cos(7q) +cos(5q) + cos(3q) + cos(q) = 4cos(q)cos(2q)cos(4q)

The left is a sum and the right is a product. If we choose to work with the
left side we will need a sum-to-product cosine identity. If we choose to work
with the right side we will need a cosine product-to-sum identity. Since
products are more complicated than sums,  I will choose to work with the
right side. 

We need to derive a cosine product-to-sum identity:

Add the two well-known identities:

          cos(A+B) = cos(A)cos(B) - sin(A)sin(B)
          cos(A-B) = cos(A)cos(B) + sin(A)sin(B)
-------------------------------------------------
 cos(A+B)+cos(A-B) = 2cos(A)cos(B)

Divide both sides by 2:

[cos(A+B)+cos(A-B)]/2 = cos(A)cos(B)

Turn it around:

cos(A)cos(B) = [cos(A+B)+cos(A-B)]/2

This is the identity we will be using in addition to the
fact that cos(-A) = cos(A)

---------------------------------------------------

Start with the right side:

4cos(q)cos(2q)cos(4q)

          Use the identity to replace cos(q)cos(2q) by
          [cos(q+2q)+cos(q-2q)]/2 or [cos(3q)+cos(-q)]/2
          or [cos(3q)+cos(q)]/2

4[[cos(3q)+cos(q)]/2]cos(4q)
2[cos(3q)+cos(q)]cos(4q)
2cos(3q)cos(4q)+2cos(q)cos(4q)

          Use the identity to replace cos(3q)cos(4q) by
          [cos(3q+4q)+cos(3q-4q)]/2 or [cos(7q)+cos(-q)]/2
          or [cos(7q)+cos(q)]/2

          Also use the identity to replace cos(q)cos(4q) by
          [cos(q+4q)+cos(q-4q)]/2 or [cos(5q)+cos(-3q)]/2
          or [cos(5q)+cos(3q)]/2

2[cos(7q)+cos(q)]/2 + 2[cos(5q)+cos(3q)]/2

Cancel the 2's

cos(7q)+cos(q)+cos(5q)+cos(3q)

Swap the two middle terms:

cos(7q)+cos(5q)+cos(q)+cos(3q)

Swap the last two terms:

cos(7q)+cos(5q)+cos(3q)+cos(q)

Edwin