trigo

I think the question is cos² na-cos²ma=sin(m+n)asin(m-n) a
Since if you substitute a=0
then
cos² na+cos²ma=2
sin(m+n)asin(m-n) a=0
A contradiction
Using product to sum formula
RHS
=sin(m+n)asin(m-n) a
=-1/2[cos(m+n+m-n)a-cos(m+n-m+n)a]
=-1/2[cos(2ma)-cos(2na)]
=1/2[cos(2na)-cos(2ma)]
=1/2[2cos² na-1-2cos² ma+1] (cos2a=2cos² a-1)
=cos² na-cos²ma
=LHS

R.H.S.
sin(m+n)asin(m-n)a = sin(ma+na)sin(ma-na)
Let x=ma & y=na
= (sinxcosy+cosxsiny)(sinxcosy-cosxsiny)
= sin²xcos²y+cosxcosysinxsiny-cosxcosysinxsiny-cos²xsin²y
= sin²xcos²y-cos²xsin²y
= (1-cos²x)cos²y-cos²x(1-cos²y)
= cos²y-cos²xcos²y-cos²x+cos²xcos²y
= cos²y+cos²x
= cos²ma+cos²na=L.H.S.

2007-03-19 11:11:35 補充：
= cos²y-cos²xcos²y-cos²x cos²xcos²y= cos²y-cos²x= cos²na-cos²ma L.H.S.

2007-03-19 11:12:23 補充：
= cos²y-cos²xcos²y-cos²x cos²xcos²y= cos²y-cos²x= cos²na-cos²ma L.H.S.