Question 1162905
sin4xcos3x
<pre>
This looks like part of the right side of the formulas for sin(A + B) and
sin(A - B), which are:

               sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
               sin(A - B) = sin(A)cos(B) - cos(A)sin(B)

So we write them both substituting A = 4x and B = 3x

               sin(4x + 3x) = sin(4x)cos(3x) + cos(4x)sin(3x)
               sin(4x - 3x) = sin(4x)cos(3x) - cos(4x)sin(3x)

Simplifying the left sides:

               sin(7x) = sin(4x)cos(3x) + cos(4x)sin(3x)
                sin(x) = sin(4x)cos(3x) - cos(4x)sin(3x)

Next we add those equations together vertically, term by term,
we see that the last terms on the right cancel out:

      sin(7x) + sin(x) = 2sin(4x)cos(3x)

We multiply both sides by 1/2 and the right side will become
what was given:

      {{{expr(1/2)(sin(7x) + sin(x)^"") = expr(1/2)(2sin(4x)cos(3x)^"")}}}

      {{{expr(1/2)sin(7x) + expr(1/2)sin(x) = sin(4x)cos(3x)}}}

So the left side is the answer since the right side is what was given
and the left side is a sum.

Edwin</pre>