Problem 41 - Precalculus with Limits

Question

In Exercises 9-50, verify the identity \( \sqrt{\dfrac{1 + \sin \theta}{1 - \sin \theta}} = \dfrac{1 + \sin \theta}{\mid \cos \theta \mid} \)

Step-by-Step Solution

Answer

Yes, \( \sqrt{\dfrac{1 + \sin \theta}{1 - \sin \theta}} = \dfrac{1 + \sin \theta}{\mid \cos \theta \mid} \) is verified to be true

Step 1: Isolate the Square Root

Starting with the left side (\(LHS\)) of the equation, we begin by isolating the square root: \(LHS = \sqrt{\dfrac{1 + \sin \theta}{1 - \sin \theta}}\)

Step 2: Square Both Sides

Next, we square both sides of the equation to get rid of the square root that is on the \(LHS\): \(LHS^2 = \dfrac{1 + sin \theta}{1 - sin \theta}\) = \( \dfrac{1 + 2\sin \theta + \sin^2 \theta }{1 - 2\sin \theta + \sin^2 \theta}\)

Step 3: Use Pythagorean Identity

We then use the Pythagorean identity to replace \(\sin^2 \theta\) with \(1 - \cos^2 \theta\) to connect \(\sin\theta\) and \(\cos \theta\), so the equation becomes: \(LHS^2 = \dfrac{1 + 2\sin \theta + 1 - \cos^2 \theta }{1 - 2\sin \theta + 1 - \cos^2 \theta} = \dfrac{2 + 2\sin \theta - \cos^2 \theta }{2 - 2\sin \theta - \cos^2 \theta}\)

Step 4: Split the fraction

Next, we split the fraction into two:\(LHS^2 = \dfrac{2 + 2\sin \theta}{2 - 2\sin \theta} - \dfrac{\cos^2 \theta}{2 - 2\sin \theta} = 1 + \dfrac{2\sin \theta}{2 - 2\sin \theta} - \dfrac{\cos^2 \theta}{2 - 2\sin \theta}\)

Step 5: Rationalize denominator, Address absolute value

Now, rationalize the denominator in second term in \(LHS^2\), and address the absolute value in the equation at the same time. This is done by using the rule \(\frac{a}{b} = \frac{a \cdot (-b)}{-b \cdot b}\), which gives us: \(LHS^2 = 1 - \sin \theta + \dfrac{\cos^2 \theta}{\cos 2\theta} = 1 + \sin \theta + \mid \cos \theta\mid = RHS\)

Key Concepts

Pythagorean Identity

The Pythagorean identity is one of the fundamental identities in trigonometry. It relates the square of sine and cosine functions. The identity is expressed as:

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

This identity is derived from the Pythagorean theorem applied to a unit circle, where the radius is always equal to 1.
Since we can substitute \( \sin^2\theta \) for \( 1 - \cos^2\theta \), this allows us to change between sine and cosine in equations.
This flexibility is particularly useful in manipulating expressions to verify trigonometric identities or solve equations.
In the provided solution, the identity was used to convert \( \sin^2 \theta \) into \( 1 - \cos^2 \theta \).
This helps in simplifying the expression and revealing a connection between the trigonometric functions involved.
When dealing with trigonometric identities, practicing the Pythagorean identities can greatly enhance your problem-solving skills.

Rationalizing Denominators

Rationalizing the denominator involves eliminating any irrational numbers or complex expressions from the bottom of a fraction. This can be useful for simplifying expressions and making calculations easier.
The technique involves multiplying the numerator and the denominator by a suitable expression that will cancel out the irrational part of the denominator.
In the given solution, rationalizing the denominator was achieved during the fraction manipulation steps.

• The equation \( \frac{a}{b} = \frac{a \cdot (-b)}{-b \cdot b} \) was used to eliminate irrational numbers.

Rationalizing is an essential skill in algebra and trigonometry for simplifying solutions and obtaining equivalent expressions in a preferred form.
In trigonometric contexts, rationalization can especially help manage expressions involving square roots or complex fractions.

Absolute Value in Trigonometry

Absolute value is a crucial concept in mathematics, often used to denote the magnitude of a number without focusing on its sign.
In trigonometry, absolute value frequently appears when simplifying expressions or verifying identities.
For example, the absolute value of cosine, \( \mid \cos \theta \mid \), represents the non-negative magnitude of the cosine of an angle.

• Because \( \cos \theta \) can be positive or negative depending on the quadrant, we use the absolute value to ensure the expression remains valid across all possible angles.

In trigonometric identity verification, absolute value helps in maintaining equivalence between expressions.
This is done by removing the ambiguity introduced by sign changes across different quadrants.
By using absolute value, trigonometric expressions become easier to work with as they remain consistent and invariant against sign influences. This leads to a reliable and standardized form as seen in the solution where \( \mid \cos \theta \mid \) is employed to match the form on the right side of the identity.