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受欢迎的三角函数 >

sin(((60))/7)=sin(((81))/a)

解答

sin (\frac{( 60 )}{7}) = sin (\frac{( 81 )}{a})

解答

a = \frac{567}{21 π - 60 + 14 πn}, a = \frac{567}{2 ( - 7 π + 30 + 7 πn )}

+1

度数

a = 0^{\circ} + 650.30975 \dots^{\circ} n, a = 0^{\circ} + 541.44511 \dots^{\circ} n

求解步骤

sin (\frac{( 60 )}{7}) = sin (\frac{( 81 )}{a})

交换两边

sin (\frac{81}{a}) = sin (\frac{60}{7})

使用反三角函数性质

sin (\frac{81}{a}) = sin (\frac{60}{7})

sin (\frac{81}{a}) = sin (\frac{60}{7})

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

\frac{81}{a} = arcsin (sin (\frac{60}{7})) + 2 πn, \frac{81}{a} = π - arcsin (sin (\frac{60}{7})) + 2 πn

\frac{81}{a} = arcsin (sin (\frac{60}{7})) + 2 πn, \frac{81}{a} = π - arcsin (sin (\frac{60}{7})) + 2 πn

解

\frac{81}{a} = arcsin (sin (\frac{60}{7})) + 2 πn : a = \frac{567}{21 π - 60 + 14 πn}; n \neq = \frac{- 21 π + 60}{14 π}

\frac{81}{a} = arcsin (sin (\frac{60}{7})) + 2 πn

在两边乘以

a

\frac{81}{a} = arcsin (sin (\frac{60}{7})) + 2 πn

在两边乘以

a

\frac{81}{a} a = arcsin (sin (\frac{60}{7})) a + 2 πna

化简

81 = \frac{21 π - 60}{7} a + 2 πna

81 = \frac{21 π - 60}{7} a + 2 πna

交换两边

\frac{21 π - 60}{7} a + 2 πna = 81

在两边乘以

7

\frac{21 π - 60}{7} a + 2 πna = 81

在两边乘以

7

\frac{21 π - 60}{7} a \cdot 7 + 2 πna \cdot 7 = 81 \cdot 7

化简

\frac{21 π - 60}{7} a \cdot 7 + 2 πna \cdot 7 = 81 \cdot 7

化简

\frac{21 π - 60}{7} a \cdot 7 : (21 π - 60) a

\frac{21 π - 60}{7} a \cdot 7

分式相乘:

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

= \frac{7 ( 21 π - 60 )}{7} a

约分:

7

= a (21 π - 60)

化简

2 πna \cdot 7 : 14 πna

2 πna \cdot 7

数字相乘:

2 \cdot 7 = 14

= 14 πna

化简

81 \cdot 7 : 567

81 \cdot 7

数字相乘:

81 \cdot 7 = 567

= 567

(21 π - 60) a + 14 πna = 567

(21 π - 60) a + 14 πna = 567

(21 π - 60) a + 14 πna = 567

乘开

a (21 π - 60) : 21 πa - 60 a

a (21 π - 60)

使用分配律:

a (b - c) = ab - a c

a = a, b = 21 π, c = 60

= a \cdot 21 π - a \cdot 60

= 21 πa - 60 a

21 πa - 60 a + 14 πna = 567

分解

21 πa - 60 a + 14 πna : a (21 π - 60 + 14 πn)

21 πa - 60 a + 14 πna

因式分解出通项

a

= a (21 π - 60 + 14 πn)

a (21 π - 60 + 14 πn) = 567

两边除以

21 π - 60 + 14 πn; n \neq = \frac{- 21 π + 60}{14 π}

a (21 π - 60 + 14 πn) = 567

两边除以

21 π - 60 + 14 πn; n \neq = \frac{- 21 π + 60}{14 π}

\frac{a ( 21 π - 60 + 14 πn )}{21 π - 60 + 14 πn} = \frac{567}{21 π - 60 + 14 πn}; n \neq = \frac{- 21 π + 60}{14 π}

化简

a = \frac{567}{21 π - 60 + 14 πn}; n \neq = \frac{- 21 π + 60}{14 π}

a = \frac{567}{21 π - 60 + 14 πn}; n \neq = \frac{- 21 π + 60}{14 π}

解

\frac{81}{a} = π - arcsin (sin (\frac{60}{7})) + 2 πn : a = \frac{567}{2 ( - 7 π + 30 + 7 πn )}; n \neq = \frac{7 π - 30}{7 π}

\frac{81}{a} = π - arcsin (sin (\frac{60}{7})) + 2 πn

在两边乘以

a

\frac{81}{a} = π - arcsin (sin (\frac{60}{7})) + 2 πn

在两边乘以

a

\frac{81}{a} a = πa - arcsin (sin (\frac{60}{7})) a + 2 πna

化简

81 = πa - \frac{21 π - 60}{7} a + 2 πna

81 = πa - \frac{21 π - 60}{7} a + 2 πna

交换两边

πa - \frac{21 π - 60}{7} a + 2 πna = 81

在两边乘以

7

πa - \frac{21 π - 60}{7} a + 2 πna = 81

在两边乘以

7

πa \cdot 7 - \frac{21 π - 60}{7} a \cdot 7 + 2 πna \cdot 7 = 81 \cdot 7

化简

πa \cdot 7 - \frac{21 π - 60}{7} a \cdot 7 + 2 πna \cdot 7 = 81 \cdot 7

化简

πa \cdot 7 : 7 πa

πa \cdot 7

使用交换律:

πa \cdot 7 = 7 πa

7 πa

化简

- \frac{21 π - 60}{7} a \cdot 7 : - (21 π - 60) a

- \frac{21 π - 60}{7} a \cdot 7

分式相乘:

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

= - \frac{7 ( 21 π - 60 )}{7} a

约分:

7

= - a (21 π - 60)

化简

2 πna \cdot 7 : 14 πna

2 πna \cdot 7

数字相乘:

2 \cdot 7 = 14

= 14 πna

化简

81 \cdot 7 : 567

81 \cdot 7

数字相乘:

81 \cdot 7 = 567

= 567

7 πa - (21 π - 60) a + 14 πna = 567

7 πa - (21 π - 60) a + 14 πna = 567

7 πa - (21 π - 60) a + 14 πna = 567

乘开

- a (21 π - 60) : - 21 πa + 60 a

- a (21 π - 60)

使用分配律:

a (b - c) = ab - a c

a = - a, b = 21 π, c = 60

= - a \cdot 21 π - (- a) \cdot 60

使用加减运算法则

- (- a) = a

= - 21 πa + 60 a

7 πa - 21 πa + 60 a + 14 πna = 567

同类项相加:

7 πa - 21 πa = - 14 πa

- 14 πa + 60 a + 14 πna = 567

分解

- 14 πa + 60 a + 14 πna : 2 a (- 7 π + 30 + 7 πn)

- 14 πa + 60 a + 14 πna

改写为

= - 7 \cdot 2 aπ + 30 \cdot 2 a + 7 \cdot 2 aπn

因式分解出通项

2 a

= 2 a (- 7 π + 30 + 7 πn)

2 a (- 7 π + 30 + 7 πn) = 567

两边除以

2 (- 7 π + 30 + 7 πn); n \neq = \frac{7 π - 30}{7 π}

2 a (- 7 π + 30 + 7 πn) = 567

两边除以

2 (- 7 π + 30 + 7 πn); n \neq = \frac{7 π - 30}{7 π}

\frac{2 a ( - 7 π + 30 + 7 πn )}{2 ( - 7 π + 30 + 7 πn )} = \frac{567}{2 ( - 7 π + 30 + 7 πn )}; n \neq = \frac{7 π - 30}{7 π}

化简

a = \frac{567}{2 ( - 7 π + 30 + 7 πn )}; n \neq = \frac{7 π - 30}{7 π}

a = \frac{567}{2 ( - 7 π + 30 + 7 πn )}; n \neq = \frac{7 π - 30}{7 π}

a = \frac{567}{21 π - 60 + 14 πn}, a = \frac{567}{2 ( - 7 π + 30 + 7 πn )}

作图

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