Question

Find the exact value of each of the remaining trigonometric functions of $\theta$

$\cos \theta =\frac{-1}{3},\frac{\mathrm{\pi }}{2}<\theta <\mathrm{\pi }$

Short answer

The values are

$\sin \theta =\frac{2\sqrt{2}}{3},csc\theta =\frac{3}{2\sqrt{2}},sec\theta =-3,\tan \theta =-2\sqrt{2},cot\theta =-\frac{1}{2\sqrt{2}}$

Step-by-step solution

.Given information

$\cos \theta =\frac{-1}{3},\frac{\mathrm{\pi }}{2}<\theta <\mathrm{\pi }$is given.

We have to find the exact value of each of the remaining trigonometric functions

Step 2. Finding the value sinθ

$$\begin{gathered} Weknowthat{\sin }^2\theta +{\cos }^2\theta =1,here\cos \theta =\frac{-1}{3} \\ \sin \theta =\pm \sqrt{1-{\cos }^2\theta },here\theta lieson\frac{\mathrm{\pi }}{2}<\theta <\mathrm{\pi }so\cos \theta isonsecondquadrantandallotherexcept\sin \theta andcsc\theta arewillbenegative. \\ hence\sin \theta =\sqrt{1-co{s}^2\theta } \\ =\sqrt{1-\frac{{(-1)}^2}{3^2}} \\ =\sqrt{1-\frac{1}{9}} \\ =\sqrt{\frac{8}{9}} \\ =\frac{2\sqrt{2}}{3} \end{gathered}$$

Step 3.Find the value of secθ and cscθ

$$\begin{gathered} Since\cos \theta =\frac{-1}{3},thensec\theta =\frac{1}{\cos \theta } \\ =\frac{1}{{\displaystyle \frac{-1}{3}}} \\ =-3 \\ csc\theta =\frac{1}{\sin \theta } \\ =\frac{1}{{\displaystyle \frac{2\sqrt{2}}{3}}} \\ =\frac{3}{2\sqrt{2}} \end{gathered}$$

Step 4. find the values of tanθ and cotθ

$$\begin{gathered} weknowthat\tan \theta =\frac{\sin \theta }{\cos \theta },\sin \theta =\frac{2\sqrt{2}}{3}and\cos \theta =\frac{-1}{3} \\ =\frac{{\displaystyle \frac{{\displaystyle 2\sqrt{2}}}{{\displaystyle 3}}}}{{\displaystyle \frac{{\displaystyle -1}}{{\displaystyle 3}}}} \\ =-2\sqrt{2} \\ cot\theta =\frac{1}{\tan \theta } \\ =\frac{1}{{\displaystyle -2\sqrt{2}}} \end{gathered}$$