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受欢迎的三角函数 >

cos(3x)+cos(5x)=cos(x)

解答

cos (3 x) + cos (5 x) = cos (x)

解答

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{12} + \frac{πn}{2}, x = \frac{5 π}{12} + \frac{πn}{2}

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 1 5^{\circ} + 9 0^{\circ} n, x = 7 5^{\circ} + 9 0^{\circ} n

求解步骤

cos (3 x) + cos (5 x) = cos (x)

两边减去

cos (x)

cos (3 x) + cos (5 x) - cos (x) = 0

使用三角恒等式改写

cos (3 x) + cos (5 x) - cos (x)

使用和差化积恒等式:

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

= - cos (x) + 2 cos (\frac{3 x + 5 x}{2}) cos (\frac{3 x - 5 x}{2})

2 cos (\frac{3 x + 5 x}{2}) cos (\frac{3 x - 5 x}{2}) = 2 cos (x) cos (4 x)

2 cos (\frac{3 x + 5 x}{2}) cos (\frac{3 x - 5 x}{2})

\frac{3 x + 5 x}{2} = 4 x

\frac{3 x + 5 x}{2}

同类项相加:

3 x + 5 x = 8 x

= \frac{8 x}{2}

数字相除:

\frac{8}{2} = 4

= 4 x

= 2 cos (4 x) cos (\frac{3 x - 5 x}{2})

\frac{3 x - 5 x}{2} = - x

\frac{3 x - 5 x}{2}

同类项相加:

3 x - 5 x = - 2 x

= \frac{- 2 x}{2}

使用分式法则:

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

= - \frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= - x

= 2 cos (4 x) cos (- x)

使用负角恒等式:

cos (- x) = cos (x)

= 2 cos (x) cos (4 x)

= - cos (x) + 2 cos (x) cos (4 x)

- cos (x) + 2 cos (4 x) cos (x) = 0

分解

- cos (x) + 2 cos (4 x) cos (x) : cos (x) (2 cos (4 x) - 1)

- cos (x) + 2 cos (4 x) cos (x)

因式分解出通项

cos (x)

= cos (x) (- 1 + 2 cos (4 x))

cos (x) (2 cos (4 x) - 1) = 0

分别求解每个部分

cos (x) = 0 or 2 cos (4 x) - 1 = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

cos (x) = 0

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

2 cos (4 x) - 1 = 0 : x = \frac{π}{12} + \frac{πn}{2}, x = \frac{5 π}{12} + \frac{πn}{2}

2 cos (4 x) - 1 = 0

将

1

到右边

2 cos (4 x) - 1 = 0

两边加上

1

2 cos (4 x) - 1 + 1 = 0 + 1

化简

2 cos (4 x) = 1

2 cos (4 x) = 1

两边除以

2

2 cos (4 x) = 1

两边除以

2

\frac{2 cos ( 4 x )}{2} = \frac{1}{2}

化简

cos (4 x) = \frac{1}{2}

cos (4 x) = \frac{1}{2}

cos (4 x) = \frac{1}{2}

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

4 x = \frac{π}{3} + 2 πn, 4 x = \frac{5 π}{3} + 2 πn

4 x = \frac{π}{3} + 2 πn, 4 x = \frac{5 π}{3} + 2 πn

解

4 x = \frac{π}{3} + 2 πn : x = \frac{π}{12} + \frac{πn}{2}

4 x = \frac{π}{3} + 2 πn

两边除以

4

4 x = \frac{π}{3} + 2 πn

两边除以

4

\frac{4 x}{4} = \frac{\frac{π}{3}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} = \frac{\frac{π}{3}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{\frac{π}{3}}{4} + \frac{2 πn}{4} : \frac{π}{12} + \frac{πn}{2}

\frac{\frac{π}{3}}{4} + \frac{2 πn}{4}

\frac{\frac{π}{3}}{4} = \frac{π}{12}

\frac{\frac{π}{3}}{4}

使用分式法则:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{3 \cdot 4}

数字相乘:

3 \cdot 4 = 12

= \frac{π}{12}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{π}{12} + \frac{πn}{2}

x = \frac{π}{12} + \frac{πn}{2}

x = \frac{π}{12} + \frac{πn}{2}

x = \frac{π}{12} + \frac{πn}{2}

解

4 x = \frac{5 π}{3} + 2 πn : x = \frac{5 π}{12} + \frac{πn}{2}

4 x = \frac{5 π}{3} + 2 πn

两边除以

4

4 x = \frac{5 π}{3} + 2 πn

两边除以

4

\frac{4 x}{4} = \frac{\frac{5 π}{3}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} = \frac{\frac{5 π}{3}}{4} + \frac{2 πn}{4}

化简

\frac{4 x}{4} : x

\frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{\frac{5 π}{3}}{4} + \frac{2 πn}{4} : \frac{5 π}{12} + \frac{πn}{2}

\frac{\frac{5 π}{3}}{4} + \frac{2 πn}{4}

\frac{\frac{5 π}{3}}{4} = \frac{5 π}{12}

\frac{\frac{5 π}{3}}{4}

使用分式法则:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{5 π}{3 \cdot 4}

数字相乘:

3 \cdot 4 = 12

= \frac{5 π}{12}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{5 π}{12} + \frac{πn}{2}

x = \frac{5 π}{12} + \frac{πn}{2}

x = \frac{5 π}{12} + \frac{πn}{2}

x = \frac{5 π}{12} + \frac{πn}{2}

x = \frac{π}{12} + \frac{πn}{2}, x = \frac{5 π}{12} + \frac{πn}{2}

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{12} + \frac{πn}{2}, x = \frac{5 π}{12} + \frac{πn}{2}

作图

流行的例子

8+4cos(2x)=12cos(x)

8 + 4 cos (2 x) = 12 cos (x)

tan (x) = - sec (x)

3tan^4(θ)+1= 2/(tan^2(θ))

3 tan^{4} (θ) + 1 = \frac{2}{tan ^{2} ( θ )}

tan(θ/2+pi/4)=1,0<= θ<= 2pi

tan (\frac{θ}{2} + \frac{π}{4}) = 1, 0 \leq θ \leq 2 π

sin(x+pi/6)+sin(x-pi/6)=(sqrt(3))/2

sin (x + \frac{π}{6}) + sin (x - \frac{π}{6}) = \frac{3}{2}