Re: sum of three sines


David Macmanus wrote:
I've been trying to find the three sum equivalent of
sinA + sinB = 2sin(A+B/2)cos(A-B/2),

To be clearer, I think you mean:

sinA + sinB = 2sin((A+B)/2)cos((A-B)/2),

i.e. what is
sinA + sinB + sinC = ??
I've looked on the web but can't seem to find it.
Does anyone know of a web-site or have the answer handy?

The analogy seems to work for 4 terms. The above identity seems to come
from the addition formula

sin(A+B) = sinAcosB + cosAsinB and sin(-A) = -sinA.

This yields

sin(A+B)-sin(A-B) = 2sinAcosB

then letting X = A+B and Y=A-B, then A = (X+Y)/2 and B = (X-Y)/2

The next one up that I could get to cancel nicely is:

sin(A+B+C)+sin(A-B-C)+sin(-A+B-C)+sin(-A-B+C) = -4sinAsinBsinC

yielding the sum identity analogous to yours:

sinX + sinY + sinZ - sin(X+Y+Z) =
4sin((X+Y)/2)sin((X+Z)/2)sin((Y+Z)/2))

This was by following the path of least resistance; there might be a
better one, or one for sum of just three sines. Any ideas?

Mitch

.