Certainly! Let's write the given product [tex]\( 12 \cos (8q) \sin (7q) \)[/tex] as a sum using trigonometric identities. We will use the product-to-sum formulas, specifically, the identity:
[tex]\[ \sin(A)\cos(B) = \frac{1}{2}[\sin(A+B) + \sin(A-B)] \][/tex]
Here, [tex]\( A = 7q \)[/tex] and [tex]\( B = 8q \)[/tex]. Therefore:
[tex]\[ \cos(8q) \sin(7q) = \frac{1}{2}[\sin(7q + 8q) + \sin(7q - 8q)] = \frac{1}{2}[\sin(15q) + \sin(-q)] \][/tex]
Since [tex]\( \sin(-x) = -\sin(x) \)[/tex], we can rewrite [tex]\( \sin(-q) \)[/tex]:
[tex]\[ \sin(-q) = -\sin(q) \][/tex]
Now, substituting this back into our formula, we get:
[tex]\[ \cos(8q) \sin(7q) = \frac{1}{2}[\sin(15q) - \sin(q)] \][/tex]
We need to multiply this expression by 12 to get the original expression:
[tex]\[ 12 \cos(8q) \sin(7q) = 12 \cdot \frac{1}{2}[\sin(15q) - \sin(q)] \][/tex]
[tex]\[ = 6[\sin(15q) - \sin(q)] \][/tex]
[tex]\[ = 6\sin(15q) - 6\sin(q) \][/tex]
Thus, the expression [tex]\( 12 \cos(8q) \sin(7q) \)[/tex] written as a sum is:
[tex]\[ 12 \cos(8q) \sin(7q) = 6\sin(15q) - 6\sin(q) \][/tex]
This is your final answer.
