Q18. - 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: $\sin \theta =\frac{9}{\sqrt{106}},\csc \theta =\frac{\sqrt{106}}{9},\cos \theta =\frac{5}{\sqrt{106}},\sec \theta =\frac{\sqrt{106}}{5},\tan \theta =\frac{9}{5},\cot \theta =\frac{5}{9}$.

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:

$\begin{array}{c}\text{Hypotenuse}=\sqrt{5^2+9^2}\\ =\sqrt{106}\end{array}$

Now, the six trigonometric functions are evaluated as:

$\begin{array}{c}\sin \theta =\frac{9}{\sqrt{106}}\csc \theta =\frac{\sqrt{106}}{9}\\ \cos \theta =\frac{5}{\sqrt{106}}\sec \theta =\frac{\sqrt{106}}{5}\\ \tan \theta =\frac{9}{5}\cot \theta =\frac{5}{9}\end{array}$

Step 4. Conclusion.

The six trigonometric functions are: $\sin \theta =\frac{9}{\sqrt{106}},\csc \theta =\frac{\sqrt{106}}{9},\cos \theta =\frac{5}{\sqrt{106}},\sec \theta =\frac{\sqrt{106}}{5},\tan \theta =\frac{9}{5},\cot \theta =\frac{5}{9}$