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受欢迎的三角函数 >

(cos(a))/(1+sin(a))=sec(a)

解答

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

解答

a = 2 πn, a = π + 2 πn

+1

度数

a = 0^{\circ} + 36 0^{\circ} n, a = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

sec (a)

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

化简

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

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

将项转换为分式:

sec (a) = \frac{sec ( a ) ( 1 + sin ( a ) )}{1 + sin ( a )}

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

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

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

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

用 sin, cos 表示

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

化简

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

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

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

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

分式相乘:

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

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

1 \cdot (1 + sin (a)) = 1 + sin (a)

1 \cdot (1 + sin (a))

乘以:

1 \cdot (1 + sin (a)) = (1 + sin (a))

= (1 + sin (a))

去除括号:

(a) = a

= 1 + sin (a)

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

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

将项转换为分式:

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

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

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

cos (a) cos (a) - (1 + sin (a))

cos (a) cos (a) = cos^{2} (a)

cos (a) cos (a)

使用指数法则:

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

cos (a) cos (a) = cos^{1 + 1} (a)

= cos^{1 + 1} (a)

数字相加:

1 + 1 = 2

= cos^{2} (a)

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

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

- (1 + sin (a)) : - 1 - sin (a)

- (1 + sin (a))

打开括号

= - (1) - (sin (a))

使用加减运算法则

+ (- a) = - a

= - 1 - sin (a)

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

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

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

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

两边加上

sin (a)

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

两边进行平方

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

两边减去

sin^{2} (a)

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

分解

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

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

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

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

整理后得

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

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

分别求解每个部分

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

cos^{2} (a) - 1 + sin (a) = 0 : a = 2 πn, a = π + 2 πn, a = \frac{π}{2} + 2 πn

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

用替代法求解

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

令

: sin (a) = u

u - u^{2} = 0

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

u - u^{2} = 0

改写成标准形式

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

- u^{2} + u = 0

使用求根公式求解

- u^{2} + u = 0

二次方程求根公式:

若

a = - 1, b = 1, c = 0

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

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 0

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 1 + 0

数字相加:

1 + 0 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

u = \frac{- 1 + 1}{2 ( - 1 )} : 0

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

去除括号:

(- a) = - a

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

数字相加/相减:

- 1 + 1 = 0

= \frac{0}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{- 2}

使用分式法则:

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

= - \frac{0}{2}

使用法则

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

= - 0

= 0

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

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

去除括号:

(- a) = - a

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

数字相减:

- 1 - 1 = - 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

使用分式法则:

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

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = 0, u = 1

u = sin (a)

代回

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

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

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

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

合并所有解

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

cos^{2} (a) - 1 - sin (a) = 0 : a = \frac{3 π}{2} + 2 πn, a = 2 πn, a = π + 2 πn

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

用替代法求解

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

令

: sin (a) = u

- u - u^{2} = 0

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

- u - u^{2} = 0

改写成标准形式

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

- u^{2} - u = 0

使用求根公式求解

- u^{2} - u = 0

二次方程求根公式:

若

a = - 1, b = - 1, c = 0

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

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

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

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

使用法则

- (- a) = a

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

(- 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 0 = 0

4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 0

= 1 + 0

数字相加:

1 + 0 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

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

去除括号:

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

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

数字相加:

1 + 1 = 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{2}{- 2}

使用分式法则:

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

= - \frac{2}{2}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 1 ) - 1}{2 ( - 1 )} : 0

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

去除括号:

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

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

数字相减:

1 - 1 = 0

= \frac{0}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{- 2}

使用分式法则:

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

= - \frac{0}{2}

使用法则

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

= - 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{3 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{3 π}{2} + 2 πn

a = \frac{3 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{3 π}{2} + 2 πn, a = 2 πn, a = π + 2 πn

合并所有解

a = 2 πn, a = π + 2 πn, a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

2 πn

的解:

真

2 πn

代入

n = 1

2 π 1

对于

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

代入

a = 2 π 1

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

整理后得

1 = 1

\Rightarrow 真

检验

π + 2 πn

的解:

真

π + 2 πn

代入

n = 1

π + 2 π 1

对于

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

代入

a = π + 2 π 1

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

整理后得

- 1 = - 1

\Rightarrow 真

检验

\frac{π}{2} + 2 πn

的解:

假

\frac{π}{2} + 2 πn

代入

n = 1

\frac{π}{2} + 2 π 1

对于

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

代入

a = \frac{π}{2} + 2 π 1

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

整理后得

0 = \infty

\Rightarrow 假

检验

\frac{3 π}{2} + 2 πn

的解:

假

\frac{3 π}{2} + 2 πn

代入

n = 1

\frac{3 π}{2} + 2 π 1

对于

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

代入

a = \frac{3 π}{2} + 2 π 1

\frac{cos ( \frac{3 π}{2} + 2 π 1 )}{1 + sin ( \frac{3 π}{2} + 2 π 1 )} = sec (\frac{3 π}{2} + 2 π 1)

未定义

\Rightarrow 假

a = 2 πn, a = π + 2 πn

作图

流行的例子

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

cos^{2} (x) + sin^{2} (2 x) = 0

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

2 cos^{2} (x) + 3 sin^{2} (x) = 2

42 - 25 cos (x) = 31

2sec^2(x)=5tan(x)

2 sec^{2} (x) = 5 tan (x)

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

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