compound angles

a) L.H.S. = sin2mx - sin2nx
= (sin mx+sin nx)(sin mx - sin nx)
= {2 sin[(m+n)/2]x cos[(m-n)/2]x}{ 2 cos[(m+n)/2]x sin[(m-n)/2]x}
= {2sin[(m+n)/2]x cos[(m+n)/2]x}{2sin[(m-n)/2]x cos[(m-n)/2]x}
= sin(m+n)x sin(m-n)x
= R.H.S.
b) sin2 9x - sin2 5x + sin4x = 0
sin (9+5)x sin (9-5)x + sin 4x = 0
sin 14x sin 4x + sin 4x = 0
sin 4x ( sin 14x + 1 ) = 0
sin 4x = 0 or sin 14x = -1
4x = 0*, 180*, 360* or 14x = 270*, 630*, 990*
x = 0*, 45*, 90* or 19.3*, 45* or 70.7* for 0 to 90* inclusively
x = 0*, 19.3*, 45*, 70.7*, 90*

2008-04-11 00:30:49 補充：
Hey, to the below, there are definitely more than 1 root in the range 0 to 90 degrees.

a.RHS
= -1/2[cos2mx-cos2nx]
= -1/2[(1-2sin^2mx)-(1-2sin^2nx)]
= sin^2mx-sin^2nx
= LHS
b. sin^2 9x - sin^2 5x + sin4x = 0
sin14xsin4x+sin4x=0
sin4x(sin14x+1)=0
sin4x=0 or sin14x= -1
4x=0,180,360 or 14x= 270,630,990
x=0,45,90 or x=19.3,45,70.7 (corr. to 3 sig. fig.)
x=0, 19.3, 45, 70.7, 90

* 答案全都是degree,不過打唔到

A) sin^2mx - sin^2nx =(sin mx)^2 - (sin nx)^2
=(sin mx+sin nx) (sin mx- sin nx)
= sin(m+n)x sin(m-n)x

B)Hence,sin^2 9x - sin^2 5x + sin4x=0
sin(9+5)x sin(9-5)x + sin4x=0
sin 14x sin 4x+ sin4x=0
sin4x (sin14x+1)=0
sin4x=0 or sin14x=-1
x=0 x=-6.43(rejected)
so X=0