Question

sa se verifice egalitatile:
[sin²(α+β)+sin² (α-β)] / 2cos² α*cos² β= tg² α + tg² β

Answer 1

Răspuns:

Am inlocuit Alfa cu x si Beta cu y

$\frac{sin^2(x+y)+sin^2(x-y)}{2cos^2x*cos^2y}=tg^2x+tg^2y$

$\frac{(sinx*cosy+siny*cosx)^2+(sinx*cosy-siny*cosx^2)}{2cos^2x*cos^2y}=tg^2x+tg^2y$

$\frac{sin^2x*cos^2y+sin^2y*cos^2x+2sinx*cosy*siny*cosx+sin^2x*cos^2y+sin^2y*cos^2x-2sinx*cosy*siny*cosx}{2cos^2x*cos^2y}$

$\frac{2sin^2x*cos^2y+2sin^2y*cos^2x}{2cos^2x*cos^2y}=tg^2x+tg^2y$

$\frac{2(sin^2x*cos&^2y+sin^2y*cos^2x)}{2cos^2x*cos^2y}=tg^2x+tg^2y$

$\frac{sin^2x*cos^2y+sin^2y*cos^2x}{cos^2x*cos^2y}=tg^2x+tg^2y$

$tg^2x+tg^2y=tg^2x+tg^2y$ "Adevarat"