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受欢迎的三角函数 >

(sec(x))/(cot(x)tan(x))=sin(x)

解答

\frac{sec ( x )}{cot ( x ) tan ( x )} = sin (x)

解答

x \in R 无解

求解步骤

\frac{sec ( x )}{cot ( x ) tan ( x )} = sin (x)

两边减去

sin (x)

\frac{sec ( x )}{cot ( x ) tan ( x )} - sin (x) = 0

化简

\frac{sec ( x )}{cot ( x ) tan ( x )} - sin (x) : \frac{sec ( x ) - sin ( x ) cot ( x ) tan ( x )}{cot ( x ) tan ( x )}

\frac{sec ( x )}{cot ( x ) tan ( x )} - sin (x)

将项转换为分式:

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

= \frac{sec ( x )}{cot ( x ) tan ( x )} - \frac{sin ( x ) cot ( x ) tan ( x )}{cot ( x ) tan ( x )}

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{sec ( x ) - sin ( x ) cot ( x ) tan ( x )}{cot ( x ) tan ( x )}

\frac{sec ( x ) - sin ( x ) cot ( x ) tan ( x )}{cot ( x ) tan ( x )} = 0

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

sec (x) - sin (x) cot (x) tan (x) = 0

用 sin, cos 表示

sec (x) - cot (x) sin (x) tan (x)

使用基本三角恒等式:

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

= \frac{1}{cos ( x )} - cot (x) sin (x) tan (x)

使用基本三角恒等式:

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

= \frac{1}{cos ( x )} - \frac{cos ( x )}{sin ( x )} sin (x) tan (x)

使用基本三角恒等式:

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

= \frac{1}{cos ( x )} - \frac{cos ( x )}{sin ( x )} sin (x) \frac{sin ( x )}{cos ( x )}

化简

\frac{1}{cos ( x )} - \frac{cos ( x )}{sin ( x )} sin (x) \frac{sin ( x )}{cos ( x )} : \frac{1 - sin ( x ) cos ( x )}{cos ( x )}

\frac{1}{cos ( x )} - \frac{cos ( x )}{sin ( x )} sin (x) \frac{sin ( x )}{cos ( x )}

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

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

分式相乘:

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

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

约分:

cos (x)

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

约分:

sin (x)

= sin (x)

= \frac{1}{cos ( x )} - sin (x)

将项转换为分式:

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

= \frac{1}{cos ( x )} - \frac{sin ( x ) cos ( x )}{cos ( x )}

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 - sin ( x ) cos ( x )}{cos ( x )}

= \frac{1 - sin ( x ) cos ( x )}{cos ( x )}

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

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

1 - cos (x) sin (x) = 0

使用三角恒等式改写

1 - cos (x) sin (x)

使用倍角公式:

2 sin (x) cos (x) = sin (2 x)

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

= 1 - \frac{sin ( 2 x )}{2}

1 - \frac{sin ( 2 x )}{2} = 0

将

1

到右边

1 - \frac{sin ( 2 x )}{2} = 0

两边减去

1

1 - \frac{sin ( 2 x )}{2} - 1 = 0 - 1

化简

- \frac{sin ( 2 x )}{2} = - 1

- \frac{sin ( 2 x )}{2} = - 1

在两边乘以

2

- \frac{sin ( 2 x )}{2} = - 1

在两边乘以

2

2 (- \frac{sin ( 2 x )}{2}) = 2 (- 1)

化简

- sin (2 x) = - 2

- sin (2 x) = - 2

两边除以

- 1

- sin (2 x) = - 2

两边除以

- 1

\frac{- sin ( 2 x )}{- 1} = \frac{- 2}{- 1}

化简

sin (2 x) = 2

sin (2 x) = 2

- 1 \leq sin (x) \leq 1

x \in R 无解

作图

流行的例子

\frac{cos ( x )}{2} = - 1

4sin(θ)+2sqrt(3)=0

4 sin (θ) + 23 = 0

400+290cos(x)+290sin(x)=0

400 + 290 cos (x) + 290 sin (x) = 0

sqrt(2)=2sin(2θ)

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4sin^2(x)+4cos(x)-5=0,(0,2pi)

4 sin^{2} (x) + 4 cos (x) - 5 = 0, (0, 2 π)