Chapter 6, Problem 45RE

To determine

Exact value of given remaining trigonometric functions.

Answer to Problem 45RE

  $\begin{array}{l}\sin \theta =\frac{1}{\sqrt{5}},\csc \theta =\sqrt{5}\\ \\ \cos \theta =-\frac{2}{\sqrt{5}},\sec \theta =-\frac{\sqrt{5}}{2}\\ \\ \tan \theta =-\frac{1}{2},\cot \theta =-2\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}\cot \theta =-2\\ \\ \frac{\pi }{2}<\theta <\pi \end{array}$

Formula used:

  $\begin{array}{l}\sin \theta =\frac{Perpendicular}{Hypotenuse},\cos \theta =\frac{Base}{Hypotenuse},\tan \theta =\frac{Perpendicular}{Base}\\ \csc \theta =\frac{Hypotenuse}{Perpendicular},\sec \theta =\frac{Hypotenuse}{Base},\cot \theta =\frac{Base}{Perpendicular}\end{array}$

  ${\left(Hypotenuse\right)}^2={\left(Perpendicular\right)}^2+{\left(Base\right)}^2$

Now we know that $\cot \theta =\frac{Base}{Perpendicular}$ , so we will find Base with the help of Hypotenuse theorem:

  $\begin{array}{l}{\left(Hypotenuse\right)}^2={\left(Perpendicular\right)}^2+{\left(Base\right)}^2\\ {\left(Hypotenuse\right)}^2={\left(1\right)}^2+{\left(2\right)}^2\\ {\left(Hypotenuse\right)}^2=1+4\\ {\left(Hypotenuse\right)}^2=5\\ Hypotenuse=\sqrt{5}\end{array}$

As $\frac{\pi }{2}<\theta <\pi$ ,so $\theta$ is in II quadrant, therefore only $\sin \theta$ and $\cos ec\theta$ will be positive:

  $\boxed{\begin{array}{l}\sin \theta =\frac{1}{\sqrt{5}},\csc \theta =\sqrt{5}\\ \\ \cos \theta =-\frac{2}{\sqrt{5}},\sec \theta =-\frac{\sqrt{5}}{2}\\ \\ \tan \theta =-\frac{1}{2},\cot \theta =-2\end{array}}$