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受欢迎的三角函数 >

cos(4θ)=cos(2θ)

解答

cos (4 θ) = cos (2 θ)

解答

θ = \frac{2 πn}{3}, θ = \frac{π}{3} + \frac{2 πn}{3}

+1

度数

θ = 0^{\circ} + 12 0^{\circ} n, θ = 6 0^{\circ} + 12 0^{\circ} n

求解步骤

cos (4 θ) = cos (2 θ)

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 : θ = \frac{2 πn}{3}, θ = \frac{π}{3} + \frac{2 πn}{3}

sin (3 θ) = 0

sin (3 θ) = 0

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} 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 + 2 πn, 3 θ = π + 2 πn

3 θ = 0 + 2 πn, 3 θ = π + 2 πn

解

3 θ = 0 + 2 πn : θ = \frac{2 πn}{3}

3 θ = 0 + 2 πn

0 + 2 πn = 2 πn

3 θ = 2 πn

两边除以

3

3 θ = 2 πn

两边除以

3

\frac{3 θ}{3} = \frac{2 πn}{3}

化简

θ = \frac{2 πn}{3}

θ = \frac{2 πn}{3}

解

3 θ = π + 2 πn : θ = \frac{π}{3} + \frac{2 πn}{3}

3 θ = π + 2 πn

两边除以

3

3 θ = π + 2 πn

两边除以

3

\frac{3 θ}{3} = \frac{π}{3} + \frac{2 πn}{3}

化简

θ = \frac{π}{3} + \frac{2 πn}{3}

θ = \frac{π}{3} + \frac{2 πn}{3}

θ = \frac{2 πn}{3}, θ = \frac{π}{3} + \frac{2 πn}{3}

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

sin (θ) = 0

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

合并所有解

θ = \frac{2 πn}{3}, θ = \frac{π}{3} + \frac{2 πn}{3}, θ = 2 πn, θ = π + 2 πn

合并重叠的区间

θ = \frac{2 πn}{3}, θ = \frac{π}{3} + \frac{2 πn}{3}

作图

流行的例子

3 cos (3 x) = 2

sec(θ)-2=-tan(θ)-3

sec (θ) - 2 = - tan (θ) - 3

sin^2(x)-sin(x)=1

sin^{2} (x) - sin (x) = 1

sqrt(2)sin(x)cos(x)+1sin(x)=0

2 sin (x) cos (x) + 1 sin (x) = 0

cos (y) = \frac{1}{2}