Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 6 - Section 6.2 - More on Trigonometric Equations - 6.2 Problem Set - Page 333: 65



Work Step by Step

Note$\;\;\;\;\;\;\;sinA=\frac{2}{3}\;\;\;;\;\;\;\;0^o\leq A\leq 90^o$ $sin^2(A)+cos^2(A)=1\;\;\;\;\;\;\;\;\;\;\;\;$ and $sin(A)=\frac{2}{3}$ $(\frac{2}{3})^2+cos^2(A)=1$ $cos^2(A)=\frac{5}{9}$ $cos(A)=\pm \frac{\sqrt{5}}{3}$ We know $0^o\leq A\leq 90^o$ $cos(A)=\frac{\sqrt{5}}{3}$ $tan(\frac{A}{2})= \frac{1+cos(A)}{sin(A)}$ $tan(\frac{A}{2})=\frac{1+\frac{\sqrt{5}}{3}}{\frac{2}{3}}=\frac{3-\sqrt{5}}{2}$ $tan(\frac{A}{2})=\frac{3-\sqrt{5}}{2}$
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.