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受欢迎的三角函数 >

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

解答

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

解答

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

+1

度数

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

求解步骤

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

使用三角恒等式改写

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

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

化简

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

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

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

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

化简

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

x + \frac{π}{2} - (x - \frac{π}{2})

将项转换为分式:

x = \frac{x 2}{2}, (x - \frac{π}{2}) = \frac{( x - \frac{π}{2} ) 2}{2}

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

乘开

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

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

= 2 x + π - 2 (x - \frac{π}{2})

乘开

- 2 (x - \frac{π}{2}) : - 2 x + π

- 2 (x - \frac{π}{2})

使用分配律:

a (b - c) = ab - a c

a = - 2, b = x, c = \frac{π}{2}

= - 2 x - (- 2) \frac{π}{2}

使用加减运算法则

- (- a) = a

= - 2 x + 2 \cdot \frac{π}{2}

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

分式相乘:

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

= \frac{π 2}{2}

约分:

2

= π

= - 2 x + π

= x \cdot 2 + π - 2 x + π

化简

x \cdot 2 + π - 2 x + π : 2 π

x \cdot 2 + π - 2 x + π

对同类项分组

= 2 x - 2 x + π + π

同类项相加:

2 x - 2 x = 0

= π + π

同类项相加:

π + π = 2 π

= 2 π

= 2 π

= \frac{2 π}{2}

数字相除:

\frac{2}{2} = 1

= π

= \frac{π}{2}

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

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

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

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

x + \frac{π}{2} + x - \frac{π}{2}

对同类项分组

= x + x + \frac{π}{2} - \frac{π}{2}

同类项相加:

x + x = 2 x

= 2 x + \frac{π}{2} - \frac{π}{2}

同类项相加:

\frac{π}{2} - \frac{π}{2} = 0

= 2 x

= \frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

= 2 sin (\frac{π}{2}) cos (x)

化简

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

使用以下普通恒等式:

sin (\frac{π}{2}) = 1

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}

= 1

= 2 \cdot 1 \cdot cos (x)

数字相乘:

2 \cdot 1 = 2

= 2 cos (x)

= 2 cos (x)

2 cos (x) = 2

两边除以

2

2 cos (x) = 2

两边除以

2

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

化简

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

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

cos (x) = \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}

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

作图

流行的例子

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