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受欢迎的三角函数 >

cos(4θ)=cos(3θ)

解答

cos (4 θ) = cos (3 θ)

解答

θ = \frac{4 πn}{7}, θ = \frac{2 π}{7} + \frac{4 πn}{7}, θ = 2 π + 4 πn

+1

度数

θ = 0^{\circ} + 102.85714 \dots^{\circ} n, θ = 51.42857 \dots^{\circ} + 102.85714 \dots^{\circ} n, θ = 36 0^{\circ} + 72 0^{\circ} n

求解步骤

cos (4 θ) = cos (3 θ)

两边减去

cos (3 θ)

cos (4 θ) - cos (3 θ) = 0

使用三角恒等式改写

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

化简

- 2 sin (\frac{4 θ + 3 θ}{2}) sin (\frac{4 θ - 3 θ}{2}) : - 2 sin (\frac{7 θ}{2}) sin (\frac{θ}{2})

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

同类项相加:

4 θ + 3 θ = 7 θ

= - 2 sin (\frac{7 θ}{2}) sin (\frac{4 θ - 3 θ}{2})

同类项相加:

4 θ - 3 θ = θ

= - 2 sin (\frac{7 θ}{2}) sin (\frac{θ}{2})

= - 2 sin (\frac{7 θ}{2}) sin (\frac{θ}{2})

- 2 sin (\frac{7 θ}{2}) sin (\frac{θ}{2}) = 0

分别求解每个部分

sin (\frac{7 θ}{2}) = 0 or sin (\frac{θ}{2}) = 0

sin (\frac{7 θ}{2}) = 0 : θ = \frac{4 πn}{7}, θ = \frac{2 π}{7} + \frac{4 πn}{7}

sin (\frac{7 θ}{2}) = 0

sin (\frac{7 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{7 θ}{2} = 0 + 2 πn, \frac{7 θ}{2} = π + 2 πn

\frac{7 θ}{2} = 0 + 2 πn, \frac{7 θ}{2} = π + 2 πn

解

\frac{7 θ}{2} = 0 + 2 πn : θ = \frac{4 πn}{7}

\frac{7 θ}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{7 θ}{2} = 2 πn

在两边乘以

2

\frac{7 θ}{2} = 2 πn

在两边乘以

2

\frac{2 \cdot 7 θ}{2} = 2 \cdot 2 πn

化简

7 θ = 4 πn

7 θ = 4 πn

两边除以

7

7 θ = 4 πn

两边除以

7

\frac{7 θ}{7} = \frac{4 πn}{7}

化简

θ = \frac{4 πn}{7}

θ = \frac{4 πn}{7}

解

\frac{7 θ}{2} = π + 2 πn : θ = \frac{2 π}{7} + \frac{4 πn}{7}

\frac{7 θ}{2} = π + 2 πn

在两边乘以

2

\frac{7 θ}{2} = π + 2 πn

在两边乘以

2

\frac{2 \cdot 7 θ}{2} = 2 π + 2 \cdot 2 πn

化简

7 θ = 2 π + 4 πn

7 θ = 2 π + 4 πn

两边除以

7

7 θ = 2 π + 4 πn

两边除以

7

\frac{7 θ}{7} = \frac{2 π}{7} + \frac{4 πn}{7}

化简

θ = \frac{2 π}{7} + \frac{4 πn}{7}

θ = \frac{2 π}{7} + \frac{4 πn}{7}

θ = \frac{4 πn}{7}, θ = \frac{2 π}{7} + \frac{4 πn}{7}

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}{7}, θ = \frac{2 π}{7} + \frac{4 πn}{7}, θ = 4 πn, θ = 2 π + 4 πn

合并重叠的区间

θ = \frac{4 πn}{7}, θ = \frac{2 π}{7} + \frac{4 πn}{7}, θ = 2 π + 4 πn

作图

流行的例子

(-2sqrt(3))/3 =sec(θ)

\frac{- 2 3}{3} = sec (θ)

cot (x) = - 8

-1+sin^2(a)+cos^2(a)=cos^2(a)

- 1 + sin^{2} (a) + cos^{2} (a) = cos^{2} (a)

(sin(x))/(1+cos(x))=3cot(x/2)

\frac{sin ( x )}{1 + cos ( x )} = 3 cot (\frac{x}{2})

sin(2x)=(sqrt(3))/2 ,0<= x<= 360

sin (2 x) = \frac{3}{2}, 0^{\circ} \leq x \leq 36 0^{\circ}