## Precalculus (10th Edition)

Published by Pearson

# Chapter 7 - Analytic Trigonometry - 7.5 Sum and Difference Formulas - 7.5 Assess Your Understanding - Page 487: 28

#### Answer

$\dfrac{\sqrt3}{2}$

#### Work Step by Step

Recall that $\cos(\alpha-\beta)=\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)$. Hence, $\cos(40^\circ)\cos(10^\circ)+\sin(40^\circ)\sin(10^\circ)\\ =\cos(40^\circ-10^\circ)\\ =\cos(30^\circ)\\ =\dfrac{\sqrt3}{2}$

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.