Trigonometry (11th Edition) Clone

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

Chapter 5 - Trigonometric Identities - Section 5.6 Half-Angle Identities - 5.6 Exercises - Page 242: 9


$$\tan67.5^\circ=\sqrt2+1$$ 9 should be matched with F.

Work Step by Step

$$\tan67.5^\circ$$ $67.5^\circ$ is the half angle of $135^\circ$ $$67.5^\circ=\frac{135^\circ}{2}$$ Therefore, $$\tan67.5^\circ=\tan\Big(\frac{135^\circ}{2}\Big)$$ - Recall the half-angle identities: $$\tan\Big(\frac{A}{2}\Big)=\frac{\sin A}{1+\cos A}$$ Apply the identity to $\tan67.5^\circ$ with $A=135^\circ$, we have $$\tan67.5^\circ=\frac{\sin135^\circ}{1+\cos135^\circ}$$ As $135^\circ+45^\circ=180^\circ$, $|\sin135^\circ|=\sin45^\circ$ and $|\cos135^\circ|=\cos45^\circ$ $135^\circ$ lies in quadrant II, where $\sin\theta\gt0$ and $\cos\theta\lt0$. Therefore, $\sin135^\circ=\sin45^\circ=\frac{\sqrt2}{2}$ and $\cos135^\circ=-\cos45^\circ=-\frac{\sqrt2}{2}$ So, $$\tan67.5^\circ=\frac{\frac{\sqrt2}{2}}{1-\frac{\sqrt2}{2}}$$ $$\tan67.5^\circ=\frac{\frac{\sqrt2}{2}}{\frac{2-\sqrt2}{2}}$$ $$\tan67.5^\circ=\frac{\sqrt2}{2-\sqrt2}$$ $$\tan67.5^\circ=\frac{\sqrt2}{\sqrt2(\sqrt2-1)}$$ $$\tan67.5^\circ=\frac{1}{\sqrt2-1}$$ Multiply both numerator and denominator with $\sqrt2+1$. - Numerator: $1\times(\sqrt2+1)=\sqrt2+1$ - Denominator: $(\sqrt2-1)(\sqrt2+1)=2-1=1$ Thus, $$\tan67.5^\circ=\frac{\sqrt2+1}{1}$$ $$\tan67.5^\circ=\sqrt2+1$$ 9 should be matched with F.
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.