Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
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.
[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.
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.