Proving Trigonometric Identities A Step By Step Guide

#1. Demonstrating the Equivalence of Trigonometric Expressions

In this section, we delve into proving the identity:

${\frac{\sec\theta - 1}{\sec\theta + 1} = \frac{1 - \cos\theta}{1 + \cos\theta} = \left(\frac{\sin\theta}{1 + \cos\theta}\right)^2}$

This identity showcases the interconnectedness of trigonometric functions and offers a valuable exercise in manipulating trigonometric expressions. To prove this identity, we will demonstrate that each expression is equivalent to the others through a series of algebraic manipulations and trigonometric substitutions.

Let's begin by focusing on the first part of the identity:

${\frac{\sec\theta - 1}{\sec\theta + 1} = \frac{1 - \cos\theta}{1 + \cos\theta}}$

Our initial goal is to transform the left-hand side (LHS) of the equation into the right-hand side (RHS). We know that ${\sec\theta}$ is the reciprocal of ${\cos\theta}$, meaning ${\sec\theta = \frac{1}{\cos\theta}}$. Substituting this into the LHS, we get:

${\frac{\frac{1}{\cos\theta} - 1}{\frac{1}{\cos\theta} + 1}}$

To simplify this complex fraction, we multiply both the numerator and the denominator by ${\cos\theta}$. This gives us:

${\frac{1 - \cos\theta}{1 + \cos\theta}}$

Notice that this is precisely the RHS of the first part of our identity. Therefore, we have successfully shown that:

${\frac{\sec\theta - 1}{\sec\theta + 1} = \frac{1 - \cos\theta}{1 + \cos\theta}}$

Now, let's move on to the second part of the identity:

${\frac{1 - \cos\theta}{1 + \cos\theta} = \left(\frac{\sin\theta}{1 + \cos\theta}\right)^2}$

Here, we will start with the middle expression, ${\frac{1 - \cos\theta}{1 + \cos\theta}}$, and manipulate it to match the rightmost expression, ${\left(\frac{\sin\theta}{1 + \cos\theta}\right)^2}$. A common technique in trigonometry is to multiply an expression by a clever form of 1. In this case, we will multiply both the numerator and the denominator by ${1 + \cos\theta}$:

${\frac{1 - \cos\theta}{1 + \cos\theta} \cdot \frac{1 + \cos\theta}{1 + \cos\theta} = \frac{(1 - \cos\theta)(1 + \cos\theta)}{(1 + \cos\theta)^2}}$

Expanding the numerator, we get:

${\frac{1 - \cos^2\theta}{(1 + \cos\theta)^2}}$

Now, we can use the Pythagorean identity, which states that ${\sin^2\theta + \cos^2\theta = 1}$. Rearranging this, we find that ${\sin^2\theta = 1 - \cos^2\theta}$. Substituting this into our expression:

${\frac{\sin^2\theta}{(1 + \cos\theta)^2}}$

Finally, we can rewrite this as:

${\left(\frac{\sin\theta}{1 + \cos\theta}\right)^2}$

This is exactly the RHS of the second part of our identity. Thus, we have proven that:

${\frac{1 - \cos\theta}{1 + \cos\theta} = \left(\frac{\sin\theta}{1 + \cos\theta}\right)^2}$

By demonstrating both parts of the identity, we have successfully proven the entire statement:

${\frac{\sec\theta - 1}{\sec\theta + 1} = \frac{1 - \cos\theta}{1 + \cos\theta} = \left(\frac{\sin\theta}{1 + \cos\theta}\right)^2}$

This identity highlights the power of trigonometric manipulation and the fundamental relationships between trigonometric functions. Understanding these relationships is crucial for solving more complex problems in trigonometry and calculus.

#2. Proving Another Trigonometric Identity Involving Powers of Sine and Cosine

In this section, we aim to prove the following trigonometric identity:

${\sin^8\theta - \cos^8\theta = (\sin^2\theta - \cos^2\theta)(1 - 2\sin^2\theta \cos^2\theta)}$

This identity involves higher powers of sine and cosine, which might seem daunting at first. However, by strategically applying algebraic techniques and trigonometric identities, we can simplify the expression and demonstrate its validity. The key to this proof lies in recognizing the difference of squares pattern and applying it repeatedly.

Let's start with the left-hand side (LHS) of the equation:

${\sin^8\theta - \cos^8\theta}$

We can recognize this as a difference of squares: ${a^2 - b^2 = (a - b)(a + b)}$, where ${a = \sin^4\theta}$ and ${b = \cos^4\theta}$. Applying this factorization, we get:

${(\sin^4\theta - \cos^4\theta)(\sin^4\theta + \cos^4\theta)}$

Notice that the first factor, ${\sin^4\theta - \cos^4\theta}$, is again a difference of squares. This time, ${a = \sin^2\theta}$ and ${b = \cos^2\theta}$. Factoring this, we have:

${(\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta)(\sin^4\theta + \cos^4\theta)}$

We know from the Pythagorean identity that ${\sin^2\theta + \cos^2\theta = 1}$. Substituting this into our expression simplifies it to:

${(\sin^2\theta - \cos^2\theta)(\sin^4\theta + \cos^4\theta)}$

Now, let's focus on the second factor, ${\sin^4\theta + \cos^4\theta}$. Our goal is to manipulate this expression to obtain the form ${1 - 2\sin^2\theta \cos^2\theta}$, which appears in the right-hand side (RHS) of our original identity. We can achieve this by adding and subtracting ${2\sin^2\theta \cos^2\theta}$:

${\sin^4\theta + \cos^4\theta = \sin^4\theta + 2\sin^2\theta \cos^2\theta + \cos^4\theta - 2\sin^2\theta \cos^2\theta}$

The first three terms now form a perfect square: ${(\sin^2\theta + \cos^2\theta)^2}$. Again, using the Pythagorean identity, we know that ${\sin^2\theta + \cos^2\theta = 1}$, so:

${(\sin^2\theta + \cos^2\theta)^2 - 2\sin^2\theta \cos^2\theta = 1 - 2\sin^2\theta \cos^2\theta}$

Substituting this back into our expression, we have:

${(\sin^2\theta - \cos^2\theta)(1 - 2\sin^2\theta \cos^2\theta)}$

This is precisely the RHS of our original identity. Therefore, we have successfully proven that:

${\sin^8\theta - \cos^8\theta = (\sin^2\theta - \cos^2\theta)(1 - 2\sin^2\theta \cos^2\theta)}$

This proof demonstrates the power of algebraic manipulation and the strategic use of trigonometric identities. By recognizing patterns and applying appropriate techniques, we can simplify complex expressions and establish their equivalence.

In conclusion, both identities showcase the beauty and interconnectedness of trigonometric functions. Mastering these types of proofs strengthens your understanding of trigonometric relationships and provides a solid foundation for tackling more advanced mathematical concepts.