Question

If sin x + cos x = a, evaluate $si{n}^{6}xco{s}^{6}x$

Answer 1

$si{n}^{6}x+co{s}^{6}x=1-3si{n}^{2}xco{s}^{2}x\frac{1}{4}[4-3si{n}^{2}2x]$
Now sin x + cos x = a squaring ,
$1+sin2x=a^2\therefore sin2x=a^2-1$
Putting in (1)
$\therefore L.H.S.=\frac{1}{4}[4-3(a^2-1{)}^2]$
$-1\ge sin2x\ge 1\Rightarrow -1\ge a^2-1\ge 1$
or $0\ge a^2\ge 2\therefore a^2\ge 0and{a}^2\ge 2$
Which implies that $-\sqrt{2}\ge a\ge \sqrt{}2$