How to prove sin^2(x) - cos^2(x) = sin^4(x) - cos^4(x)?

Start with the right side of the equation: sin^4(x) - cos^4(x) = sin^2(x)^2 - cos^2(x)^2 Using difference of squares factorization gives sin^2(x)^2 - cos^2(x)^2= sin^2(x) + cos^2(x)*sin^2(x)-cos^2(x) By Pythagorean Identities, sin^2(x) + cos^2(x) = 1, and the identity is verified.

You need only manipulate the right side of this eqn to prove the hypothesis: sin^2(x) - cos^2(x) = sin^4(x) - cos^4(x) sin^2(x) - cos^2(x) = sin^2(x) - cos^2(x)sin^2(x) + cos^2(x) We know sin^2(x) + cos^2(x) = 1 Therefore, sin^2(x) - cos^2(x) = sin^2(x) - cos^2(x)(1) = sin^2(x) - cos^2(x) QED.

Sin^2(x)-cos^2(x) = sin^4(x) - cos^4(x) sin^2(x)-cos^2(x) = sin^2(x)^2 - cos^2(x)^2.

Very simple. Sin²(x) - cos²(x) = sin?(x) - cos?(x) sin²(x) - cos²(x) = sin²(x)² - cos²(x)² sin²(x) - cos²(x) = sin²(x) - cos²(x)sin²(x) + cos²(x) But sin²(x) + cos²(x) = 1 sin²(x) - cos²(x) = sin²(x) - cos²(x)*1 sin²(x) - cos²(x) = sin²(x) - cos²(x) Demonstrated. Greetings, Roberto.

You need difference of squares for the right side and the fact that sin^2(x) + cos^2(x) = 1. Note that, sin^4(x) - cos^4(x) = (sin^2(x) + cos^2(x))(sin^2(x) - cos^2(x)) = (1)(sin^2(x) - cos^2(x)) = (sin^2(x) - cos^2(x)).

Sin^4(x) - cos^4(x) = (sin^2 x)^2 - (cos^2 x)^2 Difference of two squares... = (sin^2 x - cos^2 x)*(sin^2 x + cos^2 x) sin^2 x + cos^2 x = 1 = sin^2 x - cos^2 x.

(sin x)^4 - (cos x)^4 (sin²x - cos²x)(sin²x + cos²x) sin²x - cos²x.

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