Răspunsuri

Cel mai inteligent răspuns!
2014-04-10T15:30:12+03:00
A) Se folosesc formulele:

sin\dfrac A2=\sqrt{\dfrac{(p-b)(p-c)}{bc}};\ \ cos\dfrac A2=\sqrt{\dfrac{p(p-a)}{bc}} si analoagele si avem:

a\ sin\dfrac{B-C}{2}=a\left(sin\dfrac B2\ cos\dfrac C2-sin\dfrac C2\ cos\dfrac B2\right)=

=a\sqrt{\dfrac{(p-a)(p-c)}{ac}}\cdot\sqrt{\dfrac{p(p-c)}{ab}}-a\sqrt{\dfrac{(p-a)(p-b)}{ab}}\cdot\sqrt{\dfrac{p(p-b)}{ac}}=

=(p-c)\sqrt{\dfrac{p(p-a)}{bc}}-(p-b)\sqrt{\dfrac{p(p-a)}{bc}}=(b-c)\sqrt{\dfrac{p(p-a)}{bc}}=

=(b-c)cos\dfrac A2
Mersi, dar la al doilea stii sa faci?
Trebuie timp ceva mai mult, dar poti incerca sa folosesti formule de aici: http://www.mateonline.net/cadru.php?cap=trigonometrie_relatii_metrice.