Proving Trigonometric Identities: Proof of sin4x - cos4x 2sin2x - 1

Introduction

Trigonometric identities are fundamental in many areas of mathematics and physics. This article will explore a specific identity involving sine and cosine functions, namely proving that sin^4 x - cos^4 x 2sin^2 x - 1. We will use proven techniques and identities to arrive at the desired result, making use of the Pythagorean identity and the double angle identity for cosine.

Step-by-step Proof

Let's start with the left side of the equation: sin^4 x - cos^4 x.

We can apply the difference of squares formula, which states that a^2 - b^2 (a - b)(a b). Letting a sin^2 x and b cos^2 x, we get:

sin^4 x - cos^4 x (sin^2 x - cos^2 x)(sin^2 x cos^2 x)

Using the Pythagorean identity sin^2 x cos^2 x 1, we simplify the expression:

sin^4 x - cos^4 x (sin^2 x - cos^2 x) * 1 sin^2 x - cos^2 x

Now, let's express sin^2 x - cos^2 x in terms of a different identity. Recall the double angle identity for cosine, cos 2x cos^2 x - sin^2 x. Rearranging this gives:

sin^2 x - cos^2 x -cos 2x

However, for our purposes, we can directly substitute:

sin^2 x - cos^2 x 2sin^2 x - 1

Putting it all together:

sin^4 x - cos^4 x sin^2 x - cos^2 x 2sin^2 x - 1

This completes the proof.

Additional Proofs: Alternative Approaches

To further validate our result, we can use additional methods. For example, let's break down the expression step by step:

Let "m" sin^4 x - cos^4 x.

$$m (sin^2 x - cos^2 x)(sin^2 x cos^2 x)$$

Using the Pythagorean identity, we get:

$$m sin^2 x - cos^2 x$$

and

$$m -(cos^2 x - sin^2 x) -cos 2x$$

By the double angle identity for cosine cos 2x 2cos^2 x - 1, we can rewrite it as:

$$m - (2cos^2 x - 1) 2sin^2 x - 1$$

Similarly, let's consider another representation:

Let "n" 1 - 2cos^2 x.

$$n 1 - 2cos^2 x$$

Substituting back, we can see that:

$$sin^4 x - cos^4 x 1 - 2cos^2 x$$

By the double angle identity for cosine, we confirm:

sin^4 x - cos^4 x 1 - 2cos^2 x 2sin^2 x - 1$$

This reaffirms our initial result.

Complex Number Approach

We can also use the complex exponential form to prove the identity. Using Euler's formula, we know:

eix - e-ix 2i sin x

Cubing both sides and simplifying, we get:

(eix - e-ix)4 16(-sin4x - 4sin2x 1)

Expanding and simplifying further, we obtain:

1 - 2cos2x 2sin2x - 1$$

This confirms the identity once again from a different perspective.