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 >

(2.68)/(sin(126))=(1.2)/(sin(x))

解答

\frac{2.68}{sin ( 12 6 ^{\circ} )} = \frac{1.2}{sin ( x )}

解答

x = 0.37067 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.37067 \dots + 36 0^{\circ} n

弧度

x = 0.37067 \dots + 2 πn, x = π - 0.37067 \dots + 2 πn

求解步骤

\frac{2.68}{sin ( 12 6 ^{\circ} )} = \frac{1.2}{sin ( x )}

sin (12 6^{\circ}) = \frac{5 + 1}{4}

sin (12 6^{\circ})

使用三角恒等式改写:

cos (3 6^{\circ})

sin (12 6^{\circ})

利用以下特性:

sin (x) = cos (9 0^{\circ} - x)

= cos (9 0^{\circ} - 12 6^{\circ})

化简:

9 0^{\circ} - 12 6^{\circ} = - 3 6^{\circ}

9 0^{\circ} - 12 6^{\circ}

2, 10

的最小公倍数:

10

2, 10

最小公倍数 (LCM)

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

10

质因数分解:

2 \cdot 5

10

10

除以

210 = 5 \cdot 2

= 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 5

将每个因子乘以它在

2

或

10

中出现的最多次数

= 2 \cdot 5

数字相乘:

2 \cdot 5 = 10

= 10

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

10

对于

9 0^{\circ} :

将分母和分子乘以

5

9 0^{\circ} = \frac{18 0 ^{\circ} 5}{2 \cdot 5} = 9 0^{\circ}

= 9 0^{\circ} - 12 6^{\circ}

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{18 0 ^{\circ} 5 - 126 0 ^{\circ}}{10}

同类项相加:

90 0^{\circ} - 126 0^{\circ} = - 36 0^{\circ}

= \frac{- 36 0 ^{\circ}}{10}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - 3 6^{\circ}

约分:

2

= - 3 6^{\circ}

= cos (- 3 6^{\circ})

利用以下特性:

cos (- x) = cos (x)

cos (- 3 6^{\circ}) = cos (3 6^{\circ})

= cos (3 6^{\circ})

= cos (3 6^{\circ})

使用三角恒等式改写:

\frac{5 + 1}{4}

cos (3 6^{\circ})

显示:

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

使用以下积化和差公式:

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

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = sin (5 4^{\circ}) - sin (1 8^{\circ})

显示:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

两边除以

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

利用以下特性:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

两边除以

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

两边除以

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

代入

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

\frac{1}{2} = sin (5 4^{\circ}) - sin (1 8^{\circ})

sin (5 4^{\circ}) = cos (9 0^{\circ} - 5 4^{\circ})

\frac{1}{2} = cos (9 0^{\circ} - 5 4^{\circ}) - sin (1 8^{\circ})

\frac{1}{2} = cos (3 6^{\circ}) - sin (1 8^{\circ})

显示:

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

使用因式分解法则:

a^{2} - b^{2} = (a + b) (a - b)

a = cos (3 6^{\circ}) + sin (1 8^{\circ})

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = ((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) ((cos (3 6^{\circ}) + sin (1 8^{\circ})) - (cos (3 6^{\circ}) - sin (1 8^{\circ})))

整理后得

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 2 (2 cos (3 6^{\circ}) sin (1 8^{\circ}))

显示:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

两边除以

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

利用以下特性:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

两边除以

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

两边除以

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

代入

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 1

代入

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (\frac{1}{2})^{2} = 1

整理后得

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} = 1

两边加上

\frac{1}{4}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

整理后得

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} = \frac{5}{4}

在两侧开平方

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \pm \frac{5}{4}

cos (3 6^{\circ})

不能为负

sin (1 8^{\circ})

不能为负

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

以下方程式相加

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{2}

((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) = (\frac{5}{2} + \frac{1}{2})

整理后得

cos (3 6^{\circ}) = \frac{5 + 1}{4}

= \frac{5 + 1}{4}

= \frac{5 + 1}{4}

\frac{2.68}{\frac{5 + 1}{4}} = \frac{1.2}{sin ( x )}

使用分式交叉相乘

\frac{2.68}{\frac{5 + 1}{4}} = \frac{1.2}{sin ( x )}

使用分式交叉相乘: 若

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

则

a \cdot d = b \cdot c

2.68 sin (x) = \frac{5 + 1}{4} \cdot 1.2

化简

\frac{5 + 1}{4} \cdot 1.2 : \frac{1.2 ( 5 + 1 )}{4}

\frac{5 + 1}{4} \cdot 1.2

分式相乘:

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

= \frac{( 5 + 1 ) \cdot 1.2}{4}

2.68 sin (x) = \frac{1.2 ( 5 + 1 )}{4}

2.68 sin (x) = \frac{1.2 ( 5 + 1 )}{4}

在两边乘以

100

2.68 sin (x) = \frac{1.2 ( 5 + 1 )}{4}

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are

2

digits to the right of the decimal point, therefore multiply by

100

2.68 sin (x) \cdot 100 = \frac{1.2 ( 5 + 1 )}{4} \cdot 100

整理后得

268 sin (x) = 30 (1 + 5)

268 sin (x) = 30 (1 + 5)

两边除以

268

268 sin (x) = 30 (1 + 5)

两边除以

268

\frac{268 sin ( x )}{268} = \frac{30 ( 1 + 5 )}{268}

化简

sin (x) = \frac{15 ( 1 + 5 )}{134}

sin (x) = \frac{15 ( 1 + 5 )}{134}

验证解

找到无定义的点（奇点）:

sin (x) = 0

取

\frac{1.2}{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{15 ( 1 + 5 )}{134}

使用反三角函数性质

sin (x) = \frac{15 ( 1 + 5 )}{134}

sin (x) = \frac{15 ( 1 + 5 )}{134}

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{15 ( 1 + 5 )}{134}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{15 ( 1 + 5 )}{134}) + 36 0^{\circ} n

x = arcsin (\frac{15 ( 1 + 5 )}{134}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{15 ( 1 + 5 )}{134}) + 36 0^{\circ} n

以小数形式表示解

x = 0.37067 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.37067 \dots + 36 0^{\circ} n

作图

流行的例子

3csc^2(x)-5csc(x)-2=0

((sin(5x)))/((-1cos(5x)))=0

sin(x)=cos(40)

5cos(x)+sqrt(2)=3cos(x),0<= x<= 2pi

2tan(2x+10)=-5.5