Sum of Cubes of Sine and Cosine
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Theorem
- $\cos^3 x + \sin^3 x = \paren {\cos x + \sin x} \paren {1 - \cos x \sin x}$
Proof
\(\ds \cos^3 x + \sin^3 x\) | \(=\) | \(\ds \cos x \paren {1 - \sin^2 x} + \sin x \paren {1 - \cos^2 x}\) | Sum of Squares of Sine and Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \cos x + \sin x - \cos x \sin x \paren {\cos x + \sin x}\) | simplification | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\cos x + \sin x} \paren {1 - \cos x \sin x}\) | simplification |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Exercise $\text {XXXI}$: $5.$