Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Review Exercises - Page 707: 17


The required value is $\sqrt{3}+2$

Work Step by Step

The expression $\tan \frac{5\pi }{12}$ can be written as $\tan \left( \frac{2\pi }{12}+\frac{3\pi }{12} \right)$. Then use the identity: $\tan \left( x+y \right)=\frac{\tan x+\tan y}{1-\tan x\tan y}$ Now, replacing the values in the above equation, we get: $\begin{align} & \tan \left( \frac{2\pi }{12}+\frac{3\pi }{12} \right)=\frac{\tan \frac{2\pi }{12}+\tan \frac{3\pi }{12}}{1-\tan \frac{2\pi }{12}\tan \frac{3\pi }{12}} \\ & =\frac{\tan \frac{\pi }{6}+\tan \frac{\pi }{4}}{1-\tan \frac{\pi }{6}.\tan \frac{\pi }{4}} \\ & =\frac{\frac{\sqrt{3}}{3}+1}{1-\frac{\sqrt{3}}{3}.1} \\ & =\frac{\frac{\sqrt{3}}{3}+1}{1-\frac{\sqrt{3}}{3}}.\frac{1+\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}} \end{align}$ The equation can be further simplified as: $\begin{align} & \frac{\frac{\sqrt{3}}{3}+1}{1-\frac{\sqrt{3}}{3}}.\frac{1+\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}}=\frac{\frac{2\sqrt{3}}{3}+1+\frac{1}{3}}{1-\frac{3}{9}} \\ & =\frac{\frac{2\sqrt{3}}{3}+\frac{4}{3}}{1-\frac{1}{3}} \\ & =\frac{\frac{2\sqrt{3}}{3}+\frac{4}{3}}{\frac{2}{3}} \end{align}$ On further simplifying, we get: $\begin{align} & \frac{\frac{2\sqrt{3}}{3}+\frac{4}{3}}{\frac{2}{3}}=\left( \frac{2\sqrt{3}}{3}+\frac{4}{3} \right).\frac{3}{2} \\ & =\sqrt{3}+2 \end{align}$ Hence, the required value is $\sqrt{3}+2$
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