Question

sin^5a-cos^5a=? если sina-cosa=1/2​

Answer 1

Возводим левую и правую части равенства:

$1-sin2alpha=frac{1}{4}~~~~Leftrightarrow~~~ sin2alpha=frac{3}{4}$

$sin^5alpha-cos^5alpha=(sinalpha-cosalpha)(sin^4alpha+sin^3alphacosalpha+sin^2alphacos^2alpha+\ \ +sinalphacos^3alpha+cos^4alpha)=frac{1}{2}((sin^2alpha+cos^2alpha)^2-sin2alpha+frac{1}{2}sin^2alphasin2alpha+\ \ +frac{1}{4}sin^22alpha+frac{1}{2}cos^2alphasin2alpha)=frac{1}{2}(1-sin2alpha+frac{1}{2}sin2alpha+frac{1}{4}sin^22alpha)=\ \ \ =frac{1}{2}(1-frac{1}{2}cdotfrac{3}{4}+frac{1}{4}cdotfrac{9}{16})=frac{49}{128}$