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受欢迎的三角函数 >

cos((pit)/2) pi/2 =0

解答

cos (\frac{π t}{2}) \frac{π}{2} = 0

解答

t = 1 + 4 n, t = 3 + 4 n

+1

度数

t = 57.29577 \dots^{\circ} + 229.18311 \dots^{\circ} n, t = 171.88733 \dots^{\circ} + 229.18311 \dots^{\circ} n

求解步骤

cos (\frac{π t}{2}) \frac{π}{2} = 0

在两边乘以

2

cos (\frac{π t}{2}) \frac{π}{2} = 0

在两边乘以

2

2 cos (\frac{π t}{2}) \frac{π}{2} = 0 \cdot 2

化简

π cos (\frac{π t}{2}) = 0

π cos (\frac{π t}{2}) = 0

两边除以

π

π cos (\frac{π t}{2}) = 0

两边除以

π

\frac{π cos ( \frac{π t}{2} )}{π} = \frac{0}{π}

化简

cos (\frac{π t}{2}) = 0

cos (\frac{π t}{2}) = 0

cos (\frac{π t}{2}) = 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}

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

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

解

\frac{π t}{2} = \frac{π}{2} + 2 πn : t = 1 + 4 n

\frac{π t}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

\frac{π t}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

\frac{2 π t}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

化简

\frac{2 π t}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

化简

\frac{2 π t}{2} : π t

\frac{2 π t}{2}

数字相除:

\frac{2}{2} = 1

= π t

化简

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

2 \cdot \frac{π}{2} + 2 \cdot 2 πn

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 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= π + 4 πn

π t = π + 4 πn

π t = π + 4 πn

π t = π + 4 πn

两边除以

π

π t = π + 4 πn

两边除以

π

\frac{π t}{π} = \frac{π}{π} + \frac{4 πn}{π}

化简

\frac{π t}{π} = \frac{π}{π} + \frac{4 πn}{π}

化简

\frac{π t}{π} : t

\frac{π t}{π}

约分:

π

= t

化简

\frac{π}{π} + \frac{4 πn}{π} : 1 + 4 n

\frac{π}{π} + \frac{4 πn}{π}

使用法则

\frac{a}{a} = 1

\frac{π}{π} = 1

= 1 + \frac{4 πn}{π}

消掉

\frac{4 πn}{π} : 4 n

\frac{4 πn}{π}

约分:

π

= 4 n

= 1 + 4 n

t = 1 + 4 n

t = 1 + 4 n

t = 1 + 4 n

解

\frac{π t}{2} = \frac{3 π}{2} + 2 πn : t = 3 + 4 n

\frac{π t}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

\frac{π t}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

\frac{2 π t}{2} = 2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

化简

\frac{2 π t}{2} = 2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

化简

\frac{2 π t}{2} : π t

\frac{2 π t}{2}

数字相除:

\frac{2}{2} = 1

= π t

化简

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

2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

2 \cdot \frac{3 π}{2} = 3 π

2 \cdot \frac{3 π}{2}

分式相乘:

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

= \frac{3 π 2}{2}

约分:

2

= 3 π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= 3 π + 4 πn

π t = 3 π + 4 πn

π t = 3 π + 4 πn

π t = 3 π + 4 πn

两边除以

π

π t = 3 π + 4 πn

两边除以

π

\frac{π t}{π} = \frac{3 π}{π} + \frac{4 πn}{π}

化简

\frac{π t}{π} = \frac{3 π}{π} + \frac{4 πn}{π}

化简

\frac{π t}{π} : t

\frac{π t}{π}

约分:

π

= t

化简

\frac{3 π}{π} + \frac{4 πn}{π} : 3 + 4 n

\frac{3 π}{π} + \frac{4 πn}{π}

消掉

\frac{3 π}{π} : 3

\frac{3 π}{π}

约分:

π

= 3

= 3 + \frac{4 πn}{π}

消掉

\frac{4 πn}{π} : 4 n

\frac{4 πn}{π}

约分:

π

= 4 n

= 3 + 4 n

t = 3 + 4 n

t = 3 + 4 n

t = 3 + 4 n

t = 1 + 4 n, t = 3 + 4 n

作图

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