Answer:
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
