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受欢迎的三角函数 >

cot(-pi/5)-sec(x)=1.5

解答

cot (- \frac{π}{5}) - sec (x) = 1.5

解答

x = 1.92586 \dots + 2 πn, x = - 1.92586 \dots + 2 πn

+1

度数

x = 110.34419 \dots^{\circ} + 36 0^{\circ} n, x = - 110.34419 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cot (- \frac{π}{5}) - sec (x) = 1.5

cot (- \frac{π}{5}) = - \frac{( 3 10 + 5 2 ) 5 - 5}{20}

cot (- \frac{π}{5})

利用以下特性:

cot (- x) = - cot (x)

cot (- \frac{π}{5}) = - cot (\frac{π}{5})

= - cot (\frac{π}{5})

使用三角恒等式改写:

cot (\frac{π}{5}) = \frac{( 3 10 + 5 2 ) 5 - 5}{20}

cot (\frac{π}{5})

使用三角恒等式改写:

\frac{cos ( \frac{π}{5} )}{sin ( \frac{π}{5} )}

cot (\frac{π}{5})

使用基本三角恒等式:

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

= \frac{cos ( \frac{π}{5} )}{sin ( \frac{π}{5} )}

= \frac{cos ( \frac{π}{5} )}{sin ( \frac{π}{5} )}

使用三角恒等式改写:

cos (\frac{π}{5}) = \frac{5 + 1}{4}

cos (\frac{π}{5})

显示:

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

使用以下积化和差公式:

2 sin (x) cos (y) = sin (x + y) - sin (x - y)

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

显示:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

两边除以

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

代入

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

\frac{1}{2} = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

sin (\frac{3 π}{10}) = cos (\frac{π}{2} - \frac{3 π}{10})

\frac{1}{2} = cos (\frac{π}{2} - \frac{3 π}{10}) - sin (\frac{π}{10})

\frac{1}{2} = cos (\frac{π}{5}) - sin (\frac{π}{10})

显示:

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

使用因式分解法则:

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

a = cos (\frac{π}{5}) + sin (\frac{π}{10})

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = ((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) ((cos (\frac{π}{5}) + sin (\frac{π}{10})) - (cos (\frac{π}{5}) - sin (\frac{π}{10})))

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 2 (2 cos (\frac{π}{5}) sin (\frac{π}{10}))

显示:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

两边除以

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

代入

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 1

代入

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (\frac{1}{2})^{2} = 1

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} = 1

两边加上

\frac{1}{4}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} = \frac{5}{4}

在两侧开平方

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \pm \frac{5}{4}

cos (\frac{π}{5})

不能为负

sin (\frac{π}{10})

不能为负

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

以下方程式相加

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{2}

((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) = (\frac{5}{2} + \frac{1}{2})

整理后得

cos (\frac{π}{5}) = \frac{5 + 1}{4}

= \frac{5 + 1}{4}

使用三角恒等式改写:

sin (\frac{π}{5}) = \frac{2 5 - 5}{4}

sin (\frac{π}{5})

显示:

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

使用以下积化和差公式:

2 sin (x) cos (y) = sin (x + y) - sin (x - y)

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

显示:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

两边除以

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

代入

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

\frac{1}{2} = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

sin (\frac{3 π}{10}) = cos (\frac{π}{2} - \frac{3 π}{10})

\frac{1}{2} = cos (\frac{π}{2} - \frac{3 π}{10}) - sin (\frac{π}{10})

\frac{1}{2} = cos (\frac{π}{5}) - sin (\frac{π}{10})

显示:

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

使用因式分解法则:

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

a = cos (\frac{π}{5}) + sin (\frac{π}{10})

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = ((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) ((cos (\frac{π}{5}) + sin (\frac{π}{10})) - (cos (\frac{π}{5}) - sin (\frac{π}{10})))

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 2 (2 cos (\frac{π}{5}) sin (\frac{π}{10}))

显示:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

两边除以

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

代入

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 1

代入

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (\frac{1}{2})^{2} = 1

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} = 1

两边加上

\frac{1}{4}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} = \frac{5}{4}

在两侧开平方

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \pm \frac{5}{4}

cos (\frac{π}{5})

不能为负

sin (\frac{π}{10})

不能为负

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

以下方程式相加

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{2}

((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) = (\frac{5}{2} + \frac{1}{2})

整理后得

cos (\frac{π}{5}) = \frac{5 + 1}{4}

两边进行平方

(cos (\frac{π}{5}))^{2} = (\frac{5 + 1}{4})^{2}

利用以下特性:

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

sin^{2} (\frac{π}{5}) = 1 - cos^{2} (\frac{π}{5})

代入

cos (\frac{π}{5}) = \frac{5 + 1}{4}

sin^{2} (\frac{π}{5}) = 1 - (\frac{5 + 1}{4})^{2}

整理后得

sin^{2} (\frac{π}{5}) = \frac{5 - 5}{8}

在两侧开平方

sin (\frac{π}{5}) = \pm \frac{5 - 5}{8}

sin (\frac{π}{5})

不能为负

sin (\frac{π}{5}) = \frac{5 - 5}{8}

整理后得

sin (\frac{π}{5}) = \frac{\frac{5 - 5}{2}}{2}

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

\frac{\frac{5 - 5}{2}}{2} = \frac{2 5 - 5}{4}

\frac{\frac{5 - 5}{2}}{2}

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

\frac{5 - 5}{2}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{5 - 5}{2}

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

使用分式法则:

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

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

\frac{5 - 5}{2 2}

有理化:

\frac{2 5 - 5}{4}

\frac{5 - 5}{2 2}

乘以共轭根式

\frac{2}{2}

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

2 \cdot 22 = 4

2 \cdot 22

使用指数法则:

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

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

= 2^{1 + \frac{1}{2} + \frac{1}{2}}

同类项相加:

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

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

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

2 \cdot \frac{1}{2}

分式相乘:

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

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

约分:

2

= 1

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 5 - 5}{4}

= \frac{2 5 - 5}{4}

= \frac{2 5 - 5}{4}

= \frac{\frac{5 + 1}{4}}{\frac{2 5 - 5}{4}}

化简

\frac{\frac{5 + 1}{4}}{\frac{2 5 - 5}{4}} : \frac{( 3 10 + 5 2 ) 5 - 5}{20}

\frac{\frac{5 + 1}{4}}{\frac{2 5 - 5}{4}}

分式相除:

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

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

约分:

4

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

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

有理化:

\frac{( 3 10 + 5 2 ) 5 - 5}{20}

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

乘以共轭根式

\frac{2}{2}

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

2 5 - 5 2 = 2 5 - 5

2 5 - 5 2

使用根式运算法则:

a a = a

22 = 2

= 2 5 - 5

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

乘以共轭根式

\frac{5 - 5}{5 - 5}

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

2 5 - 5 5 - 5 = 10 - 25

2 5 - 5 5 - 5

使用根式运算法则:

a a = a

5 - 5 5 - 5 = 5 - 5

= 2 (5 - 5)

使用分配律:

a (b - c) = ab - a c

a = 2, b = 5, c = 5

= 2 \cdot 5 - 25

数字相乘:

2 \cdot 5 = 10

= 10 - 25

= \frac{2 ( 5 + 1 ) 5 - 5}{10 - 2 5}

因式分解出通项

- 2 : - 2 (5 - 5)

- 25 + 10

将

10

改写为

2 \cdot 5

= - 25 + 2 \cdot 5

因式分解出通项

- 2

= - 2 (5 - 5)

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

消掉

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

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

5 - 5 = - (5 - 5)

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

整理后得

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

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

乘以共轭根式

\frac{5 + 5}{5 + 5}

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

2 (5 + 1) 5 - 5 (5 + 5) = 610 5 - 5 + 102 5 - 5

2 (5 + 1) 5 - 5 (5 + 5)

= 2 (5 + 1) (5 + 5) 5 - 5

乘开

(5 + 1) (5 + 5) : 65 + 10

(5 + 1) (5 + 5)

使用 FOIL 方法:

(a + b) (c + d) = a c + a d + b c + b d

a = 5, b = 1, c = 5, d = 5

= 5 \cdot 5 + 55 + 1 \cdot 5 + 1 \cdot 5

= 55 + 55 + 1 \cdot 5 + 1 \cdot 5

化简

55 + 55 + 1 \cdot 5 + 1 \cdot 5 : 65 + 10

55 + 55 + 1 \cdot 5 + 1 \cdot 5

同类项相加:

55 + 1 \cdot 5 = 65

= 65 + 55 + 1 \cdot 5

使用根式运算法则:

a a = a

55 = 5

= 65 + 5 + 1 \cdot 5

数字相乘:

1 \cdot 5 = 5

= 65 + 5 + 5

数字相加:

5 + 5 = 10

= 65 + 10

= 65 + 10

= 2 5 - 5 (65 + 10)

乘开

2 5 - 5 (65 + 10) : 610 5 - 5 + 102 5 - 5

2 5 - 5 (65 + 10)

使用分配律:

a (b + c) = ab + a c

a = 2 5 - 5, b = 65, c = 10

= 2 5 - 5 \cdot 65 + 2 5 - 5 \cdot 10

= 625 5 - 5 + 102 5 - 5

625 5 - 5 = 610 5 - 5

625 5 - 5

使用根式运算法则:

a b = a \cdot b

25 5 - 5 = 2 \cdot 5 (5 - 5)

= 6 2 \cdot 5 (5 - 5)

数字相乘:

2 \cdot 5 = 10

= 6 10 (5 - 5)

使用根式运算法则:

n ab = n a n b,

假定

a \geq 0, b \geq 0

10 (5 - 5) = 10 5 - 5

= 610 5 - 5

= 610 5 - 5 + 102 5 - 5

= 610 5 - 5 + 102 5 - 5

2 (5 - 5) (5 + 5) = 40

2 (5 - 5) (5 + 5)

乘开

(5 - 5) (5 + 5) : 20

(5 - 5) (5 + 5)

使用平方差公式:

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

a = 5, b = 5

= 5^{2} - (5)^{2}

化简

5^{2} - (5)^{2} : 20

5^{2} - (5)^{2}

5^{2} = 25

5^{2}

5^{2} = 25

= 25

(5)^{2} = 5

(5)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (5^{\frac{1}{2}})^{2}

使用指数法则:

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

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

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= 5

= 25 - 5

数字相减:

25 - 5 = 20

= 20

= 20

= 2 \cdot 20

乘开

2 \cdot 20 : 40

2 \cdot 20

打开括号

= 2 \cdot 20

数字相乘:

2 \cdot 20 = 40

= 40

= 40

= \frac{6 10 5 - 5 + 10 2 5 - 5}{40}

分解

610 5 - 5 + 102 5 - 5 : 2 5 - 5 (310 + 52)

610 5 - 5 + 102 5 - 5

改写为

= 3 \cdot 2 5 - 5 10 + 5 \cdot 2 5 - 5 2

因式分解出通项

2 5 - 5

= 2 5 - 5 (310 + 52)

= \frac{2 5 - 5 ( 3 10 + 5 2 )}{40}

约分:

2

= \frac{( 3 10 + 5 2 ) 5 - 5}{20}

= \frac{( 3 10 + 5 2 ) 5 - 5}{20}

= \frac{( 3 10 + 5 2 ) 5 - 5}{20}

= - \frac{( 3 10 + 5 2 ) 5 - 5}{20}

- \frac{( 3 10 + 5 2 ) 5 - 5}{20} - sec (x) = 1.5

将

\frac{( 3 10 + 5 2 ) 5 - 5}{20}

到右边

- \frac{( 3 10 + 5 2 ) 5 - 5}{20} - sec (x) = 1.5

两边加上

\frac{( 3 10 + 5 2 ) 5 - 5}{20}

- \frac{( 3 10 + 5 2 ) 5 - 5}{20} - sec (x) + \frac{( 3 10 + 5 2 ) 5 - 5}{20} = 1.5 + \frac{( 3 10 + 5 2 ) 5 - 5}{20}

化简

- \frac{( 3 10 + 5 2 ) 5 - 5}{20} - sec (x) + \frac{( 3 10 + 5 2 ) 5 - 5}{20} = 1.5 + \frac{( 3 10 + 5 2 ) 5 - 5}{20}

化简

- \frac{( 3 10 + 5 2 ) 5 - 5}{20} - sec (x) + \frac{( 3 10 + 5 2 ) 5 - 5}{20} : - sec (x)

- \frac{( 3 10 + 5 2 ) 5 - 5}{20} - sec (x) + \frac{( 3 10 + 5 2 ) 5 - 5}{20}

同类项相加:

- \frac{( 3 10 + 5 2 ) 5 - 5}{20} + \frac{( 3 10 + 5 2 ) 5 - 5}{20} = 0

= - sec (x)

化简

1.5 + \frac{( 3 10 + 5 2 ) 5 - 5}{20} : 2.87638 \dots

1.5 + \frac{( 3 10 + 5 2 ) 5 - 5}{20}

\frac{( 3 10 + 5 2 ) 5 - 5}{20} = \frac{16.55790 \dots 5 - 5}{20}

\frac{( 3 10 + 5 2 ) 5 - 5}{20}

310 = 9.48683 \dots

310

转换为小数形式

10 = 3.16227 \dots

= 3 \cdot 3.16227 \dots

数字相乘:

3 \cdot 3.16227 \dots = 9.48683 \dots

= 9.48683 \dots

52 = 7.07106 \dots

52

转换为小数形式

2 = 1.41421 \dots

= 5 \cdot 1.41421 \dots

数字相乘:

5 \cdot 1.41421 \dots = 7.07106 \dots

= 7.07106 \dots

= \frac{( 7.07106 \dots + 9.48683 \dots ) 5 - 5}{20}

数字相加:

9.48683 \dots + 7.07106 \dots = 16.55790 \dots

= \frac{16.55790 \dots 5 - 5}{20}

= 1.5 + \frac{16.55790 \dots 5 - 5}{20}

\frac{16.55790 \dots 5 - 5}{20} = 1.37638 \dots

\frac{16.55790 \dots 5 - 5}{20}

转换为小数形式

5 - 5 = 1.66250 \dots

= \frac{1.66250 \dots \cdot 16.55790 \dots}{20}

数字相乘:

16.55790 \dots \cdot 1.66250 \dots = 27.52763 \dots

= \frac{27.52763 \dots}{20}

数字相除:

\frac{27.52763 \dots}{20} = 1.37638 \dots

= 1.37638 \dots

= 1.5 + 1.37638 \dots

数字相加:

1.5 + 1.37638 \dots = 2.87638 \dots

= 2.87638 \dots

- sec (x) = 2.87638 \dots

- sec (x) = 2.87638 \dots

- sec (x) = 2.87638 \dots

两边除以

- 1

- sec (x) = 2.87638 \dots

两边除以

- 1

\frac{- sec ( x )}{- 1} = \frac{2.87638 \dots}{- 1}

化简

sec (x) = - 2.87638 \dots

sec (x) = - 2.87638 \dots

使用反三角函数性质

sec (x) = - 2.87638 \dots

sec (x) = - 2.87638 \dots

的通解

sec (x) = - a \Rightarrow x = arcsec (- a) + 2 πn, x = - arcsec (- a) + 2 πn

x = arcsec (- 2.87638 \dots) + 2 πn, x = - arcsec (- 2.87638 \dots) + 2 πn

x = arcsec (- 2.87638 \dots) + 2 πn, x = - arcsec (- 2.87638 \dots) + 2 πn

以小数形式表示解

x = 1.92586 \dots + 2 πn, x = - 1.92586 \dots + 2 πn

作图

流行的例子

1/(cos^2(x))+(1/(2cos(x)))=0

\frac{1}{cos ^{2} ( x )} + (\frac{1}{2 cos ( x )}) = 0

cos(α)=sqrt((1+1/5)/2)

cos (α) = \frac{1 + \frac{1}{5}}{2}

(tan(x))(sin^2(x)-a)(sec(x)-a)=0,0<a<1

(tan (x)) (sin^{2} (x) - a) (sec (x) - a) = 0, 0 < a < 1

sin(pi/2 (x+1))= 2/5

sin (\frac{π}{2} (x + 1)) = \frac{2}{5}

sin(*+pi/2)=cos(x)

sin (\cdot + \frac{π}{2}) = cos (x)