Chapter 4, Problem 34RE

Solution

To find: the values of six trigonometric functions.

  $\sin \theta =\frac{2\sqrt{6}}{7},\tan \theta =\frac{2\sqrt{6}}{5},\csc \theta =\frac{7\sqrt{6}}{12},\sec \theta =\frac{7}{5},\cot \theta =\frac{5\sqrt{6}}{12}$ .

Given:

  $\cos \theta =\frac{5}{7}$

Concept used:

Pythagorean Theorem:

In any right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the hypotenuse.

In any right triangle,

  $\sin \theta =\frac{\text{opp}}{\text{hyp}},\cos \theta =\frac{\text{adj}}{\text{hyp}},\tan \theta =\frac{\text{opp}}{\text{adj}},\csc \theta =\frac{\text{hyp}}{\text{opp}},\sec \theta =\frac{\text{hyp}}{\text{adj}},\cot \theta =\frac{\text{adj}}{\text{opp}}$ .

Calculation:

The right triangle is,

  

Now by using the Pythagorean Theorem in the given triangle, it follows that

  $\begin{array}{c}{\left(AB\right)}^2={\left(AC\right)}^2+{\left(BC\right)}^2\\ 7^2=5^2+{\left(BC\right)}^2\\ {\left(BC\right)}^2=24\\ BC=2\sqrt{6}\end{array}$

Now,

  $\begin{array}{c}\sin \theta =\frac{BC}{AB}=\frac{2\sqrt{6}}{7}\\ \tan \theta =\frac{BC}{AC}=\frac{2\sqrt{6}}{5}\\ \csc \theta =\frac{AB}{BC}=\frac{7}{2\sqrt{6}}=\frac{7\sqrt{6}}{12}\\ \sec \theta =\frac{AB}{AC}=\frac{7}{5}\\ \cot \theta =\frac{AC}{BC}=\frac{5}{2\sqrt{6}}=\frac{5\sqrt{6}}{12}\end{array}$