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受欢迎的三角函数 >

cos(2x)+cos(pi/2-2x)=-sqrt(2)

解答

cos (2 x) + cos (\frac{π}{2} - 2 x) = - 2

解答

x = \frac{5 π}{8} + πn

+1

度数

x = 112. 5^{\circ} + 18 0^{\circ} n

求解步骤

cos (2 x) + cos (\frac{π}{2} - 2 x) = - 2

使用三角恒等式改写

cos (2 x) + cos (\frac{π}{2} - 2 x)

使用和差化积恒等式:

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

= 2 cos (\frac{2 x + \frac{π}{2} - 2 x}{2}) cos (\frac{2 x - ( \frac{π}{2} - 2 x )}{2})

化简

2 cos (\frac{2 x + \frac{π}{2} - 2 x}{2}) cos (\frac{2 x - ( \frac{π}{2} - 2 x )}{2}) : 2 cos (\frac{8 x - π}{4})

2 cos (\frac{2 x + \frac{π}{2} - 2 x}{2}) cos (\frac{2 x - ( \frac{π}{2} - 2 x )}{2})

\frac{2 x + \frac{π}{2} - 2 x}{2} = \frac{π}{4}

\frac{2 x + \frac{π}{2} - 2 x}{2}

2 x + \frac{π}{2} - 2 x = \frac{π}{2}

2 x + \frac{π}{2} - 2 x

对同类项分组

= 2 x - 2 x + \frac{π}{2}

同类项相加:

2 x - 2 x = 0

= \frac{π}{2}

= \frac{\frac{π}{2}}{2}

使用分式法则:

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

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

= 2 cos (\frac{π}{4}) cos (\frac{2 x - ( - 2 x + \frac{π}{2} )}{2})

\frac{2 x - ( \frac{π}{2} - 2 x )}{2} = \frac{8 x - π}{4}

\frac{2 x - ( \frac{π}{2} - 2 x )}{2}

乘开

2 x - (\frac{π}{2} - 2 x) : 4 x - \frac{π}{2}

2 x - (\frac{π}{2} - 2 x)

- (\frac{π}{2} - 2 x) : - \frac{π}{2} + 2 x

- (\frac{π}{2} - 2 x)

打开括号

= - (\frac{π}{2}) - (- 2 x)

使用加减运算法则

- (- a) = a, - (a) = - a

= - \frac{π}{2} + 2 x

= 2 x - \frac{π}{2} + 2 x

化简

2 x - \frac{π}{2} + 2 x : 4 x - \frac{π}{2}

2 x - \frac{π}{2} + 2 x

对同类项分组

= 2 x + 2 x - \frac{π}{2}

同类项相加:

2 x + 2 x = 4 x

= 4 x - \frac{π}{2}

= 4 x - \frac{π}{2}

= \frac{4 x - \frac{π}{2}}{2}

化简

4 x - \frac{π}{2} : \frac{8 x - π}{2}

4 x - \frac{π}{2}

将项转换为分式:

4 x = \frac{4 x 2}{2}

= \frac{4 x \cdot 2}{2} - \frac{π}{2}

因为分母相等，所以合并分式:

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

= \frac{4 x \cdot 2 - π}{2}

数字相乘:

4 \cdot 2 = 8

= \frac{8 x - π}{2}

= \frac{\frac{8 x - π}{2}}{2}

使用分式法则:

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

= \frac{8 x - π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{8 x - π}{4}

= 2 cos (\frac{π}{4}) cos (\frac{8 x - π}{4})

化简

cos (\frac{π}{4}) : \frac{2}{2}

cos (\frac{π}{4})

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{2}{2}

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}

= \frac{2}{2}

= 2 \cdot \frac{2}{2} cos (\frac{8 x - π}{4})

分式相乘:

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

= \frac{2 \cdot 2 cos ( \frac{8 x - π}{4} )}{2}

约分:

2

= 2 cos (\frac{8 x - π}{4})

= 2 cos (\frac{8 x - π}{4})

2 cos (\frac{8 x - π}{4}) = - 2

两边除以

2

2 cos (\frac{8 x - π}{4}) = - 2

两边除以

2

\frac{2 cos ( \frac{8 x - π}{4} )}{2} = \frac{- 2}{2}

化简

cos (\frac{8 x - π}{4}) = - 1

cos (\frac{8 x - π}{4}) = - 1

cos (\frac{8 x - π}{4}) = - 1

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}

\frac{8 x - π}{4} = π + 2 πn

\frac{8 x - π}{4} = π + 2 πn

解

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

\frac{8 x - π}{4} = π + 2 πn

在两边乘以

4

\frac{8 x - π}{4} = π + 2 πn

在两边乘以

4

\frac{4 ( 8 x - π )}{4} = 4 π + 4 \cdot 2 πn

化简

8 x - π = 4 π + 8 πn

8 x - π = 4 π + 8 πn

将

π

到右边

8 x - π = 4 π + 8 πn

两边加上

π

8 x - π + π = 4 π + 8 πn + π

化简

8 x = 5 π + 8 πn

8 x = 5 π + 8 πn

两边除以

8

8 x = 5 π + 8 πn

两边除以

8

\frac{8 x}{8} = \frac{5 π}{8} + \frac{8 πn}{8}

化简

x = \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

作图

流行的例子

cos^{2} (x) = - 1.5

2arctan(1/2)-arctan(x)= pi/4

2 arctan (\frac{1}{2}) - arctan (x) = \frac{π}{4}

cos(x)=(2-tan(x))(1+sin(x))

cos (x) = (2 - tan (x)) (1 + sin (x))

sin (θ) = \frac{13}{85}

sin(2x)=((6m-5))/8

sin (2 x) = \frac{( 6 m - 5 )}{8}