Question

Решите 2 примера 3-4sin^2α ; 3-√3tgα .

Answer 1

$3-4sin^2 alpha =3-3sin^2 alpha -sin^2 alpha =3(1-sin^2 alpha )-sin^2 alpha =\\=3cos^2 alpha -sin^2 alpha =2cos^2 alpha +(cos^2 alpha -sin^2 alpha )=2cos^2 alpha +cos2 alpha ; ;\\\3-sqrt3tg alpha =sqrt3cdot (sqrt3-tg alpha )=sqrt3cdot (tgfrac{pi}{3}-tg alpha )=\\=sqrt3cdot (frac{sinfrac{pi}{3}}{cosfrac{pi}{3}}-frac{sin alpha }{cos alpha })=sqrt3cdot frac{sinfrac{pi}{3}cos alpha -sin alpha cosfrac{pi}{3}}{cosfrac{pi}{3}cdot cos alpha }=$

$=sqrt3cdot frac{sin( frac{pi}{3}-alpha )}{frac{1}{2}cdot cos alpha }=2sqrt3cdot frac{sin( frac{pi}{3}-alpha )}{cos alpha }$

Answer 2

Спасибо большое