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: 17


$$\cos210^\circ=\cos150^\circ\cos60^\circ-\sin150^\circ\sin60^\circ$$ 17 is matched with I.

Work Step by Step

$$\cos210^\circ$$ We can write $210^\circ$ as the sum of $150^\circ$ and $60^\circ$. Thus, $$\cos210^\circ=\cos(150^\circ+60^\circ)$$ Here we can apply the sum identity for cosine for $\cos(150^\circ+60^\circ)$. $$\cos(A+B)=\cos A\cos B-\sin A\sin B$$ So, if we replace $A=150^\circ$ and $B=60^\circ$, we would have $$\cos210^\circ=\cos150^\circ\cos60^\circ-\sin150^\circ\sin60^\circ$$ This fits with choice I. Therefore, we should match 17 with I.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After 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.