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受欢迎的三角函数 >

tan(a)+sec(a)= 3/2

解答

tan (a) + sec (a) = \frac{3}{2}

解答

a = 0.39479 \dots + 2 πn

+1

度数

a = 22.61986 \dots^{\circ} + 36 0^{\circ} n

求解步骤

tan (a) + sec (a) = \frac{3}{2}

两边减去

\frac{3}{2}

tan (a) + sec (a) - \frac{3}{2} = 0

化简

tan (a) + sec (a) - \frac{3}{2} : \frac{2 tan ( a ) + 2 sec ( a ) - 3}{2}

tan (a) + sec (a) - \frac{3}{2}

将项转换为分式:

tan (a) = \frac{tan ( a ) 2}{2}, sec (a) = \frac{sec ( a ) 2}{2}

= \frac{tan ( a ) \cdot 2}{2} + \frac{sec ( a ) \cdot 2}{2} - \frac{3}{2}

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{tan ( a ) \cdot 2 + sec ( a ) \cdot 2 - 3}{2}

\frac{2 tan ( a ) + 2 sec ( a ) - 3}{2} = 0

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

2 tan (a) + 2 sec (a) - 3 = 0

用 sin, cos 表示

2 \cdot \frac{sin ( a )}{cos ( a )} + 2 \cdot \frac{1}{cos ( a )} - 3 = 0

化简

2 \cdot \frac{sin ( a )}{cos ( a )} + 2 \cdot \frac{1}{cos ( a )} - 3 : \frac{2 sin ( a ) + 2 - 3 cos ( a )}{cos ( a )}

2 \cdot \frac{sin ( a )}{cos ( a )} + 2 \cdot \frac{1}{cos ( a )} - 3

2 \cdot \frac{sin ( a )}{cos ( a )} = \frac{2 sin ( a )}{cos ( a )}

2 \cdot \frac{sin ( a )}{cos ( a )}

分式相乘:

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

= \frac{sin ( a ) \cdot 2}{cos ( a )}

2 \cdot \frac{1}{cos ( a )} = \frac{2}{cos ( a )}

2 \cdot \frac{1}{cos ( a )}

分式相乘:

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

= \frac{1 \cdot 2}{cos ( a )}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{cos ( a )}

= \frac{2 sin ( a )}{cos ( a )} + \frac{2}{cos ( a )} - 3

合并分式

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

使用法则

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

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

= \frac{2 sin ( a ) + 2}{cos ( a )} - 3

将项转换为分式:

3 = \frac{3 cos ( a )}{cos ( a )}

= \frac{sin ( a ) \cdot 2 + 2}{cos ( a )} - \frac{3 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{sin ( a ) \cdot 2 + 2 - 3 cos ( a )}{cos ( a )}

\frac{2 sin ( a ) + 2 - 3 cos ( a )}{cos ( a )} = 0

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

2 sin (a) + 2 - 3 cos (a) = 0

两边加上

3 cos (a)

2 sin (a) + 2 = 3 cos (a)

两边进行平方

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

两边减去

(3 cos (a))^{2}

(2 sin (a) + 2)^{2} - 9 cos^{2} (a) = 0

使用三角恒等式改写

(2 + 2 sin (a))^{2} - 9 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)

= (2 + 2 sin (a))^{2} - 9 (1 - sin^{2} (a))

化简

(2 + 2 sin (a))^{2} - 9 (1 - sin^{2} (a)) : 13 sin^{2} (a) + 8 sin (a) - 5

(2 + 2 sin (a))^{2} - 9 (1 - sin^{2} (a))

(2 + 2 sin (a))^{2} : 4 + 8 sin (a) + 4 sin^{2} (a)

使用完全平方公式:

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

a = 2, b = 2 sin (a)

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

化简

2^{2} + 2 \cdot 2 \cdot 2 sin (a) + (2 sin (a))^{2} : 4 + 8 sin (a) + 4 sin^{2} (a)

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

2^{2} = 4

2^{2}

2^{2} = 4

= 4

2 \cdot 2 \cdot 2 sin (a) = 8 sin (a)

2 \cdot 2 \cdot 2 sin (a)

数字相乘:

2 \cdot 2 \cdot 2 = 8

= 8 sin (a)

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

(2 sin (a))^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

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

2^{2} = 4

= 4 sin^{2} (a)

= 4 + 8 sin (a) + 4 sin^{2} (a)

= 4 + 8 sin (a) + 4 sin^{2} (a)

= 4 + 8 sin (a) + 4 sin^{2} (a) - 9 (1 - sin^{2} (a))

乘开

- 9 (1 - sin^{2} (a)) : - 9 + 9 sin^{2} (a)

- 9 (1 - sin^{2} (a))

使用分配律:

a (b - c) = ab - a c

a = - 9, b = 1, c = sin^{2} (a)

= - 9 \cdot 1 - (- 9) sin^{2} (a)

使用加减运算法则

- (- a) = a

= - 9 \cdot 1 + 9 sin^{2} (a)

数字相乘:

9 \cdot 1 = 9

= - 9 + 9 sin^{2} (a)

= 4 + 8 sin (a) + 4 sin^{2} (a) - 9 + 9 sin^{2} (a)

化简

4 + 8 sin (a) + 4 sin^{2} (a) - 9 + 9 sin^{2} (a) : 13 sin^{2} (a) + 8 sin (a) - 5

4 + 8 sin (a) + 4 sin^{2} (a) - 9 + 9 sin^{2} (a)

对同类项分组

= 8 sin (a) + 4 sin^{2} (a) + 9 sin^{2} (a) + 4 - 9

同类项相加:

4 sin^{2} (a) + 9 sin^{2} (a) = 13 sin^{2} (a)

= 8 sin (a) + 13 sin^{2} (a) + 4 - 9

数字相加/相减:

4 - 9 = - 5

= 13 sin^{2} (a) + 8 sin (a) - 5

= 13 sin^{2} (a) + 8 sin (a) - 5

= 13 sin^{2} (a) + 8 sin (a) - 5

- 5 + 13 sin^{2} (a) + 8 sin (a) = 0

用替代法求解

- 5 + 13 sin^{2} (a) + 8 sin (a) = 0

令

: sin (a) = u

- 5 + 13 u^{2} + 8 u = 0

- 5 + 13 u^{2} + 8 u = 0 : u = \frac{5}{13}, u = - 1

- 5 + 13 u^{2} + 8 u = 0

改写成标准形式

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

13 u^{2} + 8 u - 5 = 0

使用求根公式求解

13 u^{2} + 8 u - 5 = 0

二次方程求根公式:

若

a = 13, b = 8, c = - 5

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 13 ( - 5 )}{2 \cdot 13}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 13 ( - 5 )}{2 \cdot 13}

8^{2} - 4 \cdot 13 (- 5) = 18

8^{2} - 4 \cdot 13 (- 5)

使用法则

- (- a) = a

= 8^{2} + 4 \cdot 13 \cdot 5

数字相乘:

4 \cdot 13 \cdot 5 = 260

= 8^{2} + 260

8^{2} = 64

= 64 + 260

数字相加:

64 + 260 = 324

= 324

因式分解数字:

324 = 1 8^{2}

= 1 8^{2}

使用根式运算法则:

n a^{n} = a

1 8^{2} = 18

= 18

u_{1, 2} = \frac{- 8 \pm 18}{2 \cdot 13}

将解分隔开

u_{1} = \frac{- 8 + 18}{2 \cdot 13}, u_{2} = \frac{- 8 - 18}{2 \cdot 13}

u = \frac{- 8 + 18}{2 \cdot 13} : \frac{5}{13}

\frac{- 8 + 18}{2 \cdot 13}

数字相加/相减:

- 8 + 18 = 10

= \frac{10}{2 \cdot 13}

数字相乘:

2 \cdot 13 = 26

= \frac{10}{26}

约分:

2

= \frac{5}{13}

u = \frac{- 8 - 18}{2 \cdot 13} : - 1

\frac{- 8 - 18}{2 \cdot 13}

数字相减:

- 8 - 18 = - 26

= \frac{- 26}{2 \cdot 13}

数字相乘:

2 \cdot 13 = 26

= \frac{- 26}{26}

使用分式法则:

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

= - \frac{26}{26}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = \frac{5}{13}, u = - 1

u = sin (a)

代回

sin (a) = \frac{5}{13}, sin (a) = - 1

sin (a) = \frac{5}{13}, sin (a) = - 1

sin (a) = \frac{5}{13} : a = arcsin (\frac{5}{13}) + 2 πn, a = π - arcsin (\frac{5}{13}) + 2 πn

sin (a) = \frac{5}{13}

使用反三角函数性质

sin (a) = \frac{5}{13}

sin (a) = \frac{5}{13}

的通解

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

a = arcsin (\frac{5}{13}) + 2 πn, a = π - arcsin (\frac{5}{13}) + 2 πn

a = arcsin (\frac{5}{13}) + 2 πn, a = π - arcsin (\frac{5}{13}) + 2 πn

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

合并所有解

a = arcsin (\frac{5}{13}) + 2 πn, a = π - arcsin (\frac{5}{13}) + 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

tan (a) + sec (a) = \frac{3}{2}

检验解是否符合

去除与方程不符的解。

检验

arcsin (\frac{5}{13}) + 2 πn

的解:

真

arcsin (\frac{5}{13}) + 2 πn

代入

n = 1

arcsin (\frac{5}{13}) + 2 π 1

对于

tan (a) + sec (a) = \frac{3}{2}

代入

a = arcsin (\frac{5}{13}) + 2 π 1

tan (arcsin (\frac{5}{13}) + 2 π 1) + sec (arcsin (\frac{5}{13}) + 2 π 1) = \frac{3}{2}

整理后得

1.5 = 1.5

\Rightarrow 真

检验

π - arcsin (\frac{5}{13}) + 2 πn

的解:

假

π - arcsin (\frac{5}{13}) + 2 πn

代入

n = 1

π - arcsin (\frac{5}{13}) + 2 π 1

对于

tan (a) + sec (a) = \frac{3}{2}

代入

a = π - arcsin (\frac{5}{13}) + 2 π 1

tan (π - arcsin (\frac{5}{13}) + 2 π 1) + sec (π - arcsin (\frac{5}{13}) + 2 π 1) = \frac{3}{2}

整理后得

- 1.5 = 1.5

\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

tan (a) + sec (a) = \frac{3}{2}

代入

a = \frac{3 π}{2} + 2 π 1

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

未定义

\Rightarrow 假

a = arcsin (\frac{5}{13}) + 2 πn

以小数形式表示解

a = 0.39479 \dots + 2 πn

作图

流行的例子

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