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受欢迎的三角函数 >

1sin(45)=1.33sin(θ)

解答

1 sin (4 5^{\circ}) = 1.33 sin (θ)

解答

θ = 0.56055 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.56055 \dots + 36 0^{\circ} n

+1

弧度

θ = 0.56055 \dots + 2 πn, θ = π - 0.56055 \dots + 2 πn

求解步骤

1 \cdot sin (4 5^{\circ}) = 1.33 sin (θ)

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

sin (4 5^{\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}

= \frac{2}{2}

1 \cdot \frac{2}{2} = 1.33 sin (θ)

交换两边

1.33 sin (θ) = 1 \cdot \frac{2}{2}

在两边乘以

100

1.33 sin (θ) = 1 \cdot \frac{2}{2}

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

1.33 sin (θ) \cdot 100 = 1 \cdot \frac{2}{2} \cdot 100

整理后得

133 sin (θ) = 502

133 sin (θ) = 502

两边除以

133

133 sin (θ) = 502

两边除以

133

\frac{133 sin ( θ )}{133} = \frac{50 2}{133}

化简

sin (θ) = \frac{50 2}{133}

sin (θ) = \frac{50 2}{133}

使用反三角函数性质

sin (θ) = \frac{50 2}{133}

sin (θ) = \frac{50 2}{133}

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{50 2}{133}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{50 2}{133}) + 36 0^{\circ} n

θ = arcsin (\frac{50 2}{133}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{50 2}{133}) + 36 0^{\circ} n

以小数形式表示解

θ = 0.56055 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.56055 \dots + 36 0^{\circ} n

作图

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