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受欢迎的三角函数 >

tan(θ)+1=sec(θ)

解答

tan (θ) + 1 = sec (θ)

解答

θ = 2 πn

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n

求解步骤

tan (θ) + 1 = sec (θ)

两边减去

sec (θ)

tan (θ) + 1 - sec (θ) = 0

用 sin, cos 表示

1 - sec (θ) + tan (θ)

使用基本三角恒等式:

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

= 1 - \frac{1}{cos ( θ )} + tan (θ)

使用基本三角恒等式:

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

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

化简

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

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

合并分式

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

使用法则

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

= \frac{- 1 + sin ( θ )}{cos ( θ )}

= 1 + \frac{sin ( θ ) - 1}{cos ( θ )}

将项转换为分式:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

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 ( θ ) - 1 + sin ( θ )}{cos ( θ )}

乘以:

1 \cdot cos (θ) = cos (θ)

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

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

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

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

- 1 + cos (θ) + sin (θ) = 0

使用三角恒等式改写

- 1 + cos (θ) + sin (θ)

sin (θ) + cos (θ) = 2 sin (θ + \frac{π}{4})

sin (θ) + cos (θ)

改写为

= 2 (\frac{1}{2} sin (θ) + \frac{1}{2} cos (θ))

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 (θ) + sin (\frac{π}{4}) cos (θ))

使用角和恒等式:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (θ + \frac{π}{4})

= - 1 + 2 sin (θ + \frac{π}{4})

- 1 + 2 sin (θ + \frac{π}{4}) = 0

将

1

到右边

- 1 + 2 sin (θ + \frac{π}{4}) = 0

两边加上

1

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

化简

2 sin (θ + \frac{π}{4}) = 1

2 sin (θ + \frac{π}{4}) = 1

两边除以

2

2 sin (θ + \frac{π}{4}) = 1

两边除以

2

\frac{2 sin ( θ + \frac{π}{4} )}{2} = \frac{1}{2}

化简

\frac{2 sin ( θ + \frac{π}{4} )}{2} = \frac{1}{2}

化简

\frac{2 sin ( θ + \frac{π}{4} )}{2} : sin (θ + \frac{π}{4})

\frac{2 sin ( θ + \frac{π}{4} )}{2}

约分:

2

= sin (θ + \frac{π}{4})

化简

\frac{1}{2} : \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}

sin (θ + \frac{π}{4}) = \frac{2}{2}

sin (θ + \frac{π}{4}) = \frac{2}{2}

sin (θ + \frac{π}{4}) = \frac{2}{2}

sin (θ + \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}

θ + \frac{π}{4} = \frac{π}{4} + 2 πn, θ + \frac{π}{4} = \frac{3 π}{4} + 2 πn

θ + \frac{π}{4} = \frac{π}{4} + 2 πn, θ + \frac{π}{4} = \frac{3 π}{4} + 2 πn

解

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

θ + \frac{π}{4} = \frac{π}{4} + 2 πn

两边减去

\frac{π}{4}

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

化简

θ = 2 πn

解

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

θ + \frac{π}{4} = \frac{3 π}{4} + 2 πn

将

\frac{π}{4}

到右边

θ + \frac{π}{4} = \frac{3 π}{4} + 2 πn

两边减去

\frac{π}{4}

θ + \frac{π}{4} - \frac{π}{4} = \frac{3 π}{4} + 2 πn - \frac{π}{4}

化简

θ + \frac{π}{4} - \frac{π}{4} = \frac{3 π}{4} + 2 πn - \frac{π}{4}

化简

θ + \frac{π}{4} - \frac{π}{4} : θ

θ + \frac{π}{4} - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} = 0

= θ

化简

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

\frac{3 π}{4} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn - \frac{π}{4} + \frac{3 π}{4}

合并分式

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

使用法则

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

= \frac{- π + 3 π}{4}

同类项相加:

- π + 3 π = 2 π

= \frac{2 π}{4}

约分:

2

= \frac{π}{2}

= 2 πn + \frac{π}{2}

θ = 2 πn + \frac{π}{2}

θ = 2 πn + \frac{π}{2}

θ = 2 πn + \frac{π}{2}

θ = 2 πn, θ = 2 πn + \frac{π}{2}

因为方程对以下值无定义:

2 πn + \frac{π}{2}

θ = 2 πn

作图

流行的例子

2sin(θ)=sqrt(2)

2 sin (θ) = 2

5cos(x)-sqrt(3)=3cos(x)

5 cos (x) - 3 = 3 cos (x)

tan (x) + 3 = 0

sqrt(2)sin(x)-1=0

2 sin (x) - 1 = 0

tan (x) = \frac{5}{3}