Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - Review Exercises - Page 552: 47

Answer

$sin~75°=\frac{\sqrt 6+\sqrt 2}{4}$ $cos~75°=\frac{\sqrt 6-\sqrt 2}{4}$ $tan~75°=\frac{\sqrt 3+1}{\sqrt 3-1}$

Work Step by Step

$sin~75°=sin~(120°-45°)=sin~120°~cos~45°-cos~120°~sin~45°=\frac{\sqrt 3}{2}\frac{\sqrt 2}{2}-(-\frac{1}{2})\frac{\sqrt 2}{2}=\frac{\sqrt 6}{4}+\frac{\sqrt 2}{4}=\frac{\sqrt 6+\sqrt 2}{4}$ $cos~75°=cos~(120°-45°)=cos~120°~cos~45°+sin~120°~sin~45°=-\frac{1}{2}\frac{\sqrt 2}{2}+\frac{\sqrt 3}{2}\frac{\sqrt 2}{2}=-\frac{\sqrt 2}{4}+\frac{\sqrt 6}{4}=\frac{\sqrt 6-\sqrt 2}{4}$ $tan~75°=tan(120°-45°)=\frac{tan~120°-tan~45°}{1+tan~120°~tan~45°}=\frac{-\sqrt 3-1}{1+(-\sqrt 3)1}=\frac{-\sqrt 3-1}{1-\sqrt 3}=\frac{\sqrt 3+1}{\sqrt 3-1}$
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