## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.2 - Trigonometric Functions: The Unit Circle - Exercise Set - Page 547: 3

#### Answer

sin t = -$\frac{\sqrt 2}{2}$ , cos t = $\frac{\sqrt 2}{2}$ tan t =-1, csc t = -$\frac{2}{\sqrt 2}$ sec t = $\frac{2}{\sqrt 2}$ , cot t =-1

#### Work Step by Step

($\frac{\sqrt 2}{2}$ , -$\frac{\sqrt 2}{2}$) sin t corresponds to y value So sin t = -$\frac{\sqrt 2}{2}$ cos t corresponds to x value cos t = $\frac{\sqrt 2}{2}$ tan t is computed by $\frac{y}{x}$ So tan t = $\frac{\frac{-\sqrt 2}{2}}{\frac{\sqrt 2}{2}}$ = -1 csc t = $\frac{1}{sin t}$ So, csc t = -$\frac{2}{\sqrt 2}$ sec t = $\frac{1}{cos t}$ so sec t = $\frac{2}{\sqrt 2}$ cot t = $\frac{1}{tan t}$ so cot t =-1

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