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受欢迎的三角函数 >

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

解答

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

解答

\frac{2 + 3}{4}

+1

十进制

0.93301 \dots

求解步骤

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

使用三角恒等式改写:

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

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

使用积化和差恒等式:

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

= \frac{1}{2} (sin (\frac{5 π}{12} + \frac{π}{12}) + sin (\frac{5 π}{12} - \frac{π}{12}))

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}

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}

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

\frac{1}{2} (sin (\frac{π}{2}) + sin (\frac{π}{3}))

分式相乘:

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

= \frac{1 \cdot ( sin ( \frac{π}{2} ) + sin ( \frac{π}{3} ) )}{2}

1 \cdot (sin (\frac{π}{2}) + sin (\frac{π}{3})) = sin (\frac{π}{2}) + sin (\frac{π}{3})

1 \cdot (sin (\frac{π}{2}) + sin (\frac{π}{3}))

乘以:

1 \cdot (sin (\frac{π}{2}) + sin (\frac{π}{3})) = (sin (\frac{π}{2}) + sin (\frac{π}{3}))

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

去除括号:

(a) = a

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

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

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

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

使用以下普通恒等式:

sin (\frac{π}{2}) = 1

sin (\frac{π}{2})

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

= 1

使用以下普通恒等式:

sin (\frac{π}{3}) = \frac{3}{2}

sin (\frac{π}{3})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} 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}

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

化简

\frac{1 + \frac{3}{2}}{2} : \frac{2 + 3}{4}

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

化简

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

1 + \frac{3}{2}

将项转换为分式:

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

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

数字相乘:

1 \cdot 2 = 2

= \frac{2 + 3}{2}

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

使用分式法则:

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

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

数字相乘:

2 \cdot 2 = 4

= \frac{2 + 3}{4}

= \frac{2 + 3}{4}

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