Question

How do you verify the identity `(sinx+-siny)/(cosx+cosy)=tan((x+-y)/2)`?

Answer 1

We start with the trigonometric sum identities:

`sinA+sinB = \ \ \ \ 2sin((A+B)/2)cos((A-B)/2)`
`sinA-sinB = \ \ \ \ 2cos((A+B)/2)sin((A-B)/2)`
`cosA+cosB = \ \ \ \ 2cos((A+B)/2)cos((A-B)/2)`
`cosA-cosB = -2sin((A+B)/2)sin((A-B)/2)`

Consider the LHS (positive case):

`(sinx+siny)/(cosx+cosy) = ( 2sin((A+B)/2)cos((A-B)/2) ) / ( 2cos((A+B)/2)cos((A-B)/2) )`

`" " = ( sin((A+B)/2) ) / ( cos((A+B)/2) )`

`" " = tan((A+B)/2)`

Consider the LHS (negative case):

`(sinx-siny)/(cosx+cosy) = ( 2cos((A+B)/2)sin((A-B)/2) ) / ( 2cos((A+B)/2)cos((A-B)/2) )`

`" " = ( sin((A-B)/2) ) / ( cos((A-B)/2) )`

`" " = tan((A-B)/2)`

Hence, combining both cases, we have:

`(sinx+-siny)/(cosx+cosy) = tan((A+-B)/2) \ \ \ \` QED