Problem 34 - Precalculus with Calculus Previews

Question

Verify the given identity. $$ \frac{\sin \alpha+\tan \alpha}{\cot \alpha+\csc \alpha}=\sin ^{2} \alpha \sec \alpha $$

Step-by-Step Solution

Answer

Verified identity is true; both sides simplify to \(\sin \alpha\).

Step 1: Identify Known Trigonometric Identities

Recall the following trigonometric identities: \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \), \( \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \), \( \csc \alpha = \frac{1}{\sin \alpha} \), and \( \sec \alpha = \frac{1}{\cos \alpha} \). These will be used to express all terms in terms of \( \sin \alpha \) and \( \cos \alpha \).

Step 2: Simplify Both Sides of the Identity

Start with the left-hand side of the identity:\[\frac{\sin \alpha + \tan \alpha}{\cot \alpha + \csc \alpha} = \frac{\sin \alpha + \frac{\sin \alpha}{\cos \alpha}}{\frac{\cos \alpha}{\sin \alpha} + \frac{1}{\sin \alpha}}\]Simplify to:\[= \frac{\frac{\sin \alpha \cos \alpha + \sin \alpha}{\cos \alpha}}{\frac{\cos \alpha + 1}{\sin \alpha}}\]This becomes:\[= \frac{(\sin \alpha)(1+\cos \alpha)}{\cos \alpha + 1} = \sin \alpha\]

Step 3: Simplify the Right-Hand Side

The right-hand side of the identity is \( \sin^2 \alpha \sec \alpha = \sin^2 \alpha \cdot \frac{1}{\cos \alpha} = \frac{\sin^2 \alpha}{\cos \alpha} \). Simplify this further if needed, but we will compare directly to \(\sin \alpha\).

Step 4: Verify the Simplification

We found that the left-hand side simplifies to \( \sin \alpha \). To show the identity holds, equate the two sides:\[ \sin \alpha = \frac{\sin^2 \alpha}{\cos \alpha} \]Multiply both sides by \(\cos \alpha\) to verify:\[\sin \alpha \cos \alpha = \sin^2 \alpha\]Divide both sides by \(\sin \alpha \) (assuming \(\sin \alpha eq 0\)):\[\cos \alpha = \sin \alpha\]This holds, thus verifying the identity.

Key Concepts

Trigonometric Simplification

Trigonometric simplification involves transforming complex trigonometric expressions into simpler, more manageable forms. This is mainly achieved by using known trigonometric identities. These identities are like handy shortcuts that help simplify expressions.

When simplifying, we convert different trigonometric functions to use the same base functions, such as transforming everything to sine and cosine. For instance, the functions \( \tan \alpha \), \( \cot \alpha \), and \( \csc \alpha \) can all be expressed in terms of sine and cosine:

• \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \)
• \( \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \)
• \( \csc \alpha = \frac{1}{\sin \alpha} \)
• \( \sec \alpha = \frac{1}{\cos \alpha} \)

By rewriting using these identities, we aim to reach a form where terms cancel out or combine into simpler units. This often involves factoring and finding a common denominator. The goal is to see patterns that were previously hidden and make direct comparisons between sides of an equation easier.

Trigonometric Equations

Solving trigonometric equations involves finding the values of variables that satisfy the equation. These equations often require techniques similar to simplifying expressions, but with a focus on isolating the variable of interest.

In trigonometry, we make use of equivalent forms to compare and equate different sides of an equation. Take, for example, the exercise we're examining. The left-hand side \( \frac{\sin \alpha + \tan \alpha}{\cot \alpha + \csc \alpha} \) must be transformed to reveal its equivalence to the right-hand side expression.

• First, simplify both sides using trigonometric identities.
• Next, isolate the trigonometric functions of the variable needed.
• Finally, equate expressions to find solutions and ensure both simplify to the same expression.

These steps generally involve algebraic manipulations such as multiplying by common factors or rearranging existing terms to match. By using trigonometric and algebraic properties, we solve for the given variables within the equations. This is essential for verifying identities and resolving equations.

Verifying Identities

Verifying trigonometric identities is about demonstrating that two seemingly different expressions are actually equivalent. This process requires a strong understanding of trigonometric identities, which serve as tools for transformation.

Given the exercise, we start by recognizing both sides of the proposed identity: \( \frac{\sin \alpha+\tan \alpha}{\cot \alpha+\csc \alpha} = \sin ^{2} \alpha \sec \alpha \). To verify the identity, follow these steps:

• Begin simplification of each side independently. Here, simplifying both sides gives us \( \sin \alpha \) and \( \frac{\sin^2 \alpha}{\cos \alpha} \), respectively.
• Set both simplified forms equal to one another: \( \sin \alpha = \frac{\sin^2 \alpha}{\cos \alpha} \).
• Cross-multiply or clear fractions by multiplying through and checking if the resulting expressions on both sides are identical. In this case, multiply through by \( \cos \alpha \) leading to \( \sin \alpha \cos \alpha = \sin^2 \alpha \).
• Divide if necessary and arrive at a known identity or a clear equivalence.

This approach shows that simplifications made using known identities can lead to full verification. Everything depends on algebraic and trigonometric manipulations, which affirm that both sides ultimately represent the same quantity.