Question

Помогите доказать тождества. Подробно.

Answer 1

$dfrac{(1-2cos^2 alpha )(2sin^2 alpha -1)}{4sin^2 alpha cos^2 alpha }= dfrac{(2cos^2 alpha-1 )(1-2sin^2 alpha )}{4sin^2 alpha cos^2 alpha }= \ = dfrac{cos2 alphacdot cos2 alpha}{(2sin alpha cos alpha )^2}= dfrac{cos^22 alpha}{sin^22 alpha}= mathrm{ctg}^22 alpha$

$1-2sin^2left( dfrac{ pi }{4} - dfrac{ alpha }{2} right)= cos2left( dfrac{ pi }{4} - dfrac{ alpha }{2} right)= cosleft( dfrac{ pi }{2} - alpha right)=sin alpha$

$dfrac{sin alpha +sin 2 alpha }{1+cos alpha +cos 2 alpha } = dfrac{sin alpha +2sin alphacos alpha }{1+cos alpha +2cos^2 alpha-1 } = \ = dfrac{sin alpha +2sin alphacos alpha }{cos alpha +2cos^2 alpha } = dfrac{sin alpha(1 +2cos alpha) }{cos alpha(1 +2cos alpha) } = dfrac{sin alpha }{cos alpha } =mathrm{tg} alpha$