Proof
sin^27x-sin^24x = sin 7xsin 4x

sin11x • sin3x

Step-by-step explanation:

We want to prove that sin²(7x) - sin²(4x) equals sin7x sin4x.

Now, let 7x be A and let 4x be B. Thus;

Sin²A - sin²B = (sinA + sinB)(sinA - sinB)

This is because;

a² - b² = (a + b)(a - b)

Now, from double angle and half angle formula we know that;

(sinA + sinB) = 2sin((A + B)/2)cos((A - B)/2)

And

(sinA - sinB) = 2cos((A + B)/2)sin((A - B)/2)

We now have;

sin²A - sin²B = 2sin((A + B)/2)cos((A - B)/2) × 2cos((A + B)/2)sin((A - B)/2)

Now, let's put back 7x for A and let 4x for B into RHS to get;

2sin((7x + 4x)/2)cos((7x - 4x)/2) × 2cos((7x + 4x)/2)sin((7x - 4x)/2)

This gives;

2sin(11x/2)cos(3x/2) × 2cos(11x/2)sin(3x/2)

Rearranging gives;

2sin(11x/2)cos(11x/2) × 2sin(3x/2)cos(3x/2)

We now that sin 2θ = 2sin θ cos θ

Thus, we have;

sin2(11x/2) × sin2(3x/2)

This gives;

sin11x•sin3x

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