Sum of Cubes of Sine and Cosine

Theorem

$\cos^3 x + \sin^3 x = \paren {\cos x + \sin x} \paren {1 - \cos x \sin x}$

$\ds \cos^3 x + \sin^3 x$

$=$

$\ds \cos x \paren {1 - \sin^2 x} + \sin x \paren {1 - \cos^2 x}$

$\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

$\blacksquare$

1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Exercise $\text {XXXI}$: $5.$