Problem 52 - Precalculus with Calculus Previews

Question

Verify the given identity. $$ \frac{1}{1-\cos \alpha}+\frac{1}{1+\cos \alpha}=2 \csc ^{2} \alpha $$

Step-by-Step Solution

Answer

The identity is verified; both sides simplify to the same expression.

Step 1: Understanding the Problem

We need to verify the identity \( \frac{1}{1 - \cos \alpha} + \frac{1}{1 + \cos \alpha} = 2 \csc^2 \alpha \). This means we'll show that both sides are equal by transforming the left-hand side into the right-hand side.

Step 2: Simplifying the Left Side

Start by finding a common denominator for the terms on the left. The common denominator for \( 1 - \cos \alpha \) and \( 1 + \cos \alpha \) is \((1 - \cos \alpha)(1 + \cos \alpha)\).

Step 3: Combine Fractions

Combine the fractions: \[ \frac{1}{1-\cos \alpha} + \frac{1}{1+\cos \alpha} = \frac{1(1+\cos \alpha) + 1(1-\cos \alpha)}{(1-\cos \alpha)(1+\cos \alpha)}. \]This simplifies to:\[ \frac{1 + \cos \alpha + 1 - \cos \alpha}{(1-\cos \alpha)(1+\cos \alpha)} = \frac{2}{(1-\cos \alpha)(1+\cos \alpha)}. \]

Step 4: Simplifying the Denominator

Recognize that \((1 - \cos \alpha)(1 + \cos \alpha) = 1^2 - (\cos \alpha)^2 = \sin^2 \alpha\) using the Pythagorean identity \(1 - \cos^2 \alpha = \sin^2 \alpha\). So now we have:\[ \frac{2}{\sin^2 \alpha}. \]

Step 5: Recognizing Cosecant

Now note that by definition, \( \csc \alpha = \frac{1}{\sin \alpha} \). Therefore, \( \frac{1}{\sin^2 \alpha} = \csc^2 \alpha \). So:\[ \frac{2}{\sin^2 \alpha} = 2\csc^2 \alpha. \]

Step 6: Verification Complete

We have transformed the left side into \(2\csc^2 \alpha\), demonstrating that the given identity is verified: \[ \frac{1}{1-\cos \alpha} + \frac{1}{1+\cos \alpha} = 2\csc^2 \alpha. \]

Key Concepts

Cosecant

The term **cosecant** refers to one of the six fundamental trigonometric functions. It is usually denoted as \( \csc \alpha \), which stands for the cosecant of an angle \( \alpha \). Cosecant is important in trigonometry because it relates to the sine function. In fact, cosecant is the reciprocal of sine:

• The formula is \( \csc \alpha = \frac{1}{\sin \alpha} \).

This means if you know the sine of an angle, you can easily calculate the cosecant by taking the inverse of the sine value. Remember, \( \csc \alpha \) becomes undefined whenever \( \sin \alpha = 0 \), so it's crucial to consider the angle values carefully.

Cosecant arises frequently in many aspects of mathematics, especially in verifying trigonometric identities, like the one we are considering here. Since \( 2 \csc^2 \alpha = \frac{2}{\sin^2 \alpha} \), this identity shows how cosecant effectively simplifies expressions involving the sine function.

Common Denominator

Finding a **common denominator** is a crucial step in simplifying trigonometric identities, especially when dealing with compound fractions. In our given identity, we combine two separate fractions: \( \frac{1}{1 - \cos \alpha} \) and \( \frac{1}{1 + \cos \alpha} \). To add these together, we need a common denominator.

When dealing with fractions involving different denominators, we choose a common denominator that can evenly accommodate each term. For expressions \( 1 - \cos \alpha \) and \( 1 + \cos \alpha \), the simplest common denominator is their product, which is \((1 - \cos \alpha)(1 + \cos \alpha)\).


This choice is optimal because:\

• Multiplying these expressions eliminates the division between terms.
• This product is easy to simplify using another trigonometric identity: \((1 - \cos \alpha)(1 + \cos \alpha) = 1 - \cos^2 \alpha\), which equals \( \sin^2 \alpha \).

This step is key as it transforms the fraction into a simpler form, making further algebra more straightforward.

Pythagorean Identity

The **Pythagorean identity** is one of the most fundamental identities in trigonometry. It expresses the intrinsic relationship between the sine and cosine of an angle. The basic form of the Pythagorean identity is:

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

This identity is incredibly useful for simplifying expressions involving trigonometric functions because it allows you to substitute one function for another.

In our exercise, we apply a slightly altered version of this identity: \( 1 - \cos^2 \alpha = \sin^2 \alpha \). When dealing with the common denominator \((1 - \cos \alpha)(1 + \cos \alpha)\), we use this identity:

• This simplifies the denominator to \( \sin^2 \alpha \).

By simplifying \( 1 - \cos^2 \alpha \) to \( \sin^2 \alpha \), we align the expression with the cosecant function (since \( \csc \alpha = \frac{1}{\sin \alpha} \)), ultimately verifying the identity: \( \frac{2}{\sin^2 \alpha} = 2 \csc^2 \alpha \). This illustrates the identity's powerful tool in verifying and solving trigonometric equations.