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受欢迎的三角函数 >

(10)/(sin(x))=(13)/(sin(72))

解答

\frac{10}{sin ( x )} = \frac{13}{sin ( 7 2 ^{\circ} )}

解答

x = 0.82063 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.82063 \dots + 36 0^{\circ} n

+1

弧度

x = 0.82063 \dots + 2 πn, x = π - 0.82063 \dots + 2 πn

求解步骤

\frac{10}{sin ( x )} = \frac{13}{sin ( 7 2 ^{\circ} )}

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

sin (7 2^{\circ})

使用三角恒等式改写:

cos (1 8^{\circ})

sin (7 2^{\circ})

利用以下特性:

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

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

化简:

9 0^{\circ} - 7 2^{\circ} = 1 8^{\circ}

9 0^{\circ} - 7 2^{\circ}

2, 5

的最小公倍数:

10

2, 5

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

5

质因数分解:

5

5

5

是质数，因此无法因数分解

= 5

将每个因子乘以它在

2

或

5

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}

对于

7 2^{\circ} :

将分母和分子乘以

2

7 2^{\circ} = \frac{36 0 ^{\circ} 2}{5 \cdot 2} = 7 2^{\circ}

= 9 0^{\circ} - 7 2^{\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 - 72 0 ^{\circ}}{10}

同类项相加:

90 0^{\circ} - 72 0^{\circ} = 18 0^{\circ}

= 1 8^{\circ}

= cos (1 8^{\circ})

= cos (1 8^{\circ})

使用三角恒等式改写:

\frac{1 + cos ( 3 6 ^{\circ} )}{2}

cos (1 8^{\circ})

将

cos (1 8^{\circ})

写为

cos (\frac{3 6 ^{\circ}}{2})

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

使用半角公式:

cos (\frac{θ}{2}) = \frac{1 + cos ( θ )}{2}

使用倍角公式

cos (2 θ) = 2 cos^{2} (θ) - 1

用

\frac{θ}{2}

替代

θ

cos (θ) = 2 cos^{2} (\frac{θ}{2}) - 1

交换两边

2 cos^{2} (\frac{θ}{2}) = 1 + cos (θ)

两边除以

2

cos^{2} (\frac{θ}{2}) = \frac{( 1 + cos ( θ ) )}{2}

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

:

range [0, 9 0^{\circ}] [9 0^{\circ}, 18 0^{\circ}] [18 0^{\circ}, 27 0^{\circ}] [27 0^{\circ}, 36 0^{\circ}] quadrant I II III I V sin positive positive negative negative cos positive negative negative positive

cos (\frac{θ}{2}) = \frac{( 1 + cos ( θ ) )}{2}

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

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

使用三角恒等式改写:

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

化简

\frac{1 + \frac{5 + 1}{4}}{2} : \frac{2 5 + 5}{4}

\frac{1 + \frac{5 + 1}{4}}{2}

\frac{1 + \frac{5 + 1}{4}}{2} = \frac{5 + 5}{8}

\frac{1 + \frac{5 + 1}{4}}{2}

化简

1 + \frac{5 + 1}{4} : \frac{5 + 5}{4}

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

将项转换为分式:

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

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

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 4 + 5 + 1}{4}

1 \cdot 4 + 5 + 1 = 5 + 5

1 \cdot 4 + 5 + 1

数字相乘:

1 \cdot 4 = 4

= 4 + 5 + 1

数字相加:

4 + 1 = 5

= 5 + 5

= \frac{5 + 5}{4}

= \frac{\frac{5 + 5}{4}}{2}

使用分式法则:

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

= \frac{5 + 5}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{5 + 5}{8}

= \frac{5 + 5}{8}

使用根式运算法则:

假定

a \geq 0, b \geq 0

= \frac{5 + 5}{8}

8 = 22

8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

使用根式运算法则:

= 2 2^{2}

使用根式运算法则:

2^{2} = 2

= 22

= \frac{5 + 5}{2 2}

\frac{5 + 5}{2 2}

有理化:

\frac{2 5 + 5}{4}

\frac{5 + 5}{2 2}

乘以共轭根式

\frac{2}{2}

= \frac{5 + 5 2}{2 2 2}

222 = 4

222

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

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

同类项相加:

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

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

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

2 \cdot \frac{1}{2}

分式相乘:

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

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

约分:

2

= 1

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 5 + 5}{4}

= \frac{2 5 + 5}{4}

= \frac{2 5 + 5}{4}

\frac{10}{sin ( x )} = \frac{13}{\frac{2 5 + 5}{4}}

使用分式交叉相乘

\frac{10}{sin ( x )} = \frac{13}{\frac{2 5 + 5}{4}}

使用分式交叉相乘: 若

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

则

a \cdot d = b \cdot c

10 \cdot \frac{2 5 + 5}{4} = sin (x) \cdot 13

化简

10 \cdot \frac{2 5 + 5}{4} : \frac{5 2 5 + 5}{2}

10 \cdot \frac{2 5 + 5}{4}

分式相乘:

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

= \frac{2 5 + 5 \cdot 10}{4}

约分:

2

= \frac{5 2 5 + 5}{2}

\frac{5 2 5 + 5}{2} = sin (x) \cdot 13

\frac{5 2 5 + 5}{2} = sin (x) \cdot 13

交换两边

sin (x) \cdot 13 = \frac{5 2 5 + 5}{2}

两边除以

13

sin (x) \cdot 13 = \frac{5 2 5 + 5}{2}

两边除以

13

\frac{sin ( x ) \cdot 13}{13} = \frac{\frac{5 2 5 + 5}{2}}{13}

化简

\frac{sin ( x ) \cdot 13}{13} = \frac{\frac{5 2 5 + 5}{2}}{13}

化简

\frac{sin ( x ) \cdot 13}{13} : sin (x)

\frac{sin ( x ) \cdot 13}{13}

数字相除:

\frac{13}{13} = 1

= sin (x)

化简

\frac{\frac{5 2 5 + 5}{2}}{13} : \frac{5 2 5 + 5}{26}

\frac{\frac{5 2 5 + 5}{2}}{13}

使用分式法则:

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

= \frac{5 2 5 + 5}{2 \cdot 13}

数字相乘:

2 \cdot 13 = 26

= \frac{5 2 5 + 5}{26}

sin (x) = \frac{5 2 5 + 5}{26}

sin (x) = \frac{5 2 5 + 5}{26}

sin (x) = \frac{5 2 5 + 5}{26}

验证解

找到无定义的点（奇点）:

sin (x) = 0

取

\frac{10}{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{5 2 5 + 5}{26}

使用反三角函数性质

sin (x) = \frac{5 2 5 + 5}{26}

sin (x) = \frac{5 2 5 + 5}{26}

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{5 2 5 + 5}{26}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{5 2 5 + 5}{26}) + 36 0^{\circ} n

x = arcsin (\frac{5 2 5 + 5}{26}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{5 2 5 + 5}{26}) + 36 0^{\circ} n

以小数形式表示解

x = 0.82063 \dots + 36 0^{\circ} n, x = 18 0^{\circ} - 0.82063 \dots + 36 0^{\circ} n

作图

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