# Chapter 5 - Trigonometric Functions - Section 5.2 Trigonometric Functions: Unit Circle Approach - 5.2 Assess Your Understanding - Page 402: 17

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

#### Work Step by Step

With $P= \left(-\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2} \right) = (x,y)$, then $x = -\dfrac{\sqrt{2}}{2} \text{ and } \hspace{15pt} y =\dfrac{\sqrt{2}}{2}$. Thus, $\sin{t} = y$ $\sin{t} = \dfrac{\sqrt{2}}{2}$ $\cos{t} = x$ $\cos{t} = -\dfrac{\sqrt{2}}{2}$ $\tan{t} = \dfrac{y}{x}$ $\tan{t} = \dfrac{\dfrac{\sqrt{2}}{2}}{ -\dfrac{\sqrt{2}}{2} } = -1$ $\csc{t} = \dfrac{1}{y}$ $\csc{t} = \dfrac{1}{\dfrac{\sqrt{2}}{2}} = \sqrt{2}$ $\sec{t} = \dfrac{1}{x}$ $\sec{t} = \dfrac{1}{-\dfrac{\sqrt{2}}{2}}= -\sqrt{2}$ $\cot{t} = \dfrac{x}{y}$ $\cot{t} = \dfrac{-\dfrac{\sqrt{2}}{2} }{\dfrac{\sqrt{2}}{2}} = -1$

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