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受欢迎的三角函数 >

cos(4θ)=cos(2θ),180<= θ<= 360

解答

cos (4 θ) = cos (2 θ), 18 0^{\circ} \leq θ \leq 36 0^{\circ}

解答

θ = 18 0^{\circ}, θ = 24 0^{\circ}, θ = 30 0^{\circ}, θ = 36 0^{\circ}

+1

弧度

θ = π, θ = \frac{4 π}{3}, θ = \frac{5 π}{3}, θ = 2 π

求解步骤

cos (4 θ) = cos (2 θ), 18 0^{\circ} \leq θ \leq 36 0^{\circ}

两边减去

cos (2 θ)

cos (4 θ) - cos (2 θ) = 0

使用三角恒等式改写

- cos (2 θ) + cos (4 θ)

使用和差化积恒等式:

cos (s) - cos (t) = - 2 sin (\frac{s + t}{2}) sin (\frac{s - t}{2})

= - 2 sin (\frac{4 θ + 2 θ}{2}) sin (\frac{4 θ - 2 θ}{2})

化简

- 2 sin (\frac{4 θ + 2 θ}{2}) sin (\frac{4 θ - 2 θ}{2}) : - 2 sin (3 θ) sin (θ)

- 2 sin (\frac{4 θ + 2 θ}{2}) sin (\frac{4 θ - 2 θ}{2})

\frac{4 θ + 2 θ}{2} = 3 θ

\frac{4 θ + 2 θ}{2}

同类项相加:

4 θ + 2 θ = 6 θ

= \frac{6 θ}{2}

数字相除:

\frac{6}{2} = 3

= 3 θ

= - 2 sin (3 θ) sin (\frac{4 θ - 2 θ}{2})

\frac{4 θ - 2 θ}{2} = θ

\frac{4 θ - 2 θ}{2}

同类项相加:

4 θ - 2 θ = 2 θ

= \frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

= - 2 sin (3 θ) sin (θ)

= - 2 sin (3 θ) sin (θ)

- 2 sin (3 θ) sin (θ) = 0

分别求解每个部分

sin (3 θ) = 0 or sin (θ) = 0

sin (3 θ) = 0, 18 0^{\circ} \leq θ \leq 36 0^{\circ} : θ = 18 0^{\circ}, θ = 24 0^{\circ}, θ = 30 0^{\circ}, θ = 36 0^{\circ}

sin (3 θ) = 0, 18 0^{\circ} \leq θ \leq 36 0^{\circ}

sin (3 θ) = 0

的通解

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}

3 θ = 0 + 36 0^{\circ} n, 3 θ = 18 0^{\circ} + 36 0^{\circ} n

3 θ = 0 + 36 0^{\circ} n, 3 θ = 18 0^{\circ} + 36 0^{\circ} n

解

3 θ = 0 + 36 0^{\circ} n : θ = \frac{36 0 ^{\circ} n}{3}

3 θ = 0 + 36 0^{\circ} n

0 + 36 0^{\circ} n = 36 0^{\circ} n

3 θ = 36 0^{\circ} n

两边除以

3

3 θ = 36 0^{\circ} n

两边除以

3

\frac{3 θ}{3} = \frac{36 0 ^{\circ} n}{3}

化简

θ = \frac{36 0 ^{\circ} n}{3}

θ = \frac{36 0 ^{\circ} n}{3}

解

3 θ = 18 0^{\circ} + 36 0^{\circ} n : θ = 6 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

3 θ = 18 0^{\circ} + 36 0^{\circ} n

两边除以

3

3 θ = 18 0^{\circ} + 36 0^{\circ} n

两边除以

3

\frac{3 θ}{3} = 6 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

化简

θ = 6 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = 6 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = \frac{36 0 ^{\circ} n}{3}, θ = 6 0^{\circ} + \frac{36 0 ^{\circ} n}{3}

在

18 0^{\circ} \leq θ \leq 36 0^{\circ}

范围内的解

θ = 18 0^{\circ}, θ = 24 0^{\circ}, θ = 30 0^{\circ}, θ = 36 0^{\circ}

sin (θ) = 0, 18 0^{\circ} \leq θ \leq 36 0^{\circ} : θ = 18 0^{\circ}, θ = 36 0^{\circ}

sin (θ) = 0, 18 0^{\circ} \leq θ \leq 36 0^{\circ}

sin (θ) = 0

的通解

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}

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

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

解

θ = 0 + 36 0^{\circ} n : θ = 36 0^{\circ} n

θ = 0 + 36 0^{\circ} n

0 + 36 0^{\circ} n = 36 0^{\circ} n

θ = 36 0^{\circ} n

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

在

18 0^{\circ} \leq θ \leq 36 0^{\circ}

范围内的解

θ = 18 0^{\circ}, θ = 36 0^{\circ}

合并所有解

θ = 18 0^{\circ}, θ = 24 0^{\circ}, θ = 30 0^{\circ}, θ = 36 0^{\circ}

作图

流行的例子

sqrt(2)sin(x)=cos(x)

8cos(2x)=8-8cos(2x)

-(2)cos(10t)=(6)sin(10t)

tan(x)=0,0<x<2pi