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受欢迎的三角函数 >

tan(pi/(10))

解答

tan (\frac{π}{10})

解答

\frac{5 - 2 5}{5}

+1

十进制

0.32491 \dots

求解步骤

tan (\frac{π}{10})

使用三角恒等式改写:

\frac{1 - cos ( \frac{π}{5} )}{1 + cos ( \frac{π}{5} )}

tan (\frac{π}{10})

将

tan (\frac{π}{10})

写为

tan (\frac{\frac{π}{5}}{2})

= tan (\frac{\frac{π}{5}}{2})

使用半角公式:

tan (\frac{θ}{2}) = \frac{1 - cos ( θ )}{1 + cos ( θ )}

使用三角恒等式改写:

tan^{2} (θ) = \frac{1 - cos ( 2 θ )}{1 + cos ( 2 θ )}

利用以下特性

tan (θ) = \frac{sin ( θ )}{cos ( θ )}

两边进行平方

tan^{2} (θ) = \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

使用三角恒等式改写:

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

使用倍角公式

cos (2 θ) = 1 - 2 sin^{2} (θ)

交换两边

2 sin^{2} (θ) - 1 = - cos (2 θ)

两边加上

1

2 sin^{2} (θ) = 1 - cos (2 θ)

两边除以

2

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

使用三角恒等式改写:

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

使用倍角公式

cos (2 θ) = 2 cos^{2} (θ) - 1

交换两边

2 cos^{2} (θ) - 1 = cos (2 θ)

两边加上

1

2 sin^{2} (θ) = 1 + cos (2 θ)

两边除以

2

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

tan^{2} (θ) = \frac{\frac{1 - c o s ( 2 θ )}{2}}{\frac{1 + c o s ( 2 θ )}{2}}

化简

tan^{2} (θ) = \frac{1 - cos ( 2 θ )}{1 + cos ( 2 θ )}

用

\frac{θ}{2}

替代

θ

tan^{2} (\frac{θ}{2}) = \frac{1 - cos ( 2 \cdot \frac{θ}{2} )}{1 + cos ( 2 \cdot \frac{θ}{2} )}

化简

tan^{2} (\frac{θ}{2}) = \frac{1 - cos ( θ )}{1 + cos ( θ )}

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

:

range [0, \frac{π}{2}] [\frac{π}{2}, π] quadrant I II tan positive negative

tan (\frac{θ}{2}) = \frac{1 - cos ( θ )}{1 + cos ( θ )}

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

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

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

化简

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

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

\frac{1 - \frac{5 + 1}{4}}{1 + \frac{5 + 1}{4}} = \frac{3 - 5}{5 + 5}

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

化简

1 + \frac{5 + 1}{4} : \frac{5 + 5}{4}

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

将项转换为分式:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} + \frac{5 + 1}{4}

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 \cdot 4 + 5 + 1}{4}

1 \cdot 4 + 5 + 1 = 5 + 5

1 \cdot 4 + 5 + 1

数字相乘:

1 \cdot 4 = 4

= 4 + 5 + 1

数字相加:

4 + 1 = 5

= 5 + 5

= \frac{5 + 5}{4}

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

化简

1 - \frac{5 + 1}{4} : \frac{3 - 5}{4}

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

将项转换为分式:

1 = \frac{1 \cdot 4}{4}

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

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 \cdot 4 - ( 5 + 1 )}{4}

数字相乘:

1 \cdot 4 = 4

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

乘开

4 - (5 + 1) : 3 - 5

4 - (5 + 1)

- (5 + 1) : - 5 - 1

- (5 + 1)

打开括号

= - (5) - (1)

使用加减运算法则

+ (- a) = - a

= - 5 - 1

= 4 - 5 - 1

数字相减:

4 - 1 = 3

= 3 - 5

= \frac{3 - 5}{4}

= \frac{\frac{3 - 5}{4}}{\frac{5 + 5}{4}}

分式相除:

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

= \frac{( 3 - 5 ) \cdot 4}{4 ( 5 + 5 )}

约分:

4

= \frac{3 - 5}{5 + 5}

= \frac{3 - 5}{5 + 5}

\frac{3 - 5}{5 + 5} = \frac{5 - 2 5}{5}

\frac{3 - 5}{5 + 5}

乘以共轭根式

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

= \frac{( 3 - 5 ) ( 5 - 5 )}{( 5 + 5 ) ( 5 - 5 )}

(3 - 5) (5 - 5) = 20 - 85

(3 - 5) (5 - 5)

使用 FOIL 方法:

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

a = 3, b = - 5, c = 5, d = - 5

= 3 \cdot 5 + 3 (- 5) + (- 5) \cdot 5 + (- 5) (- 5)

使用加减运算法则

+ (- a) = - a, (- a) (- b) = ab

= 3 \cdot 5 - 35 - 55 + 55

化简

3 \cdot 5 - 35 - 55 + 55 : 20 - 85

3 \cdot 5 - 35 - 55 + 55

同类项相加:

- 35 - 55 = - 85

= 3 \cdot 5 - 85 + 55

数字相乘:

3 \cdot 5 = 15

= 15 - 85 + 55

使用根式运算法则:

a a = a

55 = 5

= 15 - 85 + 5

数字相加:

15 + 5 = 20

= 20 - 85

= 20 - 85

(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

= \frac{20 - 8 5}{20}

分解

20 - 85 : 4 (5 - 25)

20 - 85

改写为

= 4 \cdot 5 - 4 \cdot 25

因式分解出通项

4

= 4 (5 - 25)

= \frac{4 ( 5 - 2 5 )}{20}

约分:

4

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

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

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

流行的例子

cos(arccos(2/3))

cos (arccos (\frac{2}{3}))

sin (3 \cdot \frac{π}{2})

(706.32)/(sin(45)1.5+cos(45)0.5)

\frac{706.32}{sin ( 4 5 ^{\circ} ) 1.5 + cos ( 4 5 ^{\circ} ) 0.5}

2(cos^2(15)-sin^2(15))

2 (cos^{2} (1 5^{\circ}) - sin^{2} (1 5^{\circ}))

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