Problem 53 - Algebra and Trigonometry

Question

Verify the identity. $$ \frac{1-\cos \alpha}{\sin \alpha}=\frac{\sin \alpha}{1+\cos \alpha} $$

Step-by-Step Solution

Answer

Both sides simplify to \( \frac{\sin \alpha}{1 + \cos \alpha} \), proving the identity true.

Step 1: Understand the Identity

You need to verify that the left-hand side (LHS) of the equation \( \frac{1 - \cos \alpha}{\sin \alpha} \) equals the right-hand side (RHS) \( \frac{\sin \alpha}{1 + \cos \alpha} \). To do this, we'll simplify each side and check for equality.

Step 2: Simplify the LHS

Start with the LHS, \( \frac{1 - \cos \alpha}{\sin \alpha} \). Note that by multiplying and dividing by the same expression, \( 1 + \cos \alpha \), we can simplify: \[ \frac{1 - \cos \alpha}{\sin \alpha} \times \frac{1 + \cos \alpha}{1 + \cos \alpha} = \frac{(1 - \cos \alpha)(1 + \cos \alpha)}{\sin \alpha (1 + \cos \alpha)} \].

Step 3: Simplify the Numerator of LHS

Simplify the numerator \((1 - \cos \alpha)(1 + \cos \alpha)\) using the difference of squares formula: \(1 - \cos^2 \alpha\). Since \(1 - \cos^2 \alpha = \sin^2 \alpha\), replace and simplify: \[ \frac{\sin^2 \alpha}{\sin \alpha (1 + \cos \alpha)} \].

Step 4: Cancel Out Common Terms

Cancel \(\sin \alpha\) in the numerator and denominator: \[ \frac{\sin \alpha}{1 + \cos \alpha} \]. This is now identical to the RHS.

Key Concepts

Difference of Squares

Understanding the difference of squares can help us simplify many mathematical expressions. The fundamental formula to remember is:

• \((a - b)(a + b) = a^2 - b^2\)

This identity is very useful because it allows us to convert a product of binomials into a simple subtraction form. In our problem, we applied the difference of squares to \((1 - \cos \alpha)(1 + \cos \alpha)\). Breaking it down:

• First term (1): When squared stays 1.
• Second term (\(\cos \alpha\)): On squaring becomes \(\cos^2 \alpha\).

Using the difference of squares, we conclude that:

• \((1 - \cos \alpha)(1 + \cos \alpha) = 1^2 - (\cos \alpha)^2 = 1 - \cos^2 \alpha\)
• By employing this technique, we notice that the expression simplifies greatly to \(\sin^2 \alpha\) due to the Pythagorean identity \(1 - \cos^2 \alpha = \sin^2 \alpha\).

Simplifying Expressions

Simplifying expressions is all about making complex mathematical problems more manageable. It involves reducing the expression to its simplest form. In our task:

• We started by taking the complex-looking expression \(\frac{1-\cos \alpha}{\sin \alpha}\) and simplified it by multiplying and dividing by \((1+\cos \alpha)\).
• Doing so introduced the difference of squares into the numerator.

Where it plays a crucial role by converting \(1-\cos^2 \alpha\) into \(\sin^2 \alpha\). This provided us a straightforward way to cancel identical terms in the expression:

• The \(\sin \alpha\) in the numerator simplifies with the denominator, resulting in a simpler \(\frac{\sin \alpha}{1+\cos \alpha}\).

When simplifying expressions, remember the key is both recognizing patterns (like difference of squares) and understanding identities. This helps unravel the complex layers of any mathematical expression.

Verifying Identities

Verifying trigonometric identities tests the equivalence of expressions on both sides of an equation. Here, our mission was to check if the two sides were equal:

• First, we examined the initial LHS expression: \(\frac{1-\cos \alpha}{\sin \alpha}\).
• Through careful manipulation and simplification, using algebraic tactics and trigonometric identities, we've transformed it to see if it matches the RHS: \(\frac{\sin \alpha}{1+\cos \alpha}\).

Key steps in verifying include:

• Simplifying each side: Bring all involved terms to their simplest form. Look for possibilities of factorizations or common trigonometric identities.
• Cancelling out identical terms: Remove redundant terms that appear in both the numerator and denominator whenever valid.

Once complete, if both sides look identical, the identity is verified! This process hones your logic and understanding, encouraging looking deeply into each mathematical component.