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受欢迎的三角函数 >

(tan(315)-cos(780))/(tan(-135)-sin(210))

解答

\frac{tan ( 31 5 ^{\circ} ) - cos ( 78 0 ^{\circ} )}{tan ( - 13 5 ^{\circ} ) - sin ( 21 0 ^{\circ} )}

解答

- 1

求解步骤

\frac{tan ( 31 5 ^{\circ} ) - cos ( 78 0 ^{\circ} )}{tan ( - 13 5 ^{\circ} ) - sin ( 21 0 ^{\circ} )}

利用以下特性:

tan (- x) = - tan (x)

tan (- 13 5^{\circ}) = - tan (13 5^{\circ})

= \frac{tan ( 31 5 ^{\circ} ) - cos ( 78 0 ^{\circ} )}{- tan ( 13 5 ^{\circ} ) - sin ( 21 0 ^{\circ} )}

tan (31 5^{\circ}) = tan (13 5^{\circ})

tan (31 5^{\circ})

将

31 5^{\circ}

改写为

18 0^{\circ} + 13 5^{\circ}

= tan (18 0^{\circ} + 13 5^{\circ})

使用周期

tan

:

tan (x + 18 0^{\circ}) = tan (x)

tan (18 0^{\circ} + 13 5^{\circ}) = tan (13 5^{\circ})

= tan (13 5^{\circ})

= \frac{tan ( 13 5 ^{\circ} ) - cos ( 78 0 ^{\circ} )}{- tan ( 13 5 ^{\circ} ) - sin ( 21 0 ^{\circ} )}

cos (78 0^{\circ}) = cos (6 0^{\circ})

cos (78 0^{\circ})

将

78 0^{\circ}

改写为

36 0^{\circ} \cdot 2 + 6 0^{\circ}

= cos (36 0^{\circ} 2 + 6 0^{\circ})

使用周期

cos

:

cos (x + 36 0^{\circ} \cdot k) = cos (x)

cos (36 0^{\circ} \cdot 2 + 6 0^{\circ}) = cos (6 0^{\circ})

= cos (6 0^{\circ})

= \frac{tan ( 13 5 ^{\circ} ) - cos ( 6 0 ^{\circ} )}{- tan ( 13 5 ^{\circ} ) - sin ( 21 0 ^{\circ} )}

使用三角恒等式改写:

tan (13 5^{\circ}) = - 1

tan (13 5^{\circ})

使用三角恒等式改写:

\frac{sin ( 13 5 ^{\circ} )}{cos ( 13 5 ^{\circ} )}

tan (13 5^{\circ})

使用基本三角恒等式:

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

= \frac{sin ( 13 5 ^{\circ} )}{cos ( 13 5 ^{\circ} )}

= \frac{sin ( 13 5 ^{\circ} )}{cos ( 13 5 ^{\circ} )}

使用以下普通恒等式:

sin (13 5^{\circ}) = \frac{2}{2}

sin (13 5^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

使用以下普通恒等式:

cos (13 5^{\circ}) = - \frac{2}{2}

cos (13 5^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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{\frac{2}{2}}{- \frac{2}{2}}

化简

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

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

使用分式法则:

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

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

分式相除:

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

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

约分:

2

= - \frac{2}{2}

约分:

2

= - 1

= - 1

使用以下普通恒等式:

cos (6 0^{\circ}) = \frac{1}{2}

cos (6 0^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{1}{2}

使用三角恒等式改写:

sin (21 0^{\circ}) = - \frac{1}{2}

sin (21 0^{\circ})

使用三角恒等式改写:

sin (18 0^{\circ}) cos (3 0^{\circ}) + cos (18 0^{\circ}) sin (3 0^{\circ})

sin (21 0^{\circ})

将

sin (21 0^{\circ})

写为

sin (18 0^{\circ} + 3 0^{\circ})

= sin (18 0^{\circ} + 3 0^{\circ})

使用角和恒等式:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (18 0^{\circ}) cos (3 0^{\circ}) + cos (18 0^{\circ}) sin (3 0^{\circ})

= sin (18 0^{\circ}) cos (3 0^{\circ}) + cos (18 0^{\circ}) sin (3 0^{\circ})

使用以下普通恒等式:

sin (18 0^{\circ}) = 0

sin (18 0^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

使用以下普通恒等式:

cos (3 0^{\circ}) = \frac{3}{2}

cos (3 0^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

使用以下普通恒等式:

cos (18 0^{\circ}) = (- 1)

cos (18 0^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= (- 1)

使用以下普通恒等式:

sin (3 0^{\circ}) = \frac{1}{2}

sin (3 0^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

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

化简

= - \frac{1}{2}

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

化简

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

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

使用法则

- (- a) = a

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

化简

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

1 + \frac{1}{2}

将项转换为分式:

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

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

1 \cdot 2 + 1 = 3

1 \cdot 2 + 1

数字相乘:

1 \cdot 2 = 2

= 2 + 1

数字相加:

2 + 1 = 3

= 3

= \frac{3}{2}

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

化简

- 1 - \frac{1}{2} : - \frac{3}{2}

- 1 - \frac{1}{2}

将项转换为分式:

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

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

- 1 \cdot 2 - 1 = - 3

- 1 \cdot 2 - 1

数字相乘:

1 \cdot 2 = 2

= - 2 - 1

数字相减:

- 2 - 1 = - 3

= - 3

= \frac{- 3}{2}

使用分式法则:

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

= - \frac{3}{2}

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

使用分式法则:

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

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

分式相除:

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

= - \frac{3 \cdot 2}{2 \cdot 3}

约分:

3

= - \frac{2}{2}

约分:

2

= - 1

= - 1

流行的例子

2 - 2 cosh (3 - 2)

arctan (0.99)

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