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受欢迎的三角函数 >

sin(θ)=(10sin(105))/(24.6)

解答

sin (θ) = \frac{10 sin ( 10 5 ^{\circ} )}{24.6}

解答

θ = 0.40351 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.40351 \dots + 36 0^{\circ} n

+1

弧度

θ = 0.40351 \dots + 2 πn, θ = π - 0.40351 \dots + 2 πn

求解步骤

sin (θ) = \frac{10 sin ( 10 5 ^{\circ} )}{24.6}

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}

sin (θ) = \frac{10 \cdot \frac{2 ( 3 + 1 )}{4}}{24.6}

化简

\frac{10 \cdot \frac{2 ( 3 + 1 )}{4}}{24.6} : \frac{5 2 ( 3 + 1 )}{49.2}

\frac{10 \cdot \frac{2 ( 3 + 1 )}{4}}{24.6}

乘

10 \cdot \frac{2 ( 3 + 1 )}{4} : \frac{5 ( 1 + 3 )}{2}

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

分式相乘:

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

= \frac{2 ( 3 + 1 ) \cdot 10}{4}

约分:

2

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

使用根式运算法则:

n a = a^{\frac{1}{n}}

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

= \frac{5 \cdot 2 ^{\frac{1}{2}} ( 1 + 3 )}{2}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{5 ( 1 + 3 )}{2 ^{1 - \frac{1}{2}}}

数字相减:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{5 ( 1 + 3 )}{2 ^{\frac{1}{2}}}

使用根式运算法则:

a^{\frac{1}{n}} = n a

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

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

= \frac{\frac{5 ( 1 + 3 )}{2}}{24.6}

使用分式法则:

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

= \frac{5 ( 3 + 1 )}{2 \cdot 24.6}

\frac{5 ( 1 + 3 )}{2 \cdot 24.6}

有理化:

\frac{5 2 ( 1 + 3 )}{49.2}

\frac{5 ( 1 + 3 )}{2 \cdot 24.6}

乘以共轭根式

\frac{2}{2}

= \frac{5 ( 3 + 1 ) 2}{2 \cdot 24.6 2}

2 \cdot 24.6 2 = 49.2

2 \cdot 24.6 2

使用根式运算法则:

a a = a

22 = 2

= 2 \cdot 24.6

数字相乘:

24.6 \cdot 2 = 49.2

= 49.2

= \frac{5 2 ( 3 + 1 )}{49.2}

= \frac{5 2 ( 1 + 3 )}{49.2}

sin (θ) = \frac{5 2 ( 3 + 1 )}{49.2}

使用反三角函数性质

sin (θ) = \frac{5 2 ( 3 + 1 )}{49.2}

sin (θ) = \frac{5 2 ( 3 + 1 )}{49.2}

的通解

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

θ = arcsin (\frac{5 2 ( 3 + 1 )}{49.2}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{5 2 ( 3 + 1 )}{49.2}) + 36 0^{\circ} n

θ = arcsin (\frac{5 2 ( 3 + 1 )}{49.2}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{5 2 ( 3 + 1 )}{49.2}) + 36 0^{\circ} n

以小数形式表示解

θ = 0.40351 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.40351 \dots + 36 0^{\circ} n

作图

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