Question

prove that sin(x+y)sin(x-y) + sin(y+z)sin(y-z) + sin(z+x)sin(z-x) = 0

Answer 1

Let A = y-z, B= z-x and C= x-y, so that A+B+C=0
 
We know that,
 
Sin 2A + sin 2B + sin 2C
= 2 sin (A+B) Cos (A-B) + 2 sin C cos C
= -2 sin C cos(A-B) + 2 sinC cosC 
= -2 sin C [cos (A-B)- cos C]
= -2 sin C [cos (A-B)-cos(A+B)] 
= - 4 sin A sin B sin C ....(1) 

Now sin x sin y sin(x − y) 

=1/2[cos(x − y) − cos(x + y)] sin(x − y) 
=1/2[cos(x − y) sin(x − y) − cos(x + y) sin (x − y)]
=1/4[2cos(x − y) sin (x − y) − 2cos(x + y) sin (x − y)]
=1/4[sin 2(x − y) − sin 2x + sin 2y] ...(2)

Similarly, 

sin y sin z sin(y − z) =1/4[sin 2(y − z) − sin 2y + sin 2z]... (3) 
sin z sin x sin(z − x) =1/4[sin 2(z − x) − sin 2z + sin 2x]...(4)


Adding (2), (3) and (4) 

sin x sin y sin(x − y) + sin y sin z sin(y − z) + sin z sin x sin(z − x)
=1/4[sin 2A + sin 2B + sin 2C]
=1/4[−4 sin A sin B sin C ]
= −sin A sin B sin C ..................................(from (1)

⇒ sin x sin y sin(x − y) + sin y sin z sin(y − z) + sin z sin x sin(z − x) + sin A sin B sin C = 0

Answer 2

sin(x+y)sin(x-y) = 1/2 {2 sin(x+y)sin(x-y)}
= 1/2 { cos(x+y-x+y) - cos(x+y+x-y)}
= 1/2 { cos(2y) - cos(2x)}
Similarly,

sin(y+z)sin(y-z)= 1/2 { cos(2z) - cos(2y)}

sin(z+x)sin(z-x)= 1/2 { cos(2x) - cos(2x)}

Now, the LHS given in the question is just the sum of these.

Add these and it will come out to be zero as it is visible clearly.