Problem 38 - A Graphical Approach to Precalculus with Limits

Question

Write expression as a single trigonometric function or a power of a trigonometric function. (You may wish to use a graph to support your result.) $$\frac{1}{\tan ^{2} \alpha}+\cot \alpha \tan \alpha$$

Step-by-Step Solution

Answer

The expression simplifies to \( \csc^2 \alpha \).

Step 1: Simplify the First Term

The given expression is \( \frac{1}{\tan^2 \alpha} + \cot \alpha \tan \alpha \). First, simplify \( \frac{1}{\tan^2 \alpha} \). Remember that \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \), so \( \tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha} \). Therefore, \( \frac{1}{\tan^2 \alpha} = \frac{\cos^2 \alpha}{\sin^2 \alpha} = \cot^2 \alpha \).

Step 2: Simplify the Second Term

Now, simplify \( \cot \alpha \tan \alpha \). We know that \( \cot \alpha = \frac{1}{\tan \alpha} \), thus \( \cot \alpha \tan \alpha = 1 \).

Step 3: Combine the Terms

Combine the simplified terms from Steps 1 and 2. We have \( \cot^2 \alpha + 1 \) from the expression, since we simplified \( \frac{1}{\tan^2 \alpha} \) to \( \cot^2 \alpha \) and \( \cot \alpha \tan \alpha \) to 1.

Step 4: Recognize the Trigonometric Identity

Use the Pythagorean identity \( \cot^2 \alpha + 1 = \csc^2 \alpha \). Hence, the expression \( \frac{1}{\tan^2 \alpha} + \cot \alpha \tan \alpha \) simplifies to \( \csc^2 \alpha \).

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental in understanding relationships in right-angled triangles. They include sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)), among others. These functions help describe the angle measures and side lengths in a triangle. In terms of a unit circle:

• The sine function represents the y-coordinate of the point on the circle.
• The cosine function represents the x-coordinate.
• The tangent function is the ratio of the sine to the cosine, \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).

Using these basic functions, we can derive others like secant (\( \sec \)), cosecant (\( \csc \)), and cotangent (\( \cot \)). Understanding these relationships is key to simplifying trigonometric expressions, like in our task.

Pythagorean Identity

The Pythagorean identity is a fundamental concept in trigonometry. It relates the square of sine and cosine to one. The most common identity is:

• \( \sin^2 \theta + \cos^2 \theta = 1 \)

Other variations are derived by dividing this basic identity by either sine squared or cosine squared:

• \( 1 + \cot^2 \theta = \csc^2 \theta \)
• \( \tan^2 \theta + 1 = \sec^2 \theta \)

In the exercise, recognizing the identity \( \cot^2 \alpha + 1 = \csc^2 \alpha \) allows us to simplify the expression effectively. These identities are useful for transforming and simplifying complex trigonometric expressions.

Cotangent

The cotangent function, denoted as \( \cot \theta \), is another key trigonometric function. Cotangent is the reciprocal of tangent:

• \( \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} \)

This relation shows that the cotangent function is the ratio of the cosine function to the sine function. In our step-by-step solution, we used this property to simplify terms like \( \frac{1}{\tan^2 \alpha} \) into \( \cot^2 \alpha \). Recognizing these reciprocal relationships is crucial for manipulating and solving trigonometric equations.

Cosecant

Cosecant, represented by \( \csc \theta \), is one of the reciprocal trigonometric functions. It is the inverse of the sine function:

• \( \csc \theta = \frac{1}{\sin \theta} \)

Cosecant is particularly useful when dealing with vertical lines on the unit circle or when the sine value is small, as it magnifies this ratio. In our example, using the Pythagorean identity \( \cot^2 \alpha + 1 = \csc^2 \alpha \) allowed us to transform the original expression into a single cosecant function. Understanding how and when to utilize these reciprocal functions can simplify solving trigonometric problems, making them less daunting.