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 >

solvefor x,(17)/(sin(x))=(23)/(sin(75))

解答

求解

x, \frac{17}{sin ( x )} = \frac{23}{sin ( 7 5 ^{\circ} )}

解答

x = 0.79511 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.79511 \dots + 36 0^{\circ} n

+1

弧度

x = 0.79511 \dots + 2 πn, x = π - 0.79511 \dots + 2 πn

求解步骤

\frac{17}{sin ( x )} = \frac{23}{sin ( 7 5 ^{\circ} )}

sin (7 5^{\circ}) = \frac{6 + 2}{4}

sin (7 5^{\circ})

使用三角恒等式改写:

sin (4 5^{\circ}) cos (3 0^{\circ}) + cos (4 5^{\circ}) sin (3 0^{\circ})

sin (7 5^{\circ})

将

sin (7 5^{\circ})

写为

sin (4 5^{\circ} + 3 0^{\circ})

= sin (4 5^{\circ} + 3 0^{\circ})

使用角和恒等式:

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

= sin (4 5^{\circ}) cos (3 0^{\circ}) + cos (4 5^{\circ}) sin (3 0^{\circ})

= sin (4 5^{\circ}) cos (3 0^{\circ}) + cos (4 5^{\circ}) sin (3 0^{\circ})

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}

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 (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 (3 0^{\circ}) = \frac{1}{2}

sin (3 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{1}{2}

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

化简

\frac{2}{2} \cdot \frac{3}{2} + \frac{2}{2} \cdot \frac{1}{2} : \frac{6 + 2}{4}

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

\frac{2}{2} \cdot \frac{3}{2} = \frac{6}{4}

\frac{2}{2} \cdot \frac{3}{2}

分式相乘:

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

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

数字相乘:

2 \cdot 2 = 4

= \frac{2 3}{4}

化简

23 : 6

23

使用根式运算法则:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

数字相乘:

2 \cdot 3 = 6

= 6

= \frac{6}{4}

\frac{2}{2} \cdot \frac{1}{2} = \frac{2}{4}

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

分式相乘:

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

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

乘以:

2 \cdot 1 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

= \frac{6}{4} + \frac{2}{4}

使用法则

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

= \frac{6 + 2}{4}

= \frac{6 + 2}{4}

\frac{17}{sin ( x )} = \frac{23}{\frac{6 + 2}{4}}

使用分式交叉相乘

\frac{17}{sin ( x )} = \frac{23}{\frac{6 + 2}{4}}

使用分式交叉相乘: 若

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

则

a \cdot d = b \cdot c

17 \cdot \frac{6 + 2}{4} = sin (x) \cdot 23

化简

17 \cdot \frac{6 + 2}{4} : \frac{17 ( 6 + 2 )}{4}

17 \cdot \frac{6 + 2}{4}

分式相乘:

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

= \frac{( 6 + 2 ) \cdot 17}{4}

\frac{17 ( 6 + 2 )}{4} = sin (x) \cdot 23

\frac{17 ( 6 + 2 )}{4} = sin (x) \cdot 23

交换两边

sin (x) \cdot 23 = \frac{17 ( 6 + 2 )}{4}

两边除以

23

sin (x) \cdot 23 = \frac{17 ( 6 + 2 )}{4}

两边除以

23

\frac{sin ( x ) \cdot 23}{23} = \frac{\frac{17 ( 6 + 2 )}{4}}{23}

化简

\frac{sin ( x ) \cdot 23}{23} = \frac{\frac{17 ( 6 + 2 )}{4}}{23}

化简

\frac{sin ( x ) \cdot 23}{23} : sin (x)

\frac{sin ( x ) \cdot 23}{23}

数字相除:

\frac{23}{23} = 1

= sin (x)

化简

\frac{\frac{17 ( 6 + 2 )}{4}}{23} : \frac{17 ( 6 + 2 )}{92}

\frac{\frac{17 ( 6 + 2 )}{4}}{23}

使用分式法则:

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

= \frac{17 ( 6 + 2 )}{4 \cdot 23}

数字相乘:

4 \cdot 23 = 92

= \frac{17 ( 6 + 2 )}{92}

sin (x) = \frac{17 ( 6 + 2 )}{92}

sin (x) = \frac{17 ( 6 + 2 )}{92}

sin (x) = \frac{17 ( 6 + 2 )}{92}

验证解

找到无定义的点（奇点）:

sin (x) = 0

取

\frac{17}{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{17 ( 6 + 2 )}{92}

使用反三角函数性质

sin (x) = \frac{17 ( 6 + 2 )}{92}

sin (x) = \frac{17 ( 6 + 2 )}{92}

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{17 ( 6 + 2 )}{92}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{17 ( 6 + 2 )}{92}) + 36 0^{\circ} n

x = arcsin (\frac{17 ( 6 + 2 )}{92}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{17 ( 6 + 2 )}{92}) + 36 0^{\circ} n

以小数形式表示解

x = 0.79511 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.79511 \dots + 36 0^{\circ} n

作图

流行的例子

3-2cos^2(x)=4sin^2(x)

3 - 2 cos^{2} (x) = 4 sin^{2} (x)

4-4cos(2x)-3sin(x)=0

4 - 4 cos (2 x) - 3 sin (x) = 0

2sin^2(w)+sin(w)-1=0

2 sin^{2} (w) + sin (w) - 1 = 0

sin^2(x)+sin(x)=-cos^2(x)

sin^{2} (x) + sin (x) = - cos^{2} (x)

arctan(x)=arccos(3/4)

arctan (x) = arccos (\frac{3}{4})