zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

sin((5pi)/(12))sin(pi/(12))

解答

sin (\frac{5 π}{12}) sin (\frac{π}{12})

解答

\frac{1}{4}

+1

十进制

0.25

求解步骤

sin (\frac{5 π}{12}) sin (\frac{π}{12})

使用三角恒等式改写:

\frac{cos ( \frac{π}{3} ) - cos ( \frac{π}{2} )}{2}

sin (\frac{5 π}{12}) sin (\frac{π}{12})

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 (\frac{5 π}{12} - \frac{π}{12}) - cos (\frac{5 π}{12} + \frac{π}{12}))

化简:

\frac{5 π}{12} - \frac{π}{12} = \frac{π}{3}

\frac{5 π}{12} - \frac{π}{12}

使用法则

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

= \frac{5 π - π}{12}

同类项相加:

5 π - π = 4 π

= \frac{4 π}{12}

约分:

4

= \frac{π}{3}

化简:

\frac{5 π}{12} + \frac{π}{12} = \frac{π}{2}

\frac{5 π}{12} + \frac{π}{12}

使用法则

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

= \frac{5 π + π}{12}

同类项相加:

5 π + π = 6 π

= \frac{6 π}{12}

约分:

6

= \frac{π}{2}

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

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

分式相乘:

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

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

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

1 \cdot (cos (\frac{π}{3}) - cos (\frac{π}{2}))

乘以:

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

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

去除括号:

(a) = a

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

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

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

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

使用以下普通恒等式:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

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{1}{2}

使用以下普通恒等式:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

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}

= 0

= \frac{\frac{1}{2} - 0}{2}

化简

\frac{\frac{1}{2} - 0}{2} : \frac{1}{4}

\frac{\frac{1}{2} - 0}{2}

\frac{1}{2} - 0 = \frac{1}{2}

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

使用分式法则:

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

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

数字相乘:

2 \cdot 2 = 4

= \frac{1}{4}

= \frac{1}{4}

流行的例子

1 + tan^{2} (0)

tan (\frac{6}{4})

arctan(sqrt((1-cos(19))/(1+cos(19))))

arctan (\frac{1 - cos ( 1 9 ^{\circ} )}{1 + cos ( 1 9 ^{\circ} )})

\frac{11}{sin ( 3 4 ^{\circ} )}

arcsin (\frac{18}{25})