How do you prove #(sin2x + sin2y)/(cos2x + cos2y) = tan(x + y)#?

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Answer 1 · 2015-12-17T09:59:15

With the sum identities

we have

#(sin(2x) + sin(2y))/(cos(2x) + cos(2y)) = (2sin((2x+2y)/2)cos((2x - 2y)/2))/(2cos((2x+2y)/2)cos((2x - 2y)/2))#

#= sin(x+y)/cos(x+y)#

#= tan(x+y)#

In the case that the sum identities above are not given, here is a short proof of them using the angle-addition formulas:

Let #a = (x+y)/2# and #b = (x-y)/2#

Then

#sin(x) + sin(y) = sin(a + b) + sin(a-b)#

#= sin(a)cos(b) + cos(a)sin(b) + sin(a)cos(b) - cos(a)sin(b)#

#= 2sin(a)cos(b)#

#= 2sin((x+y)/2)cos((x-y)/2)#

And

#cos(x) + cos(y) = cos(a+b) + cos(a-b)#

#= cos(a)cos(b) - sin(a)sin(b) + cos(a)cos(b) + sin(a)sin(b)#

#= 2cos(a)cos(b)#

#= 2cos((x+y)/2)sin((x-y)/2)#