Problem 21 - Precalculus with Calculus Previews

Question

Reduce the given expression to a single trigonometric function. $$ \frac{1}{1+\sin t}+\frac{1}{1-\sin t} $$

Step-by-Step Solution

Answer

The expression simplifies to \(2 \sec^2 t\).

Step 1: Find a Common Denominator

To simplify the expression \( \frac{1}{1+\sin t} + \frac{1}{1-\sin t} \), we need a common denominator. The common denominator of \( \frac{1}{1+\sin t} \) and \( \frac{1}{1-\sin t} \) is \((1+\sin t)(1-\sin t)\).

Step 2: Express Each Fraction with the Common Denominator

Rewrite each fraction so that it has the common denominator:\[ \frac{1}{1+\sin t} = \frac{1-\sin t}{(1+\sin t)(1-\sin t)} \]\[ \frac{1}{1-\sin t} = \frac{1+\sin t}{(1+\sin t)(1-\sin t)} \]

Step 3: Add the Fractions

Combine the fractions into a single expression:\[ \frac{1-\sin t + 1+\sin t}{(1+\sin t)(1-\sin t)} \]

Step 4: Simplify the Numerator

Simplify the expression in the numerator:\[ 1 - \sin t + 1 + \sin t = 2 \]

Step 5: Simplify the Denominator

The denominator \((1+\sin t)(1-\sin t)\) simplifies using the difference of squares: \[ (1+\sin t)(1-\sin t) = 1 - \sin^2 t \]

Step 6: Apply Trigonometric Identity

Use the Pythagorean identity \(1 - \sin^2 t = \cos^2 t\), thus the expression is now:\[ \frac{2}{\cos^2 t} \]

Step 7: Express as a Single Trigonometric Function

The expression \( \frac{2}{\cos^2 t} \) can be written as \(2 \sec^2 t\). Hence, the expression reduces to a single trigonometric function.

Key Concepts

Difference of Squares

When dealing with expressions like \((1+\sin t)(1-\sin t)\), we can utilize a neat mathematical trick called the "Difference of Squares." This is an algebraic identity that states: \[(a + b)(a - b) = a^2 - b^2\]Applying this identity to \((1+\sin t)(1-\sin t)\), we set \(a = 1\) and \(b = \sin t\). Thus, \[(1+\sin t)(1-\sin t) = 1^2 - (\sin t)^2 = 1 - \sin^2 t\]This transformation is crucial in simplifying many trigonometric expressions, as you often encounter terms that fit this pattern.

• It is commonly used in calculus and algebraic manipulations.
• It simplifies the process of factoring and expanding expressions.
• It also plays a vital role in trigonometric identities.

Common Denominator

Finding a common denominator is an essential step when adding or subtracting fractions. In our original problem, we needed one to combine \(\frac{1}{1+\sin t} + \frac{1}{1-\sin t}\).A common denominator ensures that fractions can be easily added or subtracted.For the fractions given, the common denominator is \((1+\sin t)(1-\sin t)\). This allows each fraction to be rewritten as:

• \(\frac{1-\sin t}{(1+\sin t)(1-\sin t)}\)
• \(\frac{1+\sin t}{(1+\sin t)(1-\sin t)}\)

Using a common denominator aligns the fractions under a shared base, allowing seamless operation.

Pythagorean Identity

Trigonometric identities are equations involving trigonometric functions that are true for every value of the variables for which both sides of the equation are defined. A particularly important identity is the Pythagorean Identity. It states:\[\sin^2 t + \cos^2 t = 1\]From this identity, we can express \(1 - \sin^2 t\) as \(\cos^2 t\), which simplifies our denominator: \(1 - \sin^2 t = \cos^2 t\).

• This identity is fundamental in trigonometry.
• It links sine and cosine, allowing for transformation and simplification of expressions.
• It is used when converting between expressions involving squared trigonometric functions.

Secant Function

The secant function is one of the basic trigonometric functions, and it is defined in terms of the cosine function:\[\sec t = \frac{1}{\cos t}\].In the context of our problem, the reduced expression was\(\frac{2}{\cos^2 t}\).Recognizing that \(\frac{1}{\cos^2 t}\) is \(\sec^2 t\), the final result becomes:\[2 \sec^2 t\].

• The secant function is specifically useful in problems involving reciprocal trigonometric expressions.
• It is important in calculus for derivatives and integrals involving trigonometric functions.
• Understanding secant can simplify complex trigonometric solutions.