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

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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.

We notice that the left side is a difference of two squares. We'll recall the fact that a difference of two squares returns the product:

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

We'll transform the algebraic sums into products:

sin 6x - sin 4x = 2 cos [(6x+4x)/2]*sin [(6x-4x)/2]

sin 6x - sin 4x = 2 cos(5x)*sin (x)

sin 6x + sin 4x = 2 sin [(6x+4x)/2]*cos [(6x-4x)/2]

sin 6x + sin 4x = 2 sin (5x)*cos (x)

(sin 6x - sin 4x)(sin 6x + sin 4x) = 2 cos(5x)*sin (5x)*2cos (x)*sin (x)

We'll recall the double angle formula:

(sin 6x - sin 4x)(sin 6x + sin 4x) = [sin 2*(5x)]*[sin 2(x)]

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

**We notice that the given identity (sin 6x)^2 - (sin 4x)^2 = (sin 10 x)*(sin 2x) is verified.**