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受欢迎的三角函数 >

cos(θ)cos(3θ)-sin(θ)sin(3θ)=0

解答

cos (θ) cos (3 θ) - sin (θ) sin (3 θ) = 0

解答

θ = \frac{π}{8} + \frac{πn}{2}, θ = \frac{3 π}{8} + \frac{πn}{2}

+1

度数

θ = 22. 5^{\circ} + 9 0^{\circ} n, θ = 67. 5^{\circ} + 9 0^{\circ} n

求解步骤

cos (θ) cos (3 θ) - sin (θ) sin (3 θ) = 0

使用三角恒等式改写

cos (θ) cos (3 θ) - sin (θ) sin (3 θ)

使用角和恒等式:

cos (s) cos (t) - sin (s) sin (t) = cos (s + t)

= cos (θ + 3 θ)

cos (θ + 3 θ) = 0

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

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

解

θ + 3 θ = \frac{π}{2} + 2 πn : θ = \frac{π}{8} + \frac{πn}{2}

θ + 3 θ = \frac{π}{2} + 2 πn

同类项相加:

θ + 3 θ = 4 θ

4 θ = \frac{π}{2} + 2 πn

两边除以

4

4 θ = \frac{π}{2} + 2 πn

两边除以

4

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

化简

\frac{4 θ}{4} = \frac{\frac{π}{2}}{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{π}{2}}{4} + \frac{2 πn}{4} : \frac{π}{8} + \frac{πn}{2}

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

\frac{\frac{π}{2}}{4} = \frac{π}{8}

\frac{\frac{π}{2}}{4}

使用分式法则:

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

= \frac{π}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{π}{8}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

解

θ + 3 θ = \frac{3 π}{2} + 2 πn : θ = \frac{3 π}{8} + \frac{πn}{2}

θ + 3 θ = \frac{3 π}{2} + 2 πn

同类项相加:

θ + 3 θ = 4 θ

4 θ = \frac{3 π}{2} + 2 πn

两边除以

4

4 θ = \frac{3 π}{2} + 2 πn

两边除以

4

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

化简

\frac{4 θ}{4} = \frac{\frac{3 π}{2}}{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{3 π}{2}}{4} + \frac{2 πn}{4} : \frac{3 π}{8} + \frac{πn}{2}

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

\frac{\frac{3 π}{2}}{4} = \frac{3 π}{8}

\frac{\frac{3 π}{2}}{4}

使用分式法则:

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

= \frac{3 π}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{3 π}{8}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}, θ = \frac{3 π}{8} + \frac{πn}{2}

作图

流行的例子

1-2sin(θ)=cos(2θ),0<= θ<= 2pi

1 - 2 sin (θ) = cos (2 θ), 0 \leq θ \leq 2 π

2sin^2(θ)+5sin(θ)-12=0

2 sin^{2} (θ) + 5 sin (θ) - 12 = 0

cos(2θ)= 3/5 ,90<= θ<= 180

cos (2 θ) = \frac{3}{5}, 9 0^{\circ} \leq θ \leq 18 0^{\circ}

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

sin^{2} (x) - 4 cos (x) - 4 = 0, 0 \leq x \leq 2 π

solvefor x,sin(x)=0.5

so l v e f or x, sin (x) = 0.5