Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 5 - Section 5.2 - Trigonometric Functions of Real Numbers - 5.2 Exercises - Page 418: 53

Answer

$\sin t=\sqrt{1-\cos^{2}t}$

Work Step by Step

see: Signs of the Trigonometric Functions (p.412) $\left[\begin{array}{lll} Quadrant & Positive & Negative\\ I & all & none\\ II & sin, csc & cos, sec, tan, cot\\ III & tan, cot & sin, csc, cos, sec\\ IV & cos, sec & sin, csc, tan, cot \end{array}\right]$ also, on p.415, Reciprocal Identities: $\displaystyle \csc t=\frac{1}{\sin t} \qquad \displaystyle \sec t=\frac{1}{\cos t}\qquad \displaystyle \cot t=\frac{1}{\tan t}$ and: Pythagorean Identities: $\sin^{2}t+\cos^{2}t=1\qquad\tan^{2}t+1=\sec^{2}t\qquad 1 +\cot^{2}t=\csc^{2}t$ ------------------------ $\sin^{2}t+\cos^{2}t=1\qquad\Rightarrow\sin t=\pm\sqrt{1-\cos^{2}t}$ Determine the sign: in quadrant II, the sine is positive $\sin t=+\sqrt{1-\cos^{2}t}$
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