Answer
$0$
Work Step by Step
$\sin{A} \sin{B} = \dfrac{1}{2} [\cos{(A-B)} -\cos{(A+B)}]$
$\sin{4\pi} \sin{2\pi} = \dfrac{1}{2}[\cos{2\pi} - \cos{6\pi}] = 0$
Trigonometry 7th Edition
by McKeague, Charles P.; Turner, Mark D.
Published by
Cengage Learning
ISBN 10:
1111826854
ISBN 13:
978-1-11182-685-7
Chapter 5 - Section 5.5 - Additional Identities - 5.5 Problem Set - Page 311: 34
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
