Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 680: 46


The required value of $\tan \frac{3\pi }{8}$ is $\sqrt{2}+1$.

Work Step by Step

We have to find the value of $\tan \frac{3\pi }{8}$; the formula is used which represents the tan identity in term of cos and sine. And the term $\tan \frac{3\pi }{8}$ comes under the first quadrant where the value of all the trigonometric functions is positive. $\tan \frac{3\pi }{8}=\tan \frac{\frac{3\pi }{4}}{2}=\frac{1-\cos \frac{3\pi }{4}}{\sin \frac{3\pi }{4}}=\frac{1-\left( -\frac{\sqrt{2}}{2} \right)}{\frac{\sqrt{2}}{2}}=\frac{1}{\frac{\sqrt{2}}{2}}+\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$ In order to solve the equation, take the individual L.C.M. $\tan \frac{3\pi }{8}=\frac{2}{\sqrt{2}}+1=\frac{2}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}+1=\sqrt{2}+1$ And the rationalizing technique is used to compute the value in the aforementioned situation. Thus, the value of $\tan \frac{3\pi }{8}$ is $\sqrt{2}+1$.
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.