Question

Demonstrati ca in orice triunghi ABC are loc egalitatea:
$\frac{cos A}{sin B sin C}$ + $\frac{cos B}{sin A sin C}$ + $\frac{cos C}{sin A sin B}$

Answer 1

Aducem la acelasi numitor comun egalitatea si obtinem:

$\frac{sinAcosA+sinBcosB+sinCcosC}{sinAsinBsinC}$

Amplificam cu 2 toata egalitatea si tinem cont ca 2sinAcosA=sin2A (formula unghiului dublu)

$\frac{2sinAcosA+2sinBcosB+2sinCcosC}{2sinAsinBsinC} =\frac{sin2A+sin2B+sin2C}{2sinAsinBsinC}$

$sin2A+sin2B=2sin(\frac{2A+2B}{2})cos(\frac{2A-2B}{2})=2sin(A+B)cos(A-B)$

$sin2A+sin2B+sin2C=2sin(A+B)cos(A-B) +sin2C$

Stim ca A+B+C=π

A+B=π-C

$2sin(A+B)cos(A-B) +sin2C=2sinCcos(A-B)+2sinCcosC=2sinC(cos(A-B)+cos(\pi-(A+B)))$

$=2sinC(cos(A-B)+cos(A+B)=2sinC\cdot 2sinAsinB=4sinCsinAsinB$

Inlocuim la numarator si obtinem:

$\frac{4sinAsinBsinC}{2sinAsinBsinC} =2$

Un alt exercitiu gasesti aici: https://brainly.ro/tema/1031524

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