prove that sin(2a+6b) + sin(6a+2b)/sin4a + sin4b = cos(2a+2b)
sin(2a+6b) + sin(6a+2b) = 2sin(4a+4b)cos(2a-2b)
sin4a + sin4b = 2sin(2a+2b)cos(2a-2b)
Dividing, we get
sin(4a+4b)/sin(2a+2b)
= 2sin(2a+2b)cos(2a+2b)/sin(2a+2b)
= 2cos(2a+2b)
Hmmm. I have an extra factor of 2. Wolframalpha agrees. You make a typo?