Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.4 Sum and Difference Identities for Sine and Tangent - 5.4 Exercises - Page 220: 3

Answer

$$\tan15^\circ=2-\sqrt 3$$ E is the correct choice.

Work Step by Step

$$\tan15^\circ$$ $15^\circ$ can be rewritten as the difference of $45^\circ$ and $30^\circ$, which means $\tan15^\circ$ would become $$\tan(45^\circ-30^\circ)$$ Now, we can apply the identity of $\tan$ of a difference $$=\frac{\tan45^\circ-\tan30^\circ}{1+\tan45^\circ\tan30^\circ}$$ $$=\frac{1-\frac{1}{\sqrt 3}}{1+1\times\frac{1}{\sqrt 3}}$$ $$=\frac{1-\frac{\sqrt 3}{3}}{1+\frac{\sqrt 3}{3}}$$ $$=\frac{\frac{3-\sqrt 3}{3}}{\frac{3+\sqrt 3}{3}}$$ $$=\frac{3-\sqrt 3}{3+\sqrt 3}$$ $$=\frac{3-\sqrt3}{3+\sqrt 3}\times\frac{3-\sqrt 3}{3-\sqrt 3}$$ (to simplify it even further) $$=\frac{(3-\sqrt 3)^2}{(9-3)}$$ $$=\frac{12-6\sqrt 3}{6}$$ $$=2-\sqrt 3$$ Therefore, E is the correct choice.
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.