Chapter 5.1, Problem 78E

To determine

To graph: The given trigonometric expression and determine which trigonometric function is equal to the expression. Then verify the resulting identity algebraically.

Explanation of Solution

Given information:

The following trigonometric expression:

  $\frac{1}{2}(\frac{1+\sin \theta }{\cos \theta }+\frac{\cos \theta }{1+\sin \theta })$

Graph:

Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $\left(\frac{1}{2}(\frac{1+\sin \theta }{\cos \theta }+\frac{\cos \theta }{1+\sin \theta })\right)$ which is required to graph.

Step 4: Press GRAPH button to graph the function.

The graph is obtained as:

  

Interpretation:

According to the above graph, it can be observed that the given expression has the same graph as obtained for secant function. Hence, $y=\sec \theta$

Formula used:

The following identities:

The Pythagorean identity:

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

The Reciprocal identity:

  $\sec \theta =\frac{1}{\cos \theta }$

Proof:

The given trigonometric expression can be simplified as follows:

  $\begin{array}{l}y=\frac{1}{2}(\frac{1+\sin \theta }{\cos \theta }+\frac{\cos \theta }{1+\sin \theta })\text{ [Original expression]}\\ =\frac{1}{2}(\frac{(1+\sin \theta )(1+\sin \theta )+(\cos \theta )(\cos \theta )}{\cos \theta (1+\sin \theta )})\text{ [Add fractions]}\\ =\frac{1}{2}(\frac{1+{\sin }^2\theta +2\sin \theta +{\cos }^2\theta }{\cos \theta (1+\sin \theta )})\text{ [Multiply the terms in the numerator and add]}\\ =\frac{1}{2}(\frac{1+1+2\sin \theta }{\cos \theta (1+\sin \theta )})\text{ [Pythagorean identity]}\\ =\frac{1}{2}(\frac{2+2\sin \theta }{\cos \theta (1+\sin \theta )})\text{ [Add like terms]}\\ =\frac{2(1+\sin \theta )}{2\cos \theta (1+\sin \theta )}\text{ [Factor out 2 from the numerator]}\\ =\frac{1}{\cos \theta }\text{ [Divide out common factors]}\\ =\sec \theta \text{ [Reciprocal identity]}\end{array}$

Therefore, $y=\sec \theta$ .

The simplified expression is same as the expression obtained from the graph. Hence, verified.