+1

度数

a = 0^{\circ} + 36 0^{\circ} n

求解步骤

sec (a) = 1 + tan (a)

两边减去

1 + tan (a)

sec (a) - 1 - tan (a) = 0

用 sin, cos 表示

\frac{1}{cos ( a )} - 1 - \frac{sin ( a )}{cos ( a )} = 0

化简

\frac{1}{cos ( a )} - 1 - \frac{sin ( a )}{cos ( a )} : \frac{1 - sin ( a ) - cos ( a )}{cos ( a )}

\frac{1}{cos ( a )} - 1 - \frac{sin ( a )}{cos ( a )}

合并分式

\frac{1}{cos ( a )} - \frac{sin ( a )}{cos ( a )} : \frac{1 - sin ( a )}{cos ( a )}

使用法则

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

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

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

将项转换为分式:

1 = \frac{1 cos ( a )}{cos ( a )}

= \frac{1 - sin ( a )}{cos ( a )} - \frac{1 \cdot cos ( a )}{cos ( 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{1 - sin ( a ) - 1 \cdot cos ( a )}{cos ( a )}

乘以:

1 \cdot cos (a) = cos (a)

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

\frac{1 - sin ( a ) - cos ( a )}{cos ( a )} = 0

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

1 - sin (a) - cos (a) = 0

两边加上

cos (a)

1 - sin (a) = cos (a)

两边进行平方

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

两边减去

cos^{2} (a)

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

使用三角恒等式改写

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

化简

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

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

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

使用完全平方公式:

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

a = 1, b = sin (a)

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

数字相乘:

2 \cdot 1 = 2

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

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

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

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

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

打开括号

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

使用加减运算法则

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

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

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

化简

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

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

对同类项分组

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

同类项相加:

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

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

1 - 1 = 0

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

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

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

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

用替代法求解

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

令

: sin (a) = u

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

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

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

改写成标准形式

a x^{2} + b x + c = 0

2 u^{2} - 2 u = 0

使用求根公式求解

2 u^{2} - 2 u = 0

二次方程求根公式:

若

a = 2, b = - 2, c = 0

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

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

(- 2)^{2} - 4 \cdot 2 \cdot 0

使用指数法则:

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

若

n

是偶数

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

= 2^{2} - 4 \cdot 2 \cdot 0

使用法则

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 2

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

将解分隔开

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

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

\frac{- ( - 2 ) + 2}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{2 + 2}{2 \cdot 2}

数字相加:

2 + 2 = 4

= \frac{4}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 2}{2 \cdot 2} : 0

\frac{- ( - 2 ) - 2}{2 \cdot 2}

使用法则

- (- a) = a

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

数字相减:

2 - 2 = 0

= \frac{0}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{0}{4}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

二次方程组的解是:

u = 1, u = 0

u = sin (a)

代回

sin (a) = 1, sin (a) = 0

sin (a) = 1, sin (a) = 0

sin (a) = 1 : a = \frac{π}{2} + 2 πn

sin (a) = 1

sin (a) = 1

的通解

sin (x)

周期表（周期为

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

a = \frac{π}{2} + 2 πn

a = \frac{π}{2} + 2 πn

sin (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) = 0

sin (a) = 0

的通解

sin (x)

周期表（周期为

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

a = 0 + 2 πn, a = π + 2 πn

a = 0 + 2 πn, a = π + 2 πn

解

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn, a = π + 2 πn

合并所有解

a = \frac{π}{2} + 2 πn, a = 2 πn, a = π + 2 πn

将解代入原方程进行验证

将它们代入

sec (a) = 1 + tan (a)

检验解是否符合

去除与方程不符的解。

检验

\frac{π}{2} + 2 πn

的解:

真

\frac{π}{2} + 2 πn

代入

n = 1

\frac{π}{2} + 2 π 1

对于

sec (a) = 1 + tan (a)

代入

a = \frac{π}{2} + 2 π 1

sec (\frac{π}{2} + 2 π 1) = 1 + tan (\frac{π}{2} + 2 π 1)

整理后得

\infty = \infty

\Rightarrow 真

检验

2 πn

的解:

真

2 πn

代入

n = 1

2 π 1

对于

sec (a) = 1 + tan (a)

代入

a = 2 π 1

sec (2 π 1) = 1 + tan (2 π 1)

整理后得

1 = 1

\Rightarrow 真

检验

π + 2 πn

的解:

假

π + 2 πn

代入

n = 1

π + 2 π 1

对于

sec (a) = 1 + tan (a)

代入

a = π + 2 π 1

sec (π + 2 π 1) = 1 + tan (π + 2 π 1)

整理后得

- 1 = 1

\Rightarrow 假

a = \frac{π}{2} + 2 πn, a = 2 πn

因为方程对以下值无定义:

\frac{π}{2} + 2 πn

a = 2 πn

作图

流行的例子

arctan(θ)= 2/(2sqrt(3))

arctan (θ) = \frac{2}{2 3}

3sin(x)=+sin(x)

3 sin (x) = + sin (x)

solvefor x,(D^2-3D+2)y=sec^2(e^{-x})

so l v e f or x, (D^{2} - 3 D + 2) y = sec^{2} (e^{- x})

sin(θ)=(7sin(140))/(16.12288984)

sin (θ) = \frac{7 sin ( 14 0 ^{\circ} )}{16.12288984}

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