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受欢迎的三角函数 >

cos(2θ)=cos(3θ)

解答

cos (2 θ) = cos (3 θ)

解答

θ = \frac{4 πn}{5}, θ = \frac{2 π}{5} + \frac{4 πn}{5}

+1

度数

θ = 0^{\circ} + 14 4^{\circ} n, θ = 7 2^{\circ} + 14 4^{\circ} n

求解步骤

cos (2 θ) = cos (3 θ)

两边减去

cos (3 θ)

cos (2 θ) - cos (3 θ) = 0

使用三角恒等式改写

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

化简

- 2 sin (\frac{2 θ + 3 θ}{2}) sin (\frac{2 θ - 3 θ}{2}) : 2 sin (\frac{θ}{2}) sin (\frac{5 θ}{2})

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

同类项相加:

2 θ + 3 θ = 5 θ

= - 2 sin (\frac{5 θ}{2}) sin (\frac{2 θ - 3 θ}{2})

\frac{2 θ - 3 θ}{2} = - \frac{θ}{2}

\frac{2 θ - 3 θ}{2}

同类项相加:

2 θ - 3 θ = - θ

= \frac{- θ}{2}

使用分式法则:

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

= - \frac{θ}{2}

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

使用负角恒等式:

sin (- x) = - sin (x)

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

使用法则

- (- a) = a

= 2 sin (\frac{θ}{2}) sin (\frac{5 θ}{2})

= 2 sin (\frac{θ}{2}) sin (\frac{5 θ}{2})

2 sin (\frac{5 θ}{2}) sin (\frac{θ}{2}) = 0

分别求解每个部分

sin (\frac{5 θ}{2}) = 0 or sin (\frac{θ}{2}) = 0

sin (\frac{5 θ}{2}) = 0 : θ = \frac{4 πn}{5}, θ = \frac{2 π}{5} + \frac{4 πn}{5}

sin (\frac{5 θ}{2}) = 0

sin (\frac{5 θ}{2}) = 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}

\frac{5 θ}{2} = 0 + 2 πn, \frac{5 θ}{2} = π + 2 πn

\frac{5 θ}{2} = 0 + 2 πn, \frac{5 θ}{2} = π + 2 πn

解

\frac{5 θ}{2} = 0 + 2 πn : θ = \frac{4 πn}{5}

\frac{5 θ}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{5 θ}{2} = 2 πn

在两边乘以

2

\frac{5 θ}{2} = 2 πn

在两边乘以

2

\frac{2 \cdot 5 θ}{2} = 2 \cdot 2 πn

化简

5 θ = 4 πn

5 θ = 4 πn

两边除以

5

5 θ = 4 πn

两边除以

5

\frac{5 θ}{5} = \frac{4 πn}{5}

化简

θ = \frac{4 πn}{5}

θ = \frac{4 πn}{5}

解

\frac{5 θ}{2} = π + 2 πn : θ = \frac{2 π}{5} + \frac{4 πn}{5}

\frac{5 θ}{2} = π + 2 πn

在两边乘以

2

\frac{5 θ}{2} = π + 2 πn

在两边乘以

2

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

化简

5 θ = 2 π + 4 πn

5 θ = 2 π + 4 πn

两边除以

5

5 θ = 2 π + 4 πn

两边除以

5

\frac{5 θ}{5} = \frac{2 π}{5} + \frac{4 πn}{5}

化简

θ = \frac{2 π}{5} + \frac{4 πn}{5}

θ = \frac{2 π}{5} + \frac{4 πn}{5}

θ = \frac{4 πn}{5}, θ = \frac{2 π}{5} + \frac{4 πn}{5}

sin (\frac{θ}{2}) = 0 : θ = 4 πn, θ = 2 π + 4 πn

sin (\frac{θ}{2}) = 0

sin (\frac{θ}{2}) = 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}

\frac{θ}{2} = 0 + 2 πn, \frac{θ}{2} = π + 2 πn

\frac{θ}{2} = 0 + 2 πn, \frac{θ}{2} = π + 2 πn

解

\frac{θ}{2} = 0 + 2 πn : θ = 4 πn

\frac{θ}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{θ}{2} = 2 πn

在两边乘以

2

\frac{θ}{2} = 2 πn

在两边乘以

2

\frac{2 θ}{2} = 2 \cdot 2 πn

化简

θ = 4 πn

θ = 4 πn

解

\frac{θ}{2} = π + 2 πn : θ = 2 π + 4 πn

\frac{θ}{2} = π + 2 πn

在两边乘以

2

\frac{θ}{2} = π + 2 πn

在两边乘以

2

\frac{2 θ}{2} = 2 π + 2 \cdot 2 πn

化简

θ = 2 π + 4 πn

θ = 2 π + 4 πn

θ = 4 πn, θ = 2 π + 4 πn

合并所有解

θ = \frac{4 πn}{5}, θ = \frac{2 π}{5} + \frac{4 πn}{5}, θ = 4 πn, θ = 2 π + 4 πn

合并重叠的区间

θ = \frac{4 πn}{5}, θ = \frac{2 π}{5} + \frac{4 πn}{5}

作图

流行的例子

solvefor θ,2sin(θ)=1.5

so l v e f or θ, 2 sin (θ) = 1.5

cos (θ) = - 2109

solvefor x,cos(x)(cos(x)-sin(x))=0

so l v e f or x, cos (x) (cos (x) - sin (x)) = 0

sin (x) = 0.72555

15=10sin(pi/(20)(x+10))+12

15 = 10 sin (\frac{π}{20} (x + 10)) + 12