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受欢迎的三角函数 >

sin^2(2x)-cos^2(2x)= 1/2

解答

sin^{2} (2 x) - cos^{2} (2 x) = \frac{1}{2}

解答

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

+1

度数

x = 3 0^{\circ} + 9 0^{\circ} n, x = 6 0^{\circ} + 9 0^{\circ} n

求解步骤

sin^{2} (2 x) - cos^{2} (2 x) = \frac{1}{2}

使用三角恒等式改写

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

使用倍角公式:

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

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

= - cos (2 \cdot 2 x)

- cos (2 \cdot 2 x) = \frac{1}{2}

两边除以

- 1

- cos (2 \cdot 2 x) = \frac{1}{2}

两边除以

- 1

\frac{- cos ( 2 \cdot 2 x )}{- 1} = \frac{\frac{1}{2}}{- 1}

化简

\frac{- cos ( 2 \cdot 2 x )}{- 1} = \frac{\frac{1}{2}}{- 1}

化简

\frac{- cos ( 2 \cdot 2 x )}{- 1} : cos (2 \cdot 2 x)

\frac{- cos ( 2 \cdot 2 x )}{- 1}

使用分式法则:

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

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

使用法则

\frac{a}{1} = a

= cos (2 \cdot 2 x)

化简

\frac{\frac{1}{2}}{- 1} : - \frac{1}{2}

\frac{\frac{1}{2}}{- 1}

使用分式法则:

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

= - \frac{\frac{1}{2}}{1}

使用分式法则:

\frac{a}{1} = a

\frac{\frac{1}{2}}{1} = \frac{1}{2}

= - \frac{1}{2}

cos (2 \cdot 2 x) = - \frac{1}{2}

cos (2 \cdot 2 x) = - \frac{1}{2}

cos (2 \cdot 2 x) = - \frac{1}{2}

cos (2 \cdot 2 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}

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

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

解

2 \cdot 2 x = \frac{2 π}{3} + 2 πn : x = \frac{π}{6} + \frac{πn}{2}

2 \cdot 2 x = \frac{2 π}{3} + 2 πn

两边除以

4

2 \cdot 2 x = \frac{2 π}{3} + 2 πn

两边除以

4

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

化简

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

化简

\frac{2 \cdot 2 x}{4} : x

\frac{2 \cdot 2 x}{4}

数字相乘:

2 \cdot 2 = 4

= \frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

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}

x = \frac{π}{6} + \frac{πn}{2}

x = \frac{π}{6} + \frac{πn}{2}

x = \frac{π}{6} + \frac{πn}{2}

解

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

2 \cdot 2 x = \frac{4 π}{3} + 2 πn

两边除以

4

2 \cdot 2 x = \frac{4 π}{3} + 2 πn

两边除以

4

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

化简

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

化简

\frac{2 \cdot 2 x}{4} : x

\frac{2 \cdot 2 x}{4}

数字相乘:

2 \cdot 2 = 4

= \frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

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}

x = \frac{π}{3} + \frac{πn}{2}

x = \frac{π}{3} + \frac{πn}{2}

x = \frac{π}{3} + \frac{πn}{2}

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

作图

流行的例子

sin(x)=(1.2)/(sqrt(10))

sin (x) = \frac{1.2}{10}

solvefor t,x=-3cos(pit)

so l v e f or t, x = - 3 cos (π t)

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

cos(2t)-sin(t)=0.5,0<= t<= 2pi

cos (2 t) - sin (t) = 0.5, 0 \leq t \leq 2 π

sin (2 x) = 0.1