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受欢迎的三角函数 >

1sin(30)=1.7sin(θ)

解答

1 sin (3 0^{\circ}) = 1.7 sin (θ)

解答

θ = 0.29853 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.29853 \dots + 36 0^{\circ} n

+1

弧度

θ = 0.29853 \dots + 2 πn, θ = π - 0.29853 \dots + 2 πn

求解步骤

1 \cdot sin (3 0^{\circ}) = 1.7 sin (θ)

sin (3 0^{\circ}) = \frac{1}{2}

sin (3 0^{\circ})

使用以下普通恒等式:

sin (3 0^{\circ}) = \frac{1}{2}

sin (3 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{1}{2}

= \frac{1}{2}

1 \cdot \frac{1}{2} = 1.7 sin (θ)

交换两边

1.7 sin (θ) = 1 \cdot \frac{1}{2}

在两边乘以

10

1.7 sin (θ) = 1 \cdot \frac{1}{2}

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

1.7 sin (θ) \cdot 10 = 1 \cdot \frac{1}{2} \cdot 10

整理后得

17 sin (θ) = 5

17 sin (θ) = 5

两边除以

17

17 sin (θ) = 5

两边除以

17

\frac{17 sin ( θ )}{17} = \frac{5}{17}

化简

sin (θ) = \frac{5}{17}

sin (θ) = \frac{5}{17}

使用反三角函数性质

sin (θ) = \frac{5}{17}

sin (θ) = \frac{5}{17}

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}{17}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{5}{17}) + 36 0^{\circ} n

θ = arcsin (\frac{5}{17}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{5}{17}) + 36 0^{\circ} n

以小数形式表示解

θ = 0.29853 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.29853 \dots + 36 0^{\circ} n

作图

流行的例子

30=211.49-20.96cosh(0.03291765x)

30 = 211.49 - 20.96 cosh (0.03291765 x)

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(tan(x)+cos(x))/((1+sin(x)))= 1/(cos(x))

\frac{tan ( x ) + cos ( x )}{( 1 + sin ( x ) )} = \frac{1}{cos ( x )}

sin(θ)= 4/5 ,0<= θ<= pi/2

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