Problem 28 - Algebra and Trigonometry

Question

Consider the given equation. (a) Verify algebraically that the equation is an identity. (b) Confirm graphically that the equation is an identity. $$ \frac{\tan y}{\csc y}=\sec y-\cos y $$

Step-by-Step Solution

Answer

The equation is an identity, verified algebraically and graphically.

Step 1: Rewrite the left side of the equation

Start by expressing the left side of the equation \( \frac{\tan y}{\csc y} \) in terms of sine and cosine. We know that \( \tan y = \frac{\sin y}{\cos y} \) and \( \csc y = \frac{1}{\sin y} \). Substitute these into the left-hand side: \( \frac{\tan y}{\csc y} = \frac{\frac{\sin y}{\cos y}}{\frac{1}{\sin y}} = \frac{\sin^2 y}{\cos y} \).

Step 2: Simplify the right side of the equation

Now consider the right side \( \sec y - \cos y \). We know that \( \sec y = \frac{1}{\cos y} \). Therefore, the expression becomes \( \sec y - \cos y = \frac{1}{\cos y} - \cos y \). Rewrite this with a common denominator: \( \frac{1 - \cos^2 y}{\cos y} \).

Step 3: Use trigonometric identity

Recognize that \( 1 - \cos^2 y = \sin^2 y \), which is a Pythagorean identity. Substitute this into the right side: \( \frac{1 - \cos^2 y}{\cos y} = \frac{\sin^2 y}{\cos y} \).

Step 4: Compare both sides

Both the simplified left side and right side are \( \frac{\sin^2 y}{\cos y} \), showing they are identical. Therefore, the equation is an identity algebraically.

Step 5: Confirm graphically using graphing tool

To confirm graphically, plot the functions \( f(y) = \frac{\tan y}{\csc y} \) and \( g(y) = \sec y - \cos y \) using a graphing calculator or software. Both functions should yield the same graph over their domains, confirming they are identical.

Key Concepts

Algebraic verification

Algebraic verification involves proving that a given equation is universally true using algebraic manipulations. In the exercise, we begin with the equation \( \frac{\tan y}{\csc y} = \sec y - \cos y \). To confirm this is an identity, we need to manipulate both sides algebraically to show they are the same.

Firstly, transform each side into terms of sine and cosine because they are the fundamental trigonometric functions and often simplify verification.
- For \( \tan y \), use \( \frac{\sin y}{\cos y} \).
- For \( \csc y \), use \( \frac{1}{\sin y} \).
This means the left side becomes \( \frac{\frac{\sin y}{\cos y}}{\frac{1}{\sin y}} = \frac{\sin^2 y}{\cos y} \).

Next, simplify the right side \( \sec y - \cos y \). By substituting \( \sec y = \frac{1}{\cos y} \), we get \( \frac{1}{\cos y} - \cos y \). Bringing them to a common denominator results in \( \frac{1 - \cos^2 y}{\cos y} \).

Recognize from trigonometric identities that \( 1 - \cos^2 y = \sin^2 y \). Thus, the right side also becomes \( \frac{\sin^2 y}{\cos y} \). By showing both sides equal \( \frac{\sin^2 y}{\cos y} \), we've verified the identity algebraically. By understanding these steps, you can solve similar problems confidently.

Graphical confirmation

Graphical confirmation is a fantastic way to show that two expressions are indeed the same by visualizing them. To confirm the identity \( \frac{\tan y}{\csc y} = \sec y - \cos y \) graphically, we plot both functions using a graphing calculator or appropriate software.

When plotting, set up two functions:

• \( f(y) = \frac{\tan y}{\csc y} \)
• \( g(y) = \sec y - \cos y \)

As both functions are equivalent, their graphs should coincide exactly over their valid domains. This means that at each point where both functions are defined, they will overlap completely.

Be aware that every trigonometric function has limitations called domain restrictions (like points of discontinuity), which might cause some gaps in the graph. However, as long as both graphs overlap whenever they are defined, the graphical confirmation supports that the algebraic identity holds true. Using graphical methods becomes particularly useful for visual learners or those seeking additional verification beyond algebraic means.

Pythagorean identities

Pythagorean identities are one of the cornerstones of trigonometry, providing essential relationships between trigonometric functions. These identities are derived from the Pythagorean theorem and facilitate simplification and solving of trigonometric equations.

The most basic Pythagorean identity is \( \sin^2 y + \cos^2 y = 1 \). From this, we can derive two other important identities:

• \( 1 - \sin^2 y = \cos^2 y \)
• \( 1 - \cos^2 y = \sin^2 y \)

In our problem, the simplification of \( 1 - \cos^2 y \) on the right side to \( \sin^2 y \) is an application of the Pythagorean identities. This substitution allows both sides of the original equation to match, verifying the identity algebraically.

Understanding these identities enables us not only to simplify trigonometric expressions but also to solve complex problems by reducing them to these basic relationships. It’s crucial to recognize and apply these identities consistently to advance in the study of trigonometry.