cot2x = (cosx+cos3x)/(sin3x+sinx) please list all steps. It's REALLY hard .thx
cot(2x) = (cos(x) +cos(3x)) / (sin(3x) +sin(x)) -------(i)
Or, rewriting that,
cot(2x) = [cos(3x) +cos(x)] / [sin(3x) +sin(x)] -------(ii)
RHS of (ii) =
= [cos(3x) +cos(x)] / [sin(3x) +sin(x)]
Using trig identities:
cosA +cosB = 2cos((A+B)/2)cos((A-B)/2)
sinA +sinB = 2sin((A+B)/2)cos((A-B)/2),
= [2cos((3x+x)/2)cos((3x-x)/2)] / [2sin((3x+x)/2)cos((3x-x)/2)]
= cos(2x) / sin(2x)
= cot(2x)
= LHS
So, proven.