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受欢迎的三角函数 >

(50)/(sin(105))=(20)/(sin(x))

解答

\frac{50}{sin ( 10 5 ^{\circ} )} = \frac{20}{sin ( x )}

解答

x = 0.39669 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.39669 \dots + 36 0^{\circ} n

+1

弧度

x = 0.39669 \dots + 2 πn, x = π - 0.39669 \dots + 2 πn

求解步骤

\frac{50}{sin ( 10 5 ^{\circ} )} = \frac{20}{sin ( x )}

sin (10 5^{\circ}) = \frac{2 ( 3 + 1 )}{4}

sin (10 5^{\circ})

使用三角恒等式改写:

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

sin (10 5^{\circ})

将

sin (10 5^{\circ})

写为

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

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

使用角和恒等式:

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

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

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

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:

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:

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:

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

化简

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

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

因式分解出通项

\frac{2}{2}

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

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

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

使用法则

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

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

数字相乘:

2 \cdot 2 = 4

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

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

\frac{50}{\frac{2 ( 3 + 1 )}{4}} = \frac{20}{sin ( x )}

使用分式交叉相乘

\frac{50}{\frac{2 ( 3 + 1 )}{4}} = \frac{20}{sin ( x )}

使用分式交叉相乘: 若

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

则

a \cdot d = b \cdot c

50 sin (x) = \frac{2 ( 3 + 1 )}{4} \cdot 20

化简

\frac{2 ( 3 + 1 )}{4} \cdot 20 : 52 (1 + 3)

\frac{2 ( 3 + 1 )}{4} \cdot 20

分式相乘:

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

= \frac{2 ( 3 + 1 ) \cdot 20}{4}

数字相除:

\frac{20}{4} = 5

= 52 (1 + 3)

50 sin (x) = 52 (1 + 3)

50 sin (x) = 52 (1 + 3)

两边除以

50

50 sin (x) = 52 (1 + 3)

两边除以

50

\frac{50 sin ( x )}{50} = \frac{5 2 ( 1 + 3 )}{50}

化简

sin (x) = \frac{2 ( 1 + 3 )}{10}

sin (x) = \frac{2 ( 1 + 3 )}{10}

验证解

找到无定义的点（奇点）:

sin (x) = 0

取

\frac{20}{sin ( x )}

的分母，令其等于零

sin (x) = 0

以下点无定义

sin (x) = 0

将不在定义域的点与解相综合:

sin (x) = \frac{2 ( 1 + 3 )}{10}

使用反三角函数性质

sin (x) = \frac{2 ( 1 + 3 )}{10}

sin (x) = \frac{2 ( 1 + 3 )}{10}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (a) + 36 0^{\circ} n

x = arcsin (\frac{2 ( 1 + 3 )}{10}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{2 ( 1 + 3 )}{10}) + 36 0^{\circ} n

x = arcsin (\frac{2 ( 1 + 3 )}{10}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{2 ( 1 + 3 )}{10}) + 36 0^{\circ} n

以小数形式表示解

x = 0.39669 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.39669 \dots + 36 0^{\circ} n

作图

流行的例子

csc (θ) = - 1.343

csc(θ)=-3/2 ,pi<θ<(3pi)/2 ,sin(θ)

csc (θ) = - \frac{3}{2}, π < θ < \frac{3 π}{2}, sin (θ)

sin (2 x) = \frac{- 3}{2}

1 = 4 sin (x)

tan (x) = \frac{45}{67}