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受欢迎的三角函数 >

(tan((5pi)/8)-tan((3pi)/8))/(1+tan((5pi)/8)tan((3pi)/8))

解答

\frac{tan ( \frac{5 π}{8} ) - tan ( \frac{3 π}{8} )}{1 + tan ( \frac{5 π}{8} ) tan ( \frac{3 π}{8} )}

解答

1

求解步骤

\frac{tan ( \frac{5 π}{8} ) - tan ( \frac{3 π}{8} )}{1 + tan ( \frac{5 π}{8} ) tan ( \frac{3 π}{8} )}

\frac{tan ( \frac{5 π}{8} ) - tan ( \frac{3 π}{8} )}{1 + tan ( \frac{5 π}{8} ) tan ( \frac{3 π}{8} )} = \frac{tan ( \frac{3 π}{8} ) - tan ( \frac{5 π}{8} )}{- tan ( \frac{5 π}{8} ) tan ( \frac{3 π}{8} ) - 1}

\frac{tan ( \frac{5 π}{8} ) - tan ( \frac{3 π}{8} )}{1 + tan ( \frac{5 π}{8} ) tan ( \frac{3 π}{8} )}

使用分式法则:

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

\frac{tan ( \frac{5 π}{8} ) - tan ( \frac{3 π}{8} )}{1 + tan ( \frac{5 π}{8} ) tan ( \frac{3 π}{8} )} = \frac{- ( tan ( \frac{3 π}{8} ) - tan ( \frac{5 π}{8} ) )}{- ( - tan ( \frac{5 π}{8} ) tan ( \frac{3 π}{8} ) - 1 )}

= \frac{tan ( \frac{3 π}{8} ) - tan ( \frac{5 π}{8} )}{- tan ( \frac{5 π}{8} ) tan ( \frac{3 π}{8} ) - 1}

= \frac{tan ( \frac{3 π}{8} ) - tan ( \frac{5 π}{8} )}{- tan ( \frac{5 π}{8} ) tan ( \frac{3 π}{8} ) - 1}

使用三角恒等式改写:

tan (\frac{5 π}{8}) = - tan (\frac{3 π}{8})

tan (\frac{5 π}{8})

利用以下特性:

tan (x) = - tan (π - x)

tan (x)

利用以下特性:

tan (θ) = - tan (- θ)

tan (x) = - tan (- x)

= - tan (- x)

使用周期

tan

:

tan (π + θ) = tan (θ)

- tan (- x) = - tan (π - x)

= - tan (π - x)

= - tan (2 π - \frac{5 π}{8})

化简:

2 π - \frac{5 π}{8} = \frac{11 π}{8}

2 π - \frac{5 π}{8}

将项转换为分式:

2 π = \frac{2 π 8}{8}

= \frac{2 π 8}{8} - \frac{5 π}{8}

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{2 π 8 - 5 π}{8}

2 π 8 - 5 π = 11 π

2 π 8 - 5 π

数字相乘:

2 \cdot 8 = 16

= 16 π - 5 π

同类项相加:

16 π - 5 π = 11 π

= 11 π

= \frac{11 π}{8}

= - tan (\frac{11 π}{8})

tan (\frac{11 π}{8}) = tan (\frac{3 π}{8})

tan (\frac{11 π}{8})

将

\frac{11 π}{8}

改写为

π + \frac{3 π}{8}

= tan (π + \frac{3 π}{8})

使用周期

tan

:

tan (x + π) = tan (x)

tan (π + \frac{3 π}{8}) = tan (\frac{3 π}{8})

= tan (\frac{3 π}{8})

= - tan (\frac{3 π}{8})

= \frac{tan ( \frac{3 π}{8} ) - ( - tan ( \frac{3 π}{8} ) )}{- ( - tan ( \frac{3 π}{8} ) ) tan ( \frac{3 π}{8} ) - 1}

\frac{tan ( \frac{3 π}{8} ) - ( - tan ( \frac{3 π}{8} ) )}{- ( - tan ( \frac{3 π}{8} ) ) tan ( \frac{3 π}{8} ) - 1} = \frac{2 tan ( \frac{3 π}{8} )}{tan ^{2} ( \frac{3 π}{8} ) - 1}

\frac{tan ( \frac{3 π}{8} ) - ( - tan ( \frac{3 π}{8} ) )}{- ( - tan ( \frac{3 π}{8} ) ) tan ( \frac{3 π}{8} ) - 1}

使用法则

- (- a) = a

= \frac{tan ( \frac{3 π}{8} ) + tan ( \frac{3 π}{8} )}{tan ( \frac{3 π}{8} ) tan ( \frac{3 π}{8} ) - 1}

同类项相加:

tan (\frac{3 π}{8}) + tan (\frac{3 π}{8}) = 2 tan (\frac{3 π}{8})

= \frac{2 tan ( \frac{3 π}{8} )}{tan ( \frac{3 π}{8} ) tan ( \frac{3 π}{8} ) - 1}

tan (\frac{3 π}{8}) tan (\frac{3 π}{8}) - 1 = tan^{2} (\frac{3 π}{8}) - 1

tan (\frac{3 π}{8}) tan (\frac{3 π}{8}) - 1

tan (\frac{3 π}{8}) tan (\frac{3 π}{8}) = tan^{2} (\frac{3 π}{8})

tan (\frac{3 π}{8}) tan (\frac{3 π}{8})

使用指数法则:

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

tan (\frac{3 π}{8}) tan (\frac{3 π}{8}) = tan^{1 + 1} (\frac{3 π}{8})

= tan^{1 + 1} (\frac{3 π}{8})

数字相加:

1 + 1 = 2

= tan^{2} (\frac{3 π}{8})

= tan^{2} (\frac{3 π}{8}) - 1

= \frac{2 tan ( \frac{3 π}{8} )}{tan ^{2} ( \frac{3 π}{8} ) - 1}

= \frac{2 tan ( \frac{3 π}{8} )}{tan ^{2} ( \frac{3 π}{8} ) - 1}

使用三角恒等式改写:

tan (\frac{3 π}{8}) = 3 + 22

tan (\frac{3 π}{8})

使用三角恒等式改写:

\frac{1 - cos ( \frac{3 π}{4} )}{1 + cos ( \frac{3 π}{4} )}

tan (\frac{3 π}{8})

将

tan (\frac{3 π}{8})

写为

tan (\frac{\frac{3 π}{4}}{2})

= tan (\frac{\frac{3 π}{4}}{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{3 π}{4} )}{1 + cos ( \frac{3 π}{4} )}

= \frac{1 - cos ( \frac{3 π}{4} )}{1 + cos ( \frac{3 π}{4} )}

使用以下普通恒等式:

cos (\frac{3 π}{4}) = - \frac{2}{2}

cos (\frac{3 π}{4})

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - \frac{2}{2}

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

化简

\frac{1 - ( - \frac{2}{2} )}{1 - \frac{2}{2}} : 3 + 22

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

使用法则

- (- a) = a

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

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

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

化简

1 - \frac{2}{2} : \frac{2 - 2}{2}

1 - \frac{2}{2}

将项转换为分式:

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

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

数字相乘:

1 \cdot 2 = 2

= \frac{2 - 2}{2}

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

化简

1 + \frac{2}{2} : \frac{2 + 2}{2}

1 + \frac{2}{2}

将项转换为分式:

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

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

数字相乘:

1 \cdot 2 = 2

= \frac{2 + 2}{2}

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

分式相除:

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

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

约分:

2

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

分解

2 + 2 : 2 (2 + 1)

2 + 2

2 = 22

= 22 + 2

因式分解出通项

2

= 2 (2 + 1)

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

分解

2 - 2 : 2 (2 - 1)

2 - 2

2 = 22

= 22 - 2

因式分解出通项

2

= 2 (2 - 1)

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

约分:

2

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

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

\frac{2 + 1}{2 - 1} = 3 + 22

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

乘以共轭根式

\frac{2 + 1}{2 + 1}

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

(2 + 1) (2 + 1) = 3 + 22

(2 + 1) (2 + 1)

使用指数法则:

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

(2 + 1) (2 + 1) = (2 + 1)^{1 + 1}

= (2 + 1)^{1 + 1}

数字相加:

1 + 1 = 2

= (2 + 1)^{2}

使用完全平方公式:

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

a = 2, b = 1

= (2)^{2} + 22 \cdot 1 + 1^{2}

化简

(2)^{2} + 22 \cdot 1 + 1^{2} : 3 + 22

(2)^{2} + 22 \cdot 1 + 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

= (2)^{2} + 2 \cdot 1 \cdot 2 + 1

(2)^{2} = 2

(2)^{2}

使用根式运算法则:

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

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

使用指数法则:

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

= 2^{\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

= 2

22 \cdot 1 = 22

22 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 22

= 2 + 22 + 1

数字相加:

2 + 1 = 3

= 3 + 22

= 3 + 22

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

(2 - 1) (2 + 1)

使用平方差公式:

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

a = 2, b = 1

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

= (2)^{2} - 1

(2)^{2} = 2

(2)^{2}

使用根式运算法则:

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

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

使用指数法则:

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

= 2^{\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

= 2

= 2 - 1

数字相减:

2 - 1 = 1

= 1

= 1

= \frac{3 + 2 2}{1}

使用法则

\frac{a}{1} = a

= 3 + 22

= 3 + 22

= 3 + 22

= \frac{2 3 + 2 2}{( 3 + 2 2 ) ^{2} - 1}

化简

\frac{2 3 + 2 2}{( 3 + 2 2 ) ^{2} - 1} : 1

\frac{2 3 + 2 2}{( 3 + 2 2 ) ^{2} - 1}

分解

(3 + 22)^{2} - 1 : 2 (2 + 2)

(3 + 22)^{2} - 1

将

1

改写为

1^{2}

= (3 + 22)^{2} - 1^{2}

使用平方差公式:

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

(3 + 22)^{2} - 1^{2} = (3 + 22 + 1) (3 + 22 - 1)

= (3 + 22 + 1) (3 + 22 - 1)

(3 + 22 + 1) (3 + 22 - 1) = 2 (2 + 2)

(3 + 22 + 1) (3 + 22 - 1)

3 + 22 = 2 + 1

3 + 22

= 2 + 22 + 1

= (2)^{2} + 22 + (1)^{2}

1 = 1

1

使用法则

1 = 1

= 1

= (2)^{2} + 22 + 1^{2}

22 \cdot 1 = 22

22 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 22

= (2)^{2} + 22 \cdot 1 + 1^{2}

使用完全平方公式:

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

(2)^{2} + 22 \cdot 1 + 1^{2} = (2 + 1)^{2}

= (2 + 1)^{2}

使用根式运算法则:

n a^{n} = a

(2 + 1)^{2} = 2 + 1

= 2 + 1

= (1 + 1 + 2) (3 + 22 - 1)

3 + 22 = 2 + 1

3 + 22

= 2 + 22 + 1

= (2)^{2} + 22 + (1)^{2}

1 = 1

1

使用法则

1 = 1

= 1

= (2)^{2} + 22 + 1^{2}

22 \cdot 1 = 22

22 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 22

= (2)^{2} + 22 \cdot 1 + 1^{2}

使用完全平方公式:

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

(2)^{2} + 22 \cdot 1 + 1^{2} = (2 + 1)^{2}

= (2 + 1)^{2}

使用根式运算法则:

n a^{n} = a

(2 + 1)^{2} = 2 + 1

= 2 + 1

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

数字相加:

1 + 1 = 2

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

1 - 1 = 0

= 2 (2 + 2)

= 2 (2 + 2)

= \frac{2 3 + 2 2}{2 ( 2 + 2 )}

消掉

\frac{2 3 + 2 2}{2 ( 2 + 2 )} : \frac{2 3 + 2 2}{2 + 2}

\frac{2 3 + 2 2}{2 ( 2 + 2 )}

使用根式运算法则:

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

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

= \frac{2 3 + 2 2}{2 ^{\frac{1}{2}} ( 2 + 2 )}

使用指数法则:

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

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

= \frac{2 ^{- \frac{1}{2} + 1} 3 + 2 2}{2 + 2}

数字相减:

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

= \frac{2 ^{\frac{1}{2}} 3 + 2 2}{2 + 2}

使用根式运算法则:

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

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

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

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

分解

2 + 2 : 2 (2 + 1)

2 + 2

2 = 22

= 22 + 2

因式分解出通项

2

= 2 (2 + 1)

= \frac{2 3 + 2 2}{2 ( 2 + 1 )}

约分:

2

= \frac{3 + 2 2}{2 + 1}

3 + 22 = 2 + 1

3 + 22

= 2 + 22 + 1

= (2)^{2} + 22 + (1)^{2}

1 = 1

1

使用法则

1 = 1

= 1

= (2)^{2} + 22 + 1^{2}

22 \cdot 1 = 22

22 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 22

= (2)^{2} + 22 \cdot 1 + 1^{2}

使用完全平方公式:

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

(2)^{2} + 22 \cdot 1 + 1^{2} = (2 + 1)^{2}

= (2 + 1)^{2}

使用根式运算法则:

n a^{n} = a

(2 + 1)^{2} = 2 + 1

= 2 + 1

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

使用法则

\frac{a}{a} = 1

= 1

= 1

流行的例子

tan (\frac{15 π}{6})

arctan(csc(pi/2))

arctan (csc (\frac{π}{2}))

arctan (\frac{- 2}{6})

sin (0.35)

180-arctan(4/3)

180 - arctan (\frac{4}{3})