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受欢迎的三角函数 >

solvefor a,sin(a+b/2)=cos(c/2)

解答

求解

a, sin (a + \frac{b}{2}) = cos (\frac{c}{2})

解答

a = \frac{π - c - b}{2} + 2 πn, a = π + 2 πn + \frac{- π + c - b}{2}

求解步骤

sin (a + \frac{b}{2}) = cos (\frac{c}{2})

使用三角恒等式改写

cos (\frac{c}{2})

利用以下特性:

cos (x) = sin (\frac{π}{2} - x)

sin (\frac{π}{2} - \frac{c}{2})

sin (a + \frac{b}{2}) = sin (\frac{π}{2} - \frac{c}{2})

使用反三角函数性质

sin (a + \frac{b}{2}) = sin (\frac{π}{2} - \frac{c}{2})

sin (x) = sin (y) \Rightarrow x = y + 2 π n, x = π - y + 2 π n

a + \frac{b}{2} = \frac{π}{2} - \frac{c}{2} + 2 πn, a + \frac{b}{2} = π - (\frac{π}{2} - \frac{c}{2}) + 2 πn

a + \frac{b}{2} = \frac{π}{2} - \frac{c}{2} + 2 πn, a + \frac{b}{2} = π - (\frac{π}{2} - \frac{c}{2}) + 2 πn

a + \frac{b}{2} = \frac{π}{2} - \frac{c}{2} + 2 πn : a = \frac{π - c - b}{2} + 2 πn

a + \frac{b}{2} = \frac{π}{2} - \frac{c}{2} + 2 πn

将

\frac{b}{2}

到右边

a + \frac{b}{2} = \frac{π}{2} - \frac{c}{2} + 2 πn

两边减去

\frac{b}{2}

a + \frac{b}{2} - \frac{b}{2} = \frac{π}{2} - \frac{c}{2} + 2 πn - \frac{b}{2}

化简

a + \frac{b}{2} - \frac{b}{2} = \frac{π}{2} - \frac{c}{2} + 2 πn - \frac{b}{2}

化简

a + \frac{b}{2} - \frac{b}{2} : a

a + \frac{b}{2} - \frac{b}{2}

同类项相加:

\frac{b}{2} - \frac{b}{2} = 0

= a

化简

\frac{π}{2} - \frac{c}{2} + 2 πn - \frac{b}{2} : \frac{π - c - b}{2} + 2 πn

\frac{π}{2} - \frac{c}{2} + 2 πn - \frac{b}{2}

合并分式

\frac{π}{2} - \frac{c}{2} - \frac{b}{2} : \frac{π - c - b}{2}

使用法则

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

= \frac{π - c - b}{2}

= \frac{π - b - c}{2} + 2 πn

= \frac{π - c - b}{2} + 2 πn

a = \frac{π - c - b}{2} + 2 πn

a = \frac{π - c - b}{2} + 2 πn

a = \frac{π - c - b}{2} + 2 πn

a + \frac{b}{2} = π - (\frac{π}{2} - \frac{c}{2}) + 2 πn : a = π + 2 πn + \frac{- π + c - b}{2}

a + \frac{b}{2} = π - (\frac{π}{2} - \frac{c}{2}) + 2 πn

将

\frac{b}{2}

到右边

a + \frac{b}{2} = π - (\frac{π}{2} - \frac{c}{2}) + 2 πn

两边减去

\frac{b}{2}

a + \frac{b}{2} - \frac{b}{2} = π - (\frac{π}{2} - \frac{c}{2}) + 2 πn - \frac{b}{2}

化简

a + \frac{b}{2} - \frac{b}{2} = π - (\frac{π}{2} - \frac{c}{2}) + 2 πn - \frac{b}{2}

化简

a + \frac{b}{2} - \frac{b}{2} : a

a + \frac{b}{2} - \frac{b}{2}

同类项相加:

\frac{b}{2} - \frac{b}{2} = 0

= a

化简

π - (\frac{π}{2} - \frac{c}{2}) + 2 πn - \frac{b}{2} : π + 2 πn + \frac{- π + c - b}{2}

π - (\frac{π}{2} - \frac{c}{2}) + 2 πn - \frac{b}{2}

合并分式

\frac{π}{2} - \frac{c}{2} : \frac{π - c}{2}

使用法则

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

= \frac{π - c}{2}

= π - \frac{- c + π}{2} + 2 πn - \frac{b}{2}

去除括号:

(a) = a

= π - \frac{π - c}{2} + 2 πn - \frac{b}{2}

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{- ( π - c ) - b}{2}

- (π - c) : - π + c

- (π - c)

打开括号

= - π - (- c)

使用加减运算法则

- (- a) = a, - (a) = - a

= - π + c

= π + 2 πn + \frac{c - b - π}{2}

a = π + 2 πn + \frac{- π + c - b}{2}

a = π + 2 πn + \frac{- π + c - b}{2}

a = π + 2 πn + \frac{- π + c - b}{2}

a = \frac{π - c - b}{2} + 2 πn, a = π + 2 πn + \frac{- π + c - b}{2}

作图

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