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受欢迎的三角函数 >

sin(θ)+sin(3θ)=0

解答

sin (θ) + sin (3 θ) = 0

解答

θ = πn, θ = \frac{π}{2} + πn

+1

度数

θ = 0^{\circ} + 18 0^{\circ} n, θ = 9 0^{\circ} + 18 0^{\circ} n

求解步骤

sin (θ) + sin (3 θ) = 0

使用三角恒等式改写

sin (3 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})

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

化简

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

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

\frac{3 θ + θ}{2} = 2 θ

\frac{3 θ + θ}{2}

同类项相加:

3 θ + θ = 4 θ

= \frac{4 θ}{2}

数字相除:

\frac{4}{2} = 2

= 2 θ

= 2 sin (2 θ) cos (\frac{3 θ - θ}{2})

\frac{3 θ - θ}{2} = θ

\frac{3 θ - θ}{2}

同类项相加:

3 θ - θ = 2 θ

= \frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

= 2 sin (2 θ) cos (θ)

= 2 sin (2 θ) cos (θ)

2 cos (θ) sin (2 θ) = 0

分别求解每个部分

cos (θ) = 0 or sin (2 θ) = 0

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

cos (θ) = 0

cos (θ) = 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}

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

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

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

sin (2 θ) = 0

sin (2 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}

2 θ = 0 + 2 πn, 2 θ = π + 2 πn

2 θ = 0 + 2 πn, 2 θ = π + 2 πn

解

2 θ = 0 + 2 πn : θ = πn

2 θ = 0 + 2 πn

0 + 2 πn = 2 πn

2 θ = 2 πn

两边除以

2

2 θ = 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{2 πn}{2}

化简

θ = πn

θ = πn

解

2 θ = π + 2 πn : θ = \frac{π}{2} + πn

2 θ = π + 2 πn

两边除以

2

2 θ = π + 2 πn

两边除以

2

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

化简

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

θ = πn, θ = \frac{π}{2} + πn

合并所有解

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

合并重叠的区间

θ = πn, θ = \frac{π}{2} + πn

作图

流行的例子

4sin^2(x)-1=0,0<= x<2pi

4 sin^{2} (x) - 1 = 0, 0 \leq x < 2 π

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

cos^{2} (x) - 4 sin^{2} (x) = 2

sin (x) = 0.53

2cos(2θ)=sqrt(2)

2 cos (2 θ) = 2

2 + sin (x) = 0