Given that $\sin \theta =\dfrac{1}{3}$ and $\cos \theta < 0$, find the exact value of each of the remaining five trigonometric functions.

Solution  We begin by solving the identity $\sin ^{2}\theta +\cos ^{2}\theta =1$ for $\cos \theta$. $$\begin{eqnarray*} \sin ^{2}\theta +\cos ^{2}\theta &=&1 \\[3pt] \cos ^{2}\theta &=&1-\sin ^{2}\theta \\[3pt] \cos \theta &=&\pm \sqrt{1-\sin ^{2}\theta } \end{eqnarray*}$$

Because $\cos \theta <0$, we choose the negative value of the radical, and use the fact that $\sin \theta =\dfrac{1}{3}$. $$\begin{eqnarray*} \cos \theta &=&-\displaystyle\sqrt{1-\sin ^{2}\theta }\underset{\underset{\color{#0066A7}{\hbox{\(\sin { \theta =}\displaystyle\frac{1}{3}\)}}}{\color{#0066A7}{\displaystyle\uparrow }}}{=} -\displaystyle\sqrt{1-\frac{1}{9}}=-\displaystyle\sqrt{\dfrac{8}{9}}=-\dfrac{2\displaystyle\sqrt{2}}{3}\\ \end{eqnarray*}$$

A-34

The values of $\sin \theta$ and $\cos \theta$ are now known, so we use the quotient and reciprocal identities to obtain $$\begin{array}{l@{\qquad}l} \tan \theta =\dfrac{\sin \theta }{\cos \theta }=\frac{\dfrac{1}{3}}{-\dfrac{2\displaystyle\sqrt{2}}{3}}=-\dfrac{1}{2 \displaystyle\sqrt{2}}=-\dfrac{\displaystyle\sqrt{2}}{4} & \cot { \theta =}\dfrac{1}{\tan \theta }{ =-2}\displaystyle\sqrt{2} \\ \sec \theta =\dfrac{1}{\cos \theta }=\frac{1}{-\dfrac{2\displaystyle\sqrt{2}}{3} }=-\dfrac{3}{2\displaystyle\sqrt{2}}=-\dfrac{3\displaystyle\sqrt{2}}{4} & \csc { \theta =}\dfrac{1}{\sin \theta }\,{ =\,}\dfrac{1}{\dfrac{1}{3}}\,{ =3} \end{array}$$