## Trigonometry (11th Edition) Clone

Published by Pearson

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

#### Answer

$$\tan67.5^\circ=\sqrt2+1$$ 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.

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.