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受欢迎的三角函数 >

cos(4θ)-cos(6θ)=0

解答

cos (4 θ) - cos (6 θ) = 0

解答

θ = \frac{2 πn}{5}, θ = \frac{π}{5} + \frac{2 πn}{5}, θ = 2 πn, θ = π + 2 πn

+1

度数

θ = 0^{\circ} + 7 2^{\circ} n, θ = 3 6^{\circ} + 7 2^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

cos (4 θ) - cos (6 θ) = 0

使用三角恒等式改写

cos (4 θ) - cos (6 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 + 6 θ}{2}) sin (\frac{4 θ - 6 θ}{2})

化简

- 2 sin (\frac{4 θ + 6 θ}{2}) sin (\frac{4 θ - 6 θ}{2}) : 2 sin (θ) sin (5 θ)

- 2 sin (\frac{4 θ + 6 θ}{2}) sin (\frac{4 θ - 6 θ}{2})

\frac{4 θ + 6 θ}{2} = 5 θ

\frac{4 θ + 6 θ}{2}

同类项相加:

4 θ + 6 θ = 10 θ

= \frac{10 θ}{2}

数字相除:

\frac{10}{2} = 5

= 5 θ

= - 2 sin (5 θ) sin (\frac{4 θ - 6 θ}{2})

\frac{4 θ - 6 θ}{2} = - θ

\frac{4 θ - 6 θ}{2}

同类项相加:

4 θ - 6 θ = - 2 θ

= \frac{- 2 θ}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= - θ

= - 2 sin (5 θ) sin (- θ)

使用负角恒等式:

sin (- x) = - sin (x)

= - 2 (- sin (θ)) sin (5 θ)

使用法则

- (- a) = a

= 2 sin (θ) sin (5 θ)

= 2 sin (θ) sin (5 θ)

2 sin (5 θ) sin (θ) = 0

分别求解每个部分

sin (5 θ) = 0 or sin (θ) = 0

sin (5 θ) = 0 : θ = \frac{2 πn}{5}, θ = \frac{π}{5} + \frac{2 πn}{5}

sin (5 θ) = 0

sin (5 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}

5 θ = 0 + 2 πn, 5 θ = π + 2 πn

5 θ = 0 + 2 πn, 5 θ = π + 2 πn

解

5 θ = 0 + 2 πn : θ = \frac{2 πn}{5}

5 θ = 0 + 2 πn

0 + 2 πn = 2 πn

5 θ = 2 πn

两边除以

5

5 θ = 2 πn

两边除以

5

\frac{5 θ}{5} = \frac{2 πn}{5}

化简

θ = \frac{2 πn}{5}

θ = \frac{2 πn}{5}

解

5 θ = π + 2 πn : θ = \frac{π}{5} + \frac{2 πn}{5}

5 θ = π + 2 πn

两边除以

5

5 θ = π + 2 πn

两边除以

5

\frac{5 θ}{5} = \frac{π}{5} + \frac{2 πn}{5}

化简

θ = \frac{π}{5} + \frac{2 πn}{5}

θ = \frac{π}{5} + \frac{2 πn}{5}

θ = \frac{2 πn}{5}, θ = \frac{π}{5} + \frac{2 πn}{5}

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}{5}, θ = \frac{π}{5} + \frac{2 πn}{5}, θ = 2 πn, θ = π + 2 πn

作图

流行的例子

1 - sin (θ) = 0

9 sin^{2} (θ) - 1 = 0

sinh (x) = 3

cos(2θ-pi/2)=1,0<= θ<2pi

cos (2 θ - \frac{π}{2}) = 1, 0 \leq θ < 2 π

sin (2 x) = - 0.5