Write the product as a sum or difference: 14 sin(406)cos(166) =
Trigonometry (MindTap Course List)
8th Edition
ISBN:9781305652224
Author:Charles P. McKeague, Mark D. Turner
Publisher:Charles P. McKeague, Mark D. Turner
Chapter8: Complex Numbers And Polarcoordinates
Section8.3: Products And Quotients In Trigonometric Form
Problem 19PS
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![### Problem Statement:
**Write the product as a sum or difference:**
\[ 14 \sin(40b) \cos(16b) = \]
In this trigonometric problem, you are asked to convert the product of sine and cosine functions into a sum or difference. This uses the product-to-sum identities in trigonometry.
### Explanation:
According to the product-to-sum identities, the product of sine and cosine can be expressed as follows:
\[ \sin(A)\cos(B) = \frac{1}{2}[\sin(A+B) + \sin(A-B)] \]
Here, we are given:
\[ A = 40b \]
\[ B = 16b \]
Now, applying the product-to-sum formula:
\[ \sin(40b)\cos(16b) = \frac{1}{2} [\sin(40b + 16b) + \sin(40b - 16b)] \]
\[ = \frac{1}{2} [\sin(56b) + \sin(24b)] \]
Multiplying both sides by 14:
\[ 14 \sin(40b)\cos(16b) = 14 \left( \frac{1}{2}[\sin(56b) + \sin(24b)] \right) \]
\[ = 7[\sin(56b) + \sin(24b)] \]
Thus:
\[ 14 \sin(40b) \cos(16b) = 7[\sin(56b) + \sin(24b)] \]
So, the product \( 14 \sin(40b) \cos(16b) \) can be written as the sum:
\[ 14 \sin(40b) \cos(16b) = 7[\sin(56b) + \sin(24b)] \]
This solution involves using trigonometrical identities to simplify the given expression.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0e60132c-6df6-4596-95bf-3dd1dce70541%2Fc330fedd-88a0-47bc-b35a-18a9a173ac32%2Fuwo1prm_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
**Write the product as a sum or difference:**
\[ 14 \sin(40b) \cos(16b) = \]
In this trigonometric problem, you are asked to convert the product of sine and cosine functions into a sum or difference. This uses the product-to-sum identities in trigonometry.
### Explanation:
According to the product-to-sum identities, the product of sine and cosine can be expressed as follows:
\[ \sin(A)\cos(B) = \frac{1}{2}[\sin(A+B) + \sin(A-B)] \]
Here, we are given:
\[ A = 40b \]
\[ B = 16b \]
Now, applying the product-to-sum formula:
\[ \sin(40b)\cos(16b) = \frac{1}{2} [\sin(40b + 16b) + \sin(40b - 16b)] \]
\[ = \frac{1}{2} [\sin(56b) + \sin(24b)] \]
Multiplying both sides by 14:
\[ 14 \sin(40b)\cos(16b) = 14 \left( \frac{1}{2}[\sin(56b) + \sin(24b)] \right) \]
\[ = 7[\sin(56b) + \sin(24b)] \]
Thus:
\[ 14 \sin(40b) \cos(16b) = 7[\sin(56b) + \sin(24b)] \]
So, the product \( 14 \sin(40b) \cos(16b) \) can be written as the sum:
\[ 14 \sin(40b) \cos(16b) = 7[\sin(56b) + \sin(24b)] \]
This solution involves using trigonometrical identities to simplify the given expression.
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