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受欢迎的三角函数 >

sin(3θ)-sin(5θ)=sin(θ),0<= θ<= pi/2

解答

sin (3 θ) - sin (5 θ) = sin (θ), 0 \leq θ \leq \frac{π}{2}

解答

θ = 0, θ = \frac{π}{6}, θ = \frac{π}{3}

+1

度数

θ = 0^{\circ}, θ = 3 0^{\circ}, θ = 6 0^{\circ}

求解步骤

sin (3 θ) - sin (5 θ) = sin (θ), 0 \leq θ \leq \frac{π}{2}

两边减去

sin (θ)

sin (3 θ) - sin (5 θ) - sin (θ) = 0

使用三角恒等式改写

sin (3 θ) - sin (5 θ) - sin (θ)

使用和差化积恒等式:

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

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

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

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

\frac{3 θ - 5 θ}{2} = - θ

\frac{3 θ - 5 θ}{2}

同类项相加:

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

使用负角恒等式:

sin (- x) = - sin (x)

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

去除括号:

(- a) = - a

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

\frac{3 θ + 5 θ}{2} = 4 θ

\frac{3 θ + 5 θ}{2}

同类项相加:

3 θ + 5 θ = 8 θ

= \frac{8 θ}{2}

数字相除:

\frac{8}{2} = 4

= 4 θ

= - 2 cos (4 θ) sin (θ)

= - sin (θ) - 2 cos (4 θ) sin (θ)

- sin (θ) - 2 cos (4 θ) sin (θ) = 0

分解

- sin (θ) - 2 cos (4 θ) sin (θ) : - sin (θ) (2 cos (4 θ) + 1)

- sin (θ) - 2 cos (4 θ) sin (θ)

因式分解出通项

- sin (θ)

= - sin (θ) (1 + 2 cos (4 θ))

- sin (θ) (2 cos (4 θ) + 1) = 0

分别求解每个部分

sin (θ) = 0 or 2 cos (4 θ) + 1 = 0

sin (θ) = 0, 0 \leq θ \leq \frac{π}{2} : θ = 0

sin (θ) = 0, 0 \leq θ \leq \frac{π}{2}

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

在

0 \leq θ \leq \frac{π}{2}

范围内的解

θ = 0

2 cos (4 θ) + 1 = 0, 0 \leq θ \leq \frac{π}{2} : θ = \frac{π}{6}, θ = \frac{π}{3}

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

将

1

到右边

2 cos (4 θ) + 1 = 0

两边减去

1

2 cos (4 θ) + 1 - 1 = 0 - 1

化简

2 cos (4 θ) = - 1

2 cos (4 θ) = - 1

两边除以

2

2 cos (4 θ) = - 1

两边除以

2

\frac{2 cos ( 4 θ )}{2} = \frac{- 1}{2}

化简

cos (4 θ) = - \frac{1}{2}

cos (4 θ) = - \frac{1}{2}

cos (4 θ) = - \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 θ = \frac{2 π}{3} + 2 πn, 4 θ = \frac{4 π}{3} + 2 πn

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

解

4 θ = \frac{2 π}{3} + 2 πn : θ = \frac{π}{6} + \frac{πn}{2}

4 θ = \frac{2 π}{3} + 2 πn

两边除以

4

4 θ = \frac{2 π}{3} + 2 πn

两边除以

4

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

化简

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

化简

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

数字相除:

\frac{4}{4} = 1

= θ

化简

\frac{\frac{2 π}{3}}{4} + \frac{2 πn}{4} : \frac{π}{6} + \frac{πn}{2}

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

\frac{\frac{2 π}{3}}{4} = \frac{π}{6}

\frac{\frac{2 π}{3}}{4}

使用分式法则:

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

= \frac{2 π}{3 \cdot 4}

数字相乘:

3 \cdot 4 = 12

= \frac{2 π}{12}

约分:

2

= \frac{π}{6}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

解

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

4 θ = \frac{4 π}{3} + 2 πn

两边除以

4

4 θ = \frac{4 π}{3} + 2 πn

两边除以

4

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

化简

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

化简

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

数字相除:

\frac{4}{4} = 1

= θ

化简

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

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

\frac{\frac{4 π}{3}}{4} = \frac{π}{3}

\frac{\frac{4 π}{3}}{4}

使用分式法则:

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

= \frac{4 π}{3 \cdot 4}

数字相乘:

3 \cdot 4 = 12

= \frac{4 π}{12}

约分:

4

= \frac{π}{3}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}, θ = \frac{π}{3} + \frac{πn}{2}

在

0 \leq θ \leq \frac{π}{2}

范围内的解

θ = \frac{π}{6}, θ = \frac{π}{3}

合并所有解

θ = 0, θ = \frac{π}{6}, θ = \frac{π}{3}

作图

流行的例子

4 sin (2 θ) - 1 = 0

sec(w)csc(w)=sqrt(2)sec(w)

sec (w) csc (w) = 2 sec (w)

-2sin(x)-cos(2x)*2=0

- 2 sin (x) - cos (2 x) \cdot 2 = 0

-2cos(B)+4=cos(B)+6

- 2 cos (B) + 4 = cos (B) + 6

csc^3(x)+csc^2(x)=2csc(x)+2

csc^{3} (x) + csc^{2} (x) = 2 csc (x) + 2