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受欢迎的三角函数 >

tan(θ/2-pi/4)=1

解答

tan (\frac{θ}{2} - \frac{π}{4}) = 1

解答

θ = π + 2 πn

+1

度数

θ = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

tan (\frac{θ}{2} - \frac{π}{4}) = 1

tan (\frac{θ}{2} - \frac{π}{4}) = 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

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

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

解

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

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

将

\frac{π}{4}

到右边

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

两边加上

\frac{π}{4}

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

化简

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

化简

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

\frac{θ}{2} - \frac{π}{4} + \frac{π}{4}

同类项相加:

- \frac{π}{4} + \frac{π}{4} = 0

= \frac{θ}{2}

化简

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

\frac{π}{4} + πn + \frac{π}{4}

对同类项分组

= \frac{π}{4} + \frac{π}{4} + πn

合并分式

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

使用法则

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

= \frac{π + π}{4}

同类项相加:

π + π = 2 π

= \frac{2 π}{4}

约分:

2

= \frac{π}{2}

= \frac{π}{2} + πn

\frac{θ}{2} = \frac{π}{2} + πn

\frac{θ}{2} = \frac{π}{2} + πn

\frac{θ}{2} = \frac{π}{2} + πn

在两边乘以

2

\frac{θ}{2} = \frac{π}{2} + πn

在两边乘以

2

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

化简

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

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

2 \cdot \frac{π}{2} + 2 πn : π + 2 πn

2 \cdot \frac{π}{2} + 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

= π + 2 πn

θ = π + 2 πn

θ = π + 2 πn

θ = π + 2 πn

θ = π + 2 πn

作图

流行的例子

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