show-that-cos-6-x-sin-6-x-1-8-5-3cos4x-

Question Number 75983 by mathocean1 last updated on 21/Dec/19

show that

\cos^{6} x + \sin^{6} x = \frac{1}{8} (5 + 3 \cos 4 x)

Commented by MJS last updated on 21/Dec/19

answer is part of the solution of the other

question

Commented by mathmax by abdo last updated on 22/Dec/19

{c o s}^{6} x + {s i n}^{6} x = {({c o s}^{2} x)}^{3} + {({s i n}^{2} x)}^{3}

= ({c o s}^{2} x + {s i n}^{2} x) ({c o s}^{4} x - {c o s}^{2} x {s i n}^{2} x + {s i n}^{4} x)

= {c o s}^{4} x + {s i n}^{4} x - {c o s}^{2} x {s i n}^{2} x = {({c o s}^{2} x + {s i n}^{2} x)}^{2} - 3 {c o s}^{2} x {s i n}^{2} x

= 1 - \frac{3}{4} {s i n}^{2} (2 x) = 1 - \frac{3}{4} (\frac{1 - c o s (4 x)}{2}) = 1 - \frac{3}{8} + \frac{3}{8} c o s (4 x)

= \frac{5}{8} + \frac{3}{8} c o s (4 x) s o t h e r e s u l t i s p r o v e d .