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Proving the Trigonometric Identity: cos^4x - sin^4x cos^2x - sin^2x

January 07, 2025Workplace1559
Proving the Trigonometric Identity: cos4x - sin4x cos2x - sin2x To pr

Proving the Trigonometric Identity: cos4x - sin4x cos2x - sin2x

To prove the trigonometric identity (cos^4 x - sin^4 x cos^2 x - sin^2 x), we can utilize the difference of squares formula. Let's go through this step by step.

Step 1: Factoring Using the Difference of Squares Formula

The difference of squares identity is given by:

[text{a}^2 - text{b}^2 (text{a} - text{b})(text{a} text{b})]

We can apply this identity to (cos^4 x - sin^4 x), which can be rewritten as:

[cos^4 x - sin^4 x (cos^2 x)^2 - (sin^2 x)^2 (cos^2 x - sin^2 x)(cos^2 x sin^2 x).]

Step 2: Simplification Using the Pythagorean Identity

From the Pythagorean identity, we know that:

[cos^2 x sin^2 x 1]

Substituting this into our factored expression, we get:

[(cos^2 x - sin^2 x)(cos^2 x sin^2 x) (cos^2 x - sin^2 x) times 1 cos^2 x - sin^2 x.]

Step 3: Final Simplification

Thus, we have:

[cos^4 x - sin^4 x cos^2 x - sin^2 x.]

Conclusion

This concludes the proof that the left-hand side equals the right-hand side:

[cos^4 x - sin^4 x cos^2 x - sin^2 x.]

Therefore, the identity is proven. The key steps in the proof involved recognizing the difference of squares and applying the Pythagorean identity.

Expanding the Proof

Consider another way to write the expression:

[cos^4 x - sin^4 x (cos^2 x)^2 - (sin^2 x)^2.]

Using the difference of squares identity again:

[(cos^2 x)^2 - (sin^2 x)^2 (cos^2 x - sin^2 x)(cos^2 x sin^2 x).]

Substituting (cos^2 x sin^2 x 1) into the expression, we get:

[(cos^2 x - sin^2 x)(1) cos^2 x - sin^2 x.]

Hence, we have shown that:

[cos^4 x - sin^4 x cos^2 x - sin^2 x.]

Remembering Essential Relationships

One must remember the following relationships:

(text{a}^2 - text{b}^2 (text{a} - text{b})(text{a} text{b}))[Difference of Squares formula] (cos^2 x sin^2 x 1)[Pythagorean Identity]

Additionally, note that (cos^4 x (cos^2 x)^2) and (sin^4 x (sin^2 x)^2). Let (a cos^2 x) and (b sin^2 x). Then:

[(cos^2 x)^2 - (sin^2 x)^2 a^2 - b^2 (a - b)(a b) cos^2 x - sin^2 x (cos^2 x sin^2 x) cos^2 x - sin^2 x.]

This reaffirms that:

[cos^4 x - sin^4 x cos^2 x - sin^2 x.]

And therefore, we have provided a detailed proof that the given identity holds true.