Question

sin x  - cos x= n
Найти sin2x и наименьшее значение n

Answer 1

$n^2=(sinx-cosx)^2=sin^2x+cos^2x-2sinxcdot cosx=1-sin2x;$

$sin2x=1-n^2; ; to \\-1 leq 1-n^2 leq 1,\\-2 leq -n^2 leq 0\\0 leq n^2 leq 2; ; to ; ; n^2-2 leq 0,; ; (n -sqrt2)(n+sqrt2) leq 0\\+ + + + + + [-sqrt2]- - - - - - [sqrt2]+ + + + + + \\-sqrt2 leq n leq sqrt2,nin Z(celoe!!!); to ; ; -1 leq n leq 1$