Q4. - Algebra 2

Question

Find the values of six trigonometric functions for angle $\left(\theta \right)$ shown below.

Step-by-Step Solution

Answer

The six trigonometric functions are: $\sin \theta =\frac{8}{17},\csc \theta =\frac{17}{8},\cos \theta =\frac{15}{17},\sec \theta =\frac{17}{15},\tan \theta =\frac{8}{15},\cot \theta =\frac{15}{8}$.

Step 1. Write down the given information.

The given figure is shown below.

Step 2. Concept used.

The six trigonometric functions are $\sin \theta , \csc \theta , \cos \theta , \sec \theta , \tan \theta , \cot \theta$.  

Let $(x,y)$ be a point in X-Y plane and (r) be the length of a line segment drawn from the origin to the point.

In, $\Delta OAB,OA=r, OB=x, AB=y$

Therefore, six trigonometric functions are;

 $\begin{array}{c}\sin \theta =\frac{AB}{OA}=\frac{y}{r},\csc \theta =\frac{1}{\sin \theta }=\frac{OA}{AB}=\frac{r}{y}\\ \cos \theta =\frac{OB}{OA}=\frac{x}{r},\sec \theta =\frac{1}{\cos \theta }=\frac{OA}{OB}=\frac{r}{x}\\ \tan \theta =\frac{AB}{OB}=\frac{y}{x},\cot \theta =\frac{1}{\tan \theta }=\frac{OB}{AB}=\frac{x}{y}\end{array}$                          

Step 3. Calculation.

From the given figure, the hypotenuse is evaluated as:

 $\begin{array}{c}\text{Hypotenuse}=\sqrt{8^2+{15}^2}\\ =\sqrt{64+225}\\ =17\end{array}$

Now, the six trigonometric functions are evaluated as:

$\begin{array}{c}\sin \theta =\frac{8}{17}\csc \theta =\frac{17}{8}\\ \cos \theta =\frac{15}{17}\sec \theta =\frac{17}{15}\\ \tan \theta =\frac{8}{15}\cot \theta =\frac{15}{8}\end{array}$

Step 4. Conclusion.

The six trigonometric functions are: $\sin \theta =\frac{8}{17},\csc \theta =\frac{17}{8},\cos \theta =\frac{15}{17},\sec \theta =\frac{17}{15},\tan \theta =\frac{8}{15},\cot \theta =\frac{15}{8}$