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受欢迎的三角函数 >

sin(195)cos(75)

解答

sin (19 5^{\circ}) cos (7 5^{\circ})

解答

\frac{- 2 + 3}{4}

+1

十进制

- 0.06698 \dots

求解步骤

sin (19 5^{\circ}) cos (7 5^{\circ})

使用三角恒等式改写:

\frac{sin ( 27 0 ^{\circ} ) + sin ( 12 0 ^{\circ} )}{2}

sin (19 5^{\circ}) cos (7 5^{\circ})

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 (19 5^{\circ} + 7 5^{\circ}) + sin (19 5^{\circ} - 7 5^{\circ}))

化简

= \frac{sin ( 27 0 ^{\circ} ) + sin ( 12 0 ^{\circ} )}{2}

= \frac{sin ( 27 0 ^{\circ} ) + sin ( 12 0 ^{\circ} )}{2}

使用三角恒等式改写:

sin (27 0^{\circ}) = - 1

sin (27 0^{\circ})

使用三角恒等式改写:

sin (18 0^{\circ}) cos (9 0^{\circ}) + cos (18 0^{\circ}) sin (9 0^{\circ})

sin (27 0^{\circ})

将

sin (27 0^{\circ})

写为

sin (18 0^{\circ} + 9 0^{\circ})

= sin (18 0^{\circ} + 9 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 (9 0^{\circ}) + cos (18 0^{\circ}) sin (9 0^{\circ})

= sin (18 0^{\circ}) cos (9 0^{\circ}) + cos (18 0^{\circ}) sin (9 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 (9 0^{\circ}) = 0

cos (9 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}

= 0

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 (9 0^{\circ}) = 1

sin (9 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}

= 1

= 0 \cdot 0 + (- 1) \cdot 1

化简

= - 1

使用以下普通恒等式:

sin (12 0^{\circ}) = \frac{3}{2}

sin (12 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}

= \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}

流行的例子

(8.3)/(cos(32.11))

\frac{8.3}{cos ( 32.1 1 ^{\circ} )}

tan^{2} (4 5^{\circ} - \frac{40}{2})

arccos(-3/(sqrt(13)))

arccos (- \frac{3}{13})

tan (4 0^{\circ}) \cdot 15

arctan (\frac{28}{19})