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受欢迎的三角函数 >

r=a(1-sin(x))

解答

r = a (1 - sin (x))

解答

x = arcsin (- \frac{r - a}{a}) + 2 πn, x = π + arcsin (\frac{r - a}{a}) + 2 πn

求解步骤

r = a (1 - sin (x))

交换两边

a (1 - sin (x)) = r

两边除以

a; a \neq = 0

a (1 - sin (x)) = r

两边除以

a; a \neq = 0

\frac{a ( 1 - sin ( x ) )}{a} = \frac{r}{a}; a \neq = 0

化简

1 - sin (x) = \frac{r}{a}; a \neq = 0

1 - sin (x) = \frac{r}{a}; a \neq = 0

将

1

到右边

1 - sin (x) = \frac{r}{a}; a \neq = 0

两边减去

1

1 - sin (x) - 1 = \frac{r}{a} - 1; a \neq = 0

化简

- sin (x) = \frac{r}{a} - 1; a \neq = 0

- sin (x) = \frac{r}{a} - 1; a \neq = 0

两边除以

- 1; a \neq = 0

- sin (x) = \frac{r}{a} - 1; a \neq = 0

两边除以

- 1; a \neq = 0

\frac{- sin ( x )}{- 1} = \frac{\frac{r}{a}}{- 1} - \frac{1}{- 1}; a \neq = 0

化简

\frac{- sin ( x )}{- 1} = \frac{\frac{r}{a}}{- 1} - \frac{1}{- 1}

化简

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

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

使用分式法则:

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

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

使用法则

\frac{a}{1} = a

= sin (x)

化简

\frac{\frac{r}{a}}{- 1} - \frac{1}{- 1} : - \frac{r - a}{a}

\frac{\frac{r}{a}}{- 1} - \frac{1}{- 1}

使用法则

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

= \frac{\frac{r}{a} - 1}{- 1}

使用分式法则:

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

= - \frac{\frac{r}{a} - 1}{1}

化简

\frac{r}{a} - 1 : \frac{r - a}{a}

\frac{r}{a} - 1

将项转换为分式:

1 = \frac{1 a}{a}

= \frac{r}{a} - \frac{1 \cdot a}{a}

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{r - 1 \cdot a}{a}

乘以:

1 \cdot a = a

= \frac{r - a}{a}

= - \frac{\frac{r - a}{a}}{1}

使用分式法则:

\frac{a}{1} = a

= - \frac{r - a}{a}

sin (x) = - \frac{r - a}{a}; a \neq = 0

sin (x) = - \frac{r - a}{a}; a \neq = 0

sin (x) = - \frac{r - a}{a}; a \neq = 0

使用反三角函数性质

sin (x) = - \frac{r - a}{a}

sin (x) = - \frac{r - a}{a}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{r - a}{a}) + 2 πn, x = π + arcsin (\frac{r - a}{a}) + 2 πn

x = arcsin (- \frac{r - a}{a}) + 2 πn, x = π + arcsin (\frac{r - a}{a}) + 2 πn

作图

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