Trigonometrie relatii

AnaMaria2323 · Mesaj de **AnaMaria2323** » 12 Iun 2018, 18:35

Să se arate ca intr-un triunghi oarecare ABC au loc relațiile : a) $tg\frac{A}{2}*tg\frac{B}{2} + tg\frac{B}{2}*tg\frac{C}{2}+ tg\frac{C}{2}* tg\frac{A}{2}=1$
b) $\frac{a}{sin^{2}\frac{A}{2}} + \frac{b}{sin^{2}\frac{B}{2}} + \frac{c}{sin^{2}\frac{C}{2}}= \frac{4pR}{r}$

Felixx · Mesaj de **Felixx** » 12 Iun 2018, 20:05

a)
Aplicam :
$tg\frac{A}{2}=\sqrt{\frac{\left ( p-b \right )\left ( p-c \right )}{p\left ( p-a \right )}}$
Analog pentru $tg\frac{B}{2}$ si $tg\frac{C}{2}$
Atunci membrul stang devine :
$\frac{p-c}{p}+\frac{p-a}{p}+\frac{p-b}{p}=\frac{3p-\left ( a+b+c \right )}{p}=\frac{3p-2p}{p}=\frac{p}{p}=1$

Felixx · Mesaj de **Felixx** » 12 Iun 2018, 20:45

b) Folosim relatia:
$\sin\frac{A}{2}=\sqrt{\frac{\left ( p-b \right )\left ( p-c \right )}{bc}}$ si analoagele.
Atunci membrul stang devine:
$\frac{abc}{\left ( p-b \right )\left ( p-c \right )}+\frac{abc}{\left ( p-a \right )\left ( p-c \right )}+\frac{abc}{\left ( p-a \right )\left ( p-b \right )}=abc\left ( \frac{1}{\left ( p-b \right )\left ( p-c \right )}+\frac{1}{\left ( p-a \right )\left ( p-c \right )}+\frac{1}{\left ( p-a \right )\left ( p-b \right )} \right )=abc\frac{3p-\left ( a+b+c \right )}{\left ( p-a \right )\left ( p-b \right )\left ( p-c \right )}=abc\frac{3p-2p}{\left ( p-a \right )\left ( p-b \right )\left ( p-c \right )}=abc\frac{p}{\left ( p-a \right )\left ( p-b \right )\left ( p-c \right )}=abc\frac{p^{2}}{p\left ( \left ( p-a \right )\left ( p-b \right )\left ( p-c \right ) \right )}=abc\frac{p^{2}}{S^{2}}=abc\frac{p^{2}}{p^{2}r^{2}}=\frac{abc}{r^{2}}=\frac{4RS}{r^{2}}=\frac{4Rpr}{r^{2}}=\frac{4Rp}{r}$
Am folosit:
$S=pr$ , $S=\frac{abc}{4R}$ , $p=\frac{a+b+c}{2}$ si $S=\sqrt{p\left ( p-a \right )\left ( p-b \right )\left ( p-c \right )}$

Forum AniDeȘcoală.ro

Trigonometrie relatii

Trigonometrie relatii

Re: Trigonometrie relatii

Re: Trigonometrie relatii