Question

Penultima egalitate urgent!!!
Mercii. 20 pct :)

Answer 1

$\displaystyle~Avem~de~demonstrat~identitatea \\ \\ tg \frac{x}{2} tg \frac{y}{2}+tg \frac{z}{2} \cdot \frac{sin \frac{x+y}{2}}{cos \frac{x}{2}cos \frac{y}{2}}= \frac{1}{cos \frac{x}{2} cos \frac{y}{2}} \left(cos \frac{x+y}{2}+sin \frac{x}{2}sin \frac{y}{2} \right). \\ \\ Voi~nota~cu~E~expresia~din~membrul~stang.$

$\displaystyle Avem~E= \frac{sin \frac{x}{2}sin\frac{y}{2}}{cos \frac{x}{2} cos \frac{y}{2}}+tg \frac{z}{2} \cdot \frac{sin \frac{x+y}{2}}{cos \frac{x}{2}cos \frac{y}{2}}= \\ \\ = \frac{1}{cos \frac{x}{2} cos \frac{y}{2}} \left( sin \frac{x}{2}sin\frac{y}{2}+tg \frac{z}{2} \cdot sin \frac{x+y}{2} \right). \\ \\ Mai~ramane~de~demonstrat~ca~ tg \frac{z}{2} \cdot sin \frac{x+y}{2}=cos \frac{x+y}{2}.$

$\displaystyle Demonstratie:~x+y+z= \pi \Rightarrow tg \frac{z}{2}=tg \left( \frac{\pi}{2}- \frac{x+y}{2} \right)= \\ \\ =ctg \frac{x+y}{2}. \\ \\ Deci~tg \frac{z}{2} \cdot sin \frac{x+y}{2}=ctg \frac{x+y}{2} \cdot sin \frac{x+y}{2}=cos \frac{x+y}{2},~q.e.d.$

$\displaystyle ----------------------------- \\ \\ O~alta~abordare:~Cum~x,y,z \in (0, \pi)~si~x+y+z= \pi,~rezulta \\ \\ ca~x,y,z~sunt~unghiurile~unui~triunghi. \\ \\ Notam~x=A,~y=B,~z=C. \\ \\ Atunci,~in~triunghiul~ABC~are~loc~relatia \\ \\ \boxed{ tg \frac{A}{2}= \sqrt{\frac{(p-b)(p-c)}{p(p-a)}} },~precum~si~relatiile~analoage. \\ \\ Dupa~inlocuire~si~calculare~se~obtine~\sum tg \frac{A}{2} tg \frac{B}{2}=1,~q.e.d.$