Prove that sin^2 6x - sin^2 4x = sin 10x * sin 2x .

Starting with the other side of the equation
sin(10x)sin(2x)

Since the left side has 6x and 4x,  10x =
6x + 4x and 2x = 4x - 2x

Then we use the addition and
subtraction formulas

sin(10x) = sin(6x+4x) =
sin(6x)cos(4x)+sin(4x)cos(6x)

sin(2x) = sin(6x - 4x) =
sin(6x)cos(4x) - sin(4x)cos(6x)

Now combine the above
formulas

sin(10x)sin(2x) =
(sin(6x)cos(4x)+sin(4x)cos(6x))(sin(6x)cos(4x) -
sin(4x)cos(6x))

Multiply to
get

= sin^(6x)cos^2(4x) + sin(4x)cos(6x)sin(6x)cos(4x) -
sin(6x)cos(4x)sin(4x)cos(6x) - sin^(4x)cos^(6x)

The middle
terms sin(4x)cos(6x)sin(6x)cos(4x) - sin(6x)cos(4x)sin(4x)cos(6x) is zero so we
eliminate them to get

= sin^(6x)cos^2(4x) -
sin^(4x)cos^2(6x)

Now since cos^2(a) = 1 - sin^2(a) we
get

= sin^2(6x)(1-sin^2(4x)) - sin^2(4x)(1 -
sin^2(6x))

Multiplying

=
sin^2(6x) - sin^2(6x)sin^2(4x) - sin^2(4x) +
sin^2(4x)sin^2(6x)

The second and fourth terms are
identical but opposite in sign so they are eleiminated and we get the expression we are
expecting.

= sin^2(6x) -
sin^2(4x)

So

sin(10x)sin(2x) =
sin^2(6x) - sin^2(4x) which is the identity we wanted to verify.