Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Review Exercises - Page 248: 24

Answer

$$\sin300^\circ=2\sin150^\circ\cos150^\circ$$ 24 is with D.

Work Step by Step

$$\sin300^\circ$$ We rewrite $300^\circ$ as double the angle $150^\circ$. In detail, $$\sin300^\circ=\sin(2\times150^\circ)$$ This fact points to the use of the double identity for sine, stating that $$\sin(2A)=2\sin A\cos A$$ Therefore, for $A=150^\circ$: $$\sin300^\circ=2\sin150^\circ\cos150^\circ$$ The equation indicates that 24 should be matched with D.
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