Chapter 10.3, Problem 1GP

Solution

To find: the values of other five trigonometric functions of $\theta$ .

  $\sin \theta =\frac{\sqrt{35}}{6},\csc \theta =\frac{6\sqrt{35}}{35},\sec \theta =6,\tan \theta =\sqrt{35},\cot \theta =\frac{\sqrt{35}}{35}$ .

Given:

  $\cos \theta =\frac{1}{6},0<\theta <\frac{\pi }{2}$ .

Concept used:

Pythagorean identity:

  ${\sin }^2\theta +{\cos }^2\theta =1$

Reciprocal identities:

  $\csc \theta =\frac{1}{\sin \theta },\sec \theta =\frac{1}{\cos \theta }$

Tangent and Cotangent identities:

  $\tan \theta =\frac{\sin \theta }{\cos \theta },\cot \theta =\frac{\cos \theta }{\sin \theta }$

Calculation:

By using the Pythagorean identity ${\sin }^2\theta +{\cos }^2\theta =1$ , it follows that

  $\begin{array}{c}{\sin }^2\theta +{\left(\frac{1}{6}\right)}^2=1\\ {\sin }^2\theta +\frac{1}{36}=1\\ {\sin }^2\theta =1-\frac{1}{36}\\ {\sin }^2\theta =\frac{35}{36}\\ \sin \theta =\pm \frac{\sqrt{35}}{6}\end{array}$

Now since $\theta$ is in quadrant I, $\sin \theta$ is positive.

  $\therefore \sin \theta =\frac{\sqrt{35}}{6}$

Now by using the reciprocal identities, it follows that

  $\begin{array}{c}\csc \theta =\frac{1}{\raisebox{1ex}{$\sqrt{35}$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\&\sec \theta =\frac{1}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\\ \csc \theta =\frac{6}{\sqrt{35}}\&\sec \theta =6\\ \csc \theta =\frac{6\sqrt{35}}{35}\&\sec \theta =6\end{array}$

Now by using the tangent and cotangent identities, it follows that

  $\begin{array}{c}\tan \theta =\frac{\raisebox{1ex}{$\sqrt{35}$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\&\cot \theta =\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\raisebox{1ex}{$\sqrt{35}$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\\ \tan \theta =\sqrt{35}\&\cot \theta =\frac{1}{\sqrt{35}}\\ \tan \theta =\sqrt{35}\&\cot \theta =\frac{\sqrt{35}}{35}\end{array}$