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

1 Answer
Steve M
Aug 18, 2017

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

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