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受欢迎的三角函数 >

tan(t)+sec(t)=1

解答

tan (t) + sec (t) = 1

解答

t = 2 πn + 2 π

+1

度数

t = 36 0^{\circ} + 36 0^{\circ} n

求解步骤

tan (t) + sec (t) = 1

两边减去

1

tan (t) + sec (t) - 1 = 0

用 sin, cos 表示

- 1 + sec (t) + tan (t)

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

= - 1 + \frac{1}{cos ( t )} + tan (t)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - 1 + \frac{1}{cos ( t )} + \frac{sin ( t )}{cos ( t )}

化简

- 1 + \frac{1}{cos ( t )} + \frac{sin ( t )}{cos ( t )} : \frac{- cos ( t ) + 1 + sin ( t )}{cos ( t )}

- 1 + \frac{1}{cos ( t )} + \frac{sin ( t )}{cos ( t )}

合并分式

\frac{1}{cos ( t )} + \frac{sin ( t )}{cos ( t )} : \frac{1 + sin ( t )}{cos ( t )}

使用法则

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

= \frac{1 + sin ( t )}{cos ( t )}

= - 1 + \frac{sin ( t ) + 1}{cos ( t )}

将项转换为分式:

1 = \frac{1 cos ( t )}{cos ( t )}

= - \frac{1 \cdot cos ( t )}{cos ( t )} + \frac{1 + sin ( t )}{cos ( t )}

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{- 1 \cdot cos ( t ) + 1 + sin ( t )}{cos ( t )}

乘以:

1 \cdot cos (t) = cos (t)

= \frac{- cos ( t ) + 1 + sin ( t )}{cos ( t )}

= \frac{- cos ( t ) + 1 + sin ( t )}{cos ( t )}

\frac{1 - cos ( t ) + sin ( t )}{cos ( t )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

1 - cos (t) + sin (t) = 0

使用三角恒等式改写

1 - cos (t) + sin (t)

sin (t) - cos (t) = 2 sin (t - \frac{π}{4})

sin (t) - cos (t)

改写为

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

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{1}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (t) - sin (\frac{π}{4}) cos (t))

使用角和恒等式:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

= 2 sin (t - \frac{π}{4})

= 1 + 2 sin (t - \frac{π}{4})

1 + 2 sin (t - \frac{π}{4}) = 0

将

1

到右边

1 + 2 sin (t - \frac{π}{4}) = 0

两边减去

1

1 + 2 sin (t - \frac{π}{4}) - 1 = 0 - 1

化简

2 sin (t - \frac{π}{4}) = - 1

2 sin (t - \frac{π}{4}) = - 1

两边除以

2

2 sin (t - \frac{π}{4}) = - 1

两边除以

2

\frac{2 sin ( t - \frac{π}{4} )}{2} = \frac{- 1}{2}

化简

\frac{2 sin ( t - \frac{π}{4} )}{2} = \frac{- 1}{2}

化简

\frac{2 sin ( t - \frac{π}{4} )}{2} : sin (t - \frac{π}{4})

\frac{2 sin ( t - \frac{π}{4} )}{2}

约分:

2

= sin (t - \frac{π}{4})

化简

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

\frac{- 1}{2}

使用分式法则:

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

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

sin (t - \frac{π}{4}) = - \frac{2}{2}

sin (t - \frac{π}{4}) = - \frac{2}{2}

sin (t - \frac{π}{4}) = - \frac{2}{2}

sin (t - \frac{π}{4}) = - \frac{2}{2}

的通解

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}

t - \frac{π}{4} = \frac{5 π}{4} + 2 πn, t - \frac{π}{4} = \frac{7 π}{4} + 2 πn

t - \frac{π}{4} = \frac{5 π}{4} + 2 πn, t - \frac{π}{4} = \frac{7 π}{4} + 2 πn

解

t - \frac{π}{4} = \frac{5 π}{4} + 2 πn : t = 2 πn + \frac{3 π}{2}

t - \frac{π}{4} = \frac{5 π}{4} + 2 πn

将

\frac{π}{4}

到右边

t - \frac{π}{4} = \frac{5 π}{4} + 2 πn

两边加上

\frac{π}{4}

t - \frac{π}{4} + \frac{π}{4} = \frac{5 π}{4} + 2 πn + \frac{π}{4}

化简

t - \frac{π}{4} + \frac{π}{4} = \frac{5 π}{4} + 2 πn + \frac{π}{4}

化简

t - \frac{π}{4} + \frac{π}{4} : t

t - \frac{π}{4} + \frac{π}{4}

同类项相加:

- \frac{π}{4} + \frac{π}{4} = 0

= t

化简

\frac{5 π}{4} + 2 πn + \frac{π}{4} : 2 πn + \frac{3 π}{2}

\frac{5 π}{4} + 2 πn + \frac{π}{4}

对同类项分组

= 2 πn + \frac{π}{4} + \frac{5 π}{4}

合并分式

\frac{π}{4} + \frac{5 π}{4} : \frac{3 π}{2}

使用法则

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

= \frac{π + 5 π}{4}

同类项相加:

π + 5 π = 6 π

= \frac{6 π}{4}

约分:

2

= \frac{3 π}{2}

= 2 πn + \frac{3 π}{2}

t = 2 πn + \frac{3 π}{2}

t = 2 πn + \frac{3 π}{2}

t = 2 πn + \frac{3 π}{2}

解

t - \frac{π}{4} = \frac{7 π}{4} + 2 πn : t = 2 πn + 2 π

t - \frac{π}{4} = \frac{7 π}{4} + 2 πn

将

\frac{π}{4}

到右边

t - \frac{π}{4} = \frac{7 π}{4} + 2 πn

两边加上

\frac{π}{4}

t - \frac{π}{4} + \frac{π}{4} = \frac{7 π}{4} + 2 πn + \frac{π}{4}

化简

t - \frac{π}{4} + \frac{π}{4} = \frac{7 π}{4} + 2 πn + \frac{π}{4}

化简

t - \frac{π}{4} + \frac{π}{4} : t

t - \frac{π}{4} + \frac{π}{4}

同类项相加:

- \frac{π}{4} + \frac{π}{4} = 0

= t

化简

\frac{7 π}{4} + 2 πn + \frac{π}{4} : 2 πn + 2 π

\frac{7 π}{4} + 2 πn + \frac{π}{4}

对同类项分组

= 2 πn + \frac{π}{4} + \frac{7 π}{4}

合并分式

\frac{π}{4} + \frac{7 π}{4} : 2 π

使用法则

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

= \frac{π + 7 π}{4}

同类项相加:

π + 7 π = 8 π

= \frac{8 π}{4}

数字相除:

\frac{8}{4} = 2

= 2 π

= 2 πn + 2 π

t = 2 πn + 2 π

t = 2 πn + 2 π

t = 2 πn + 2 π

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

因为方程对以下值无定义:

2 πn + \frac{3 π}{2}

t = 2 πn + 2 π

作图

流行的例子

sin(5x-10)=cos(x-8)

sin (5 x - 10) = cos (x - 8)

sin^2(x)+5sin(x)=3

sin^{2} (x) + 5 sin (x) = 3

14+7cos(2x)=21cos(x)

14 + 7 cos (2 x) = 21 cos (x)

sin(3x)=0,0<= x<2pi

sin (3 x) = 0, 0 \leq x < 2 π

sin(θ)=-1/5 ,pi<θ<(3pi)/2 ,cos(θ)

sin (θ) = - \frac{1}{5}, π < θ < \frac{3 π}{2}, cos (θ)