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受欢迎的三角函数 >

cos(2θ)= 3/5 ,90<= θ<= 180

解答

cos (2 θ) = \frac{3}{5}, 9 0^{\circ} \leq θ \leq 18 0^{\circ}

解答

θ = \frac{36 0 ^{\circ} - 0.92729 \dots}{2}

+1

弧度

θ = \frac{2 π - 0.92729 \dots}{2}

求解步骤

cos (2 θ) = \frac{3}{5}, 9 0^{\circ} \leq θ \leq 18 0^{\circ}

使用反三角函数性质

cos (2 θ) = \frac{3}{5}

cos (2 θ) = \frac{3}{5}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

2 θ = arccos (\frac{3}{5}) + 36 0^{\circ} n, 2 θ = 36 0^{\circ} - arccos (\frac{3}{5}) + 36 0^{\circ} n

2 θ = arccos (\frac{3}{5}) + 36 0^{\circ} n, 2 θ = 36 0^{\circ} - arccos (\frac{3}{5}) + 36 0^{\circ} n

解

2 θ = arccos (\frac{3}{5}) + 36 0^{\circ} n : θ = \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

2 θ = arccos (\frac{3}{5}) + 36 0^{\circ} n

两边除以

2

2 θ = arccos (\frac{3}{5}) + 36 0^{\circ} n

两边除以

2

\frac{2 θ}{2} = \frac{arccos ( \frac{3}{5} )}{2} + \frac{36 0 ^{\circ} n}{2}

化简

θ = \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

θ = \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

解

2 θ = 36 0^{\circ} - arccos (\frac{3}{5}) + 36 0^{\circ} n : θ = 18 0^{\circ} - \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

2 θ = 36 0^{\circ} - arccos (\frac{3}{5}) + 36 0^{\circ} n

两边除以

2

2 θ = 36 0^{\circ} - arccos (\frac{3}{5}) + 36 0^{\circ} n

两边除以

2

\frac{2 θ}{2} = 18 0^{\circ} - \frac{arccos ( \frac{3}{5} )}{2} + \frac{36 0 ^{\circ} n}{2}

化简

θ = 18 0^{\circ} - \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

θ = 18 0^{\circ} - \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

θ = \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n, θ = 18 0^{\circ} - \frac{arccos ( \frac{3}{5} )}{2} + 18 0^{\circ} n

在

9 0^{\circ} \leq θ \leq 18 0^{\circ}

范围内的解

θ = \frac{36 0 ^{\circ} - arccos ( \frac{3}{5} )}{2}

以小数形式表示解

θ = \frac{36 0 ^{\circ} - 0.92729 \dots}{2}

作图

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