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受欢迎的三角函数 >

(sin(45)sin(15))/(cos(135)cos(105))

解答

\frac{sin ( 4 5 ^{\circ} ) sin ( 1 5 ^{\circ} )}{cos ( 13 5 ^{\circ} ) cos ( 10 5 ^{\circ} )}

解答

1

求解步骤

\frac{sin ( 4 5 ^{\circ} ) sin ( 1 5 ^{\circ} )}{cos ( 13 5 ^{\circ} ) cos ( 10 5 ^{\circ} )}

使用三角恒等式改写:

sin (4 5^{\circ}) sin (1 5^{\circ}) = \frac{cos ( 3 0 ^{\circ} ) - cos ( 6 0 ^{\circ} )}{2}

sin (4 5^{\circ}) sin (1 5^{\circ})

使用积化和差恒等式:

sin (s) sin (t) = \frac{1}{2} (cos (s - t) - cos (s + t))

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

化简

= \frac{cos ( 3 0 ^{\circ} ) - cos ( 6 0 ^{\circ} )}{2}

= \frac{\frac{c o s ( 3 0 ^{\circ} ) - c o s ( 6 0 ^{\circ} )}{2}}{cos ( 13 5 ^{\circ} ) cos ( 10 5 ^{\circ} )}

化简

= \frac{cos ( 3 0 ^{\circ} ) - cos ( 6 0 ^{\circ} )}{2 cos ( 13 5 ^{\circ} ) cos ( 10 5 ^{\circ} )}

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 (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}

使用三角恒等式改写:

cos (10 5^{\circ}) = \frac{2 ( 1 - 3 )}{4}

cos (10 5^{\circ})

使用三角恒等式改写:

cos (6 0^{\circ}) cos (4 5^{\circ}) - sin (6 0^{\circ}) sin (4 5^{\circ})

cos (10 5^{\circ})

将

cos (10 5^{\circ})

写为

cos (6 0^{\circ} + 4 5^{\circ})

= cos (6 0^{\circ} + 4 5^{\circ})

使用角和恒等式:

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

= cos (6 0^{\circ}) cos (4 5^{\circ}) - sin (6 0^{\circ}) sin (4 5^{\circ})

= cos (6 0^{\circ}) cos (4 5^{\circ}) - sin (6 0^{\circ}) sin (4 5^{\circ})

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}

使用以下普通恒等式:

cos (4 5^{\circ}) = \frac{2}{2}

cos (4 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}

使用以下普通恒等式:

sin (6 0^{\circ}) = \frac{3}{2}

sin (6 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{3}{2}

使用以下普通恒等式:

sin (4 5^{\circ}) = \frac{2}{2}

sin (4 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}

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

化简

\frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2} : \frac{2 ( 1 - 3 )}{4}

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

因式分解出通项

\frac{2}{2}

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

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

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

使用法则

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

= \frac{1 - 3}{2}

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

分式相乘:

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

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

数字相乘:

2 \cdot 2 = 4

= \frac{2 ( 1 - 3 )}{4}

= \frac{2 ( 1 - 3 )}{4}

= \frac{\frac{3}{2} - \frac{1}{2}}{2 ( - \frac{2}{2} ) \frac{2 ( 1 - 3 )}{4}}

化简

\frac{\frac{3}{2} - \frac{1}{2}}{2 ( - \frac{2}{2} ) \frac{2 ( 1 - 3 )}{4}} : 1

\frac{\frac{3}{2} - \frac{1}{2}}{2 ( - \frac{2}{2} ) \frac{2 ( 1 - 3 )}{4}}

合并分式

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

使用法则

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

= \frac{3 - 1}{2}

= \frac{\frac{3 - 1}{2}}{2 ( - \frac{2}{2} ) \frac{2 ( 1 - 3 )}{4}}

去除括号:

(- a) = - a

= \frac{\frac{3 - 1}{2}}{- 2 \cdot \frac{2}{2} \cdot \frac{2 ( 1 - 3 )}{4}}

使用分式法则:

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

= - \frac{\frac{3 - 1}{2}}{2 \cdot \frac{2}{2} \cdot \frac{2 ( 1 - 3 )}{4}}

使用分式法则:

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

\frac{\frac{3 - 1}{2}}{2 \cdot \frac{2}{2} \cdot \frac{2 ( 1 - 3 )}{4}} = \frac{3 - 1}{2 \cdot 2 \cdot \frac{2}{2} \cdot \frac{2 ( 1 - 3 )}{4}}

= - \frac{3 - 1}{2 \cdot 2 \cdot \frac{2}{2} \cdot \frac{2 ( 1 - 3 )}{4}}

数字相乘:

2 \cdot 2 = 4

= - \frac{3 - 1}{4 \cdot \frac{2}{2} \cdot \frac{2 ( 1 - 3 )}{4}}

乘

4 \cdot \frac{2}{2} \cdot \frac{2 ( 1 - 3 )}{4} : 1 - 3

4 \cdot \frac{2}{2} \cdot \frac{2 ( 1 - 3 )}{4}

分式相乘:

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

= \frac{2 2 ( 1 - 3 ) \cdot 4}{2 \cdot 4}

约分:

4

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

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

22 (1 - 3)

使用根式运算法则:

a a = a

22 = 2

= 2 (1 - 3)

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

数字相除:

\frac{2}{2} = 1

= 1 - 3

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

1 - 3 = - (3 - 1)

= \frac{3 - 1}{- ( 3 - 1 )}

整理后得

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

约分:

3 - 1

= - (- 1)

使用法则

- (- a) = a

= 1

= 1

流行的例子

(sec^2(0)-tan^2(0))/(1+cot^2(0))

\frac{sec ^{2} ( 0 ) - tan ^{2} ( 0 )}{1 + cot ^{2} ( 0 )}

cos (20 5^{\circ})

sin (6 0^{\circ}) 6

sin^2(15)+cos^2(15)

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

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