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受欢迎的三角函数 >

cos(2θ)=-(sqrt(2))/2

解答

cos (2 θ) = - \frac{2}{2}

解答

θ = \frac{3 π}{8} + πn, θ = \frac{5 π}{8} + πn

+1

度数

θ = 67. 5^{\circ} + 18 0^{\circ} n, θ = 112. 5^{\circ} + 18 0^{\circ} n

求解步骤

cos (2 θ) = - \frac{2}{2}

cos (2 θ) = - \frac{2}{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}

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

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

解

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

2 θ = \frac{3 π}{4} + 2 πn

两边除以

2

2 θ = \frac{3 π}{4} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

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

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

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

\frac{\frac{3 π}{4}}{2}

使用分式法则:

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

= \frac{3 π}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{3 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{3 π}{8} + πn

θ = \frac{3 π}{8} + πn

θ = \frac{3 π}{8} + πn

θ = \frac{3 π}{8} + πn

解

2 θ = \frac{5 π}{4} + 2 πn : θ = \frac{5 π}{8} + πn

2 θ = \frac{5 π}{4} + 2 πn

两边除以

2

2 θ = \frac{5 π}{4} + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} = \frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2} : \frac{5 π}{8} + πn

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

\frac{\frac{5 π}{4}}{2} = \frac{5 π}{8}

\frac{\frac{5 π}{4}}{2}

使用分式法则:

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

= \frac{5 π}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{5 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{5 π}{8} + πn

θ = \frac{5 π}{8} + πn

θ = \frac{5 π}{8} + πn

θ = \frac{5 π}{8} + πn

θ = \frac{3 π}{8} + πn, θ = \frac{5 π}{8} + πn

作图

流行的例子

tan(θ/2+pi/3)=-1

tan (\frac{θ}{2} + \frac{π}{3}) = - 1

sin(2x)=sqrt(3)cos(2x)

sin (2 x) = 3 cos (2 x)

sec(x)=-sqrt(2),0<= x<= 2pi

sec (x) = - 2, 0 \leq x \leq 2 π

solvefor t,g=sin(2pift)

so l v e f or t, g = sin (2 π f t)

sqrt(3)sin(x)-cos(x)=-sqrt(3)

3 sin (x) - cos (x) = - 3