a.maths (compound angles)

(a)
sinAsinB=-1/2[cos(A+B)-cos(A-B)]
So
sin (m+n)x sin(m-n)x
=-1/2[cos[(m+n)x+(m-n)x]-cos[(m+n)x-(m-n)x]
=-1/2[cos[2mx]-cos[2nx]]
=-1/2[1-2sin^2[mx]+2sin^2[nx]-1]
=sin^2[mx]-sin^2[nx]
(b)
sin^2 9x - sin^2 5x+ sin 4x = 0

sin (9x+5x)sin(9x-5x)+sin4x=0
sin4x(sin14x+1)=0
sin4x=0 or sin14x=-1
when sin4x=0
4x=0,180,360
x=0,45,90
when sin14x=-1
14x=270,630,990,1350
x=135/7,45,495/7,

R.H.S.
= sin (m+n)x sin (m-n)x
= ( sinmx cosnx + sinnx cosmx )( sinmx cosnx - sinnx cosmx )
= sin^2 mx cos^2 nx - sin^2 nx cos^2 mx
= sin^2 mx - sin^2 mx sin^2 nx - sin^2 nx + sin^2 nx sin^2 mx
= L.H.S.

sin^2 9x - sin^2 5x + sin4x = 0
sin14x sin4x + sin4x = 0
sin4x = 0 or sin14x = -1(rejected)
x = 0

(A)
LHS:sin^2 mx - sin^2 nx
=(sin mx+sin nx)(sin mx-sin nx)
=(2sin((mx+nx)/2)cos((mx-nx)/2))*(2cos((mx+nx)/2)sin((mx-nx)/2))
=2sin((x(m+n))/2)cos((x(m+n))/2)*2sin((x(m-n))/2)cos((x(m-n))/2)
=sin(m+n)xsin(m-n)x
=RHS
(B)
sin^2 9x - sin^2 5x+ sin 4x = 0
sin((9+5)x)sin((9-5)x)+sin4x=0
sin14x * sin4x+sin4x=0
sin4x(sin14x + 1)=0
sin4x=0 or sin14x+1=0
4x=0 or sin14x= -1
x=0 or 14x= -90
x=0 or x=-6.428......(rej.)