What is the general solution for solvefor a,sin(a+b/2)=cos(c/2) ?

The general solution for solvefor a,sin(a+b/2)=cos(c/2) is a=(pi-c-b)/2+2pin,a=pi+2pin+(-pi+c-b)/2

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

Solution

solve for

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

Solution

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

Solution steps

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

Rewrite using trig identities

cos (\frac{c}{2})

Use the following identity:

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})

Apply trig inverse properties

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

Move

\frac{b}{2}

to the right side

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

Subtract

\frac{b}{2}

from both sides

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

Simplify

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

Simplify

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

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

Add similar elements:

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

= a

Simplify

\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}

Combine the fractions

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

Apply rule

\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

Move

\frac{b}{2}

to the right side

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

Subtract

\frac{b}{2}

from both sides

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

Simplify

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

Simplify

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

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

Add similar elements:

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

= a

Simplify

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}

Combine the fractions

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

Apply rule

\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}

Remove parentheses:

(a) = a

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

Since the denominators are equal, combine the fractions:

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

= \frac{- ( π - c ) - b}{2}

- (π - c) : - π + c

- (π - c)

Distribute parentheses

= - π - (- c)

Apply minus-plus rules

- (- 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}

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor a,sin(a+b/2)=cos(c/2) ?
The general solution for solvefor a,sin(a+b/2)=cos(c/2) is a=(pi-c-b)/2+2pin,a=pi+2pin+(-pi+c-b)/2