Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 2 - Acute Angles and Right Triangles - Section 2.1 Trigonometric Functions of Acute Angles - 2.1 Exercises: 57

Answer

$\cos$ 45$^{\circ}$ = $\frac{\sqrt2}{2}$

Work Step by Step

$\cos$ 45$^{\circ}$ We must find side $x$ of the 45$^{\circ}$ - 45$^{\circ}$ Right Triangle. Pythagorean Theorem: $c$$^{2}$ = $a$$^{2}$ + $b$$^{2}$ x$^{2}$ = 1$^{2}$ + 1$^{2}$ x$^{2}$ = 2 x = $\sqrt2$ Now consider the triangle from the perspective of a 45$^{\circ}$ angle. Hypotenuse = $\sqrt2$ Opposite = 1 Adjacent = 1 $\csc$ 45$^{\circ}$ = $\frac{Adjacent}{Hypotenuse}$ = $\frac{1}{\sqrt2}$ = $\frac{\sqrt2}{2}$ Therefore: $\cos$ 45$^{\circ}$ = $\frac{\sqrt2}{2}$
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