Općinsko natjecanje 2006 SŠ3 4
Dodao/la:
arhiva2. travnja 2012. Ako za kutove trokuta
![\alpha](/media/m/f/c/3/fc35d340e96ae7906bf381cae06e4d59.png)
i
![\beta](/media/m/c/e/f/cef1e3bcf491ef3475085d09fd7d291e.png)
vrijedi relacija
![\sin^2 \left(\dfrac{\alpha}{2}\right) \cdot \cos^3 \left(\dfrac{\beta}{2}\right)
= \sin^2 \left(\dfrac{\beta}{2}\right) \cdot \cos^3 \left(\dfrac{\alpha}{2}\right),](/media/m/a/c/1/ac10296d87ff68597aeef78d3b8a4143.png)
dokaži da je
![\alpha=\beta](/media/m/0/7/2/072d2adae664e15adf7b9de77ad54709.png)
.
%V0
Ako za kutove trokuta $\alpha$ i $\beta$ vrijedi relacija $$
\sin^2 \left(\dfrac{\alpha}{2}\right) \cdot \cos^3 \left(\dfrac{\beta}{2}\right)
= \sin^2 \left(\dfrac{\beta}{2}\right) \cdot \cos^3 \left(\dfrac{\alpha}{2}\right),
$$ dokaži da je $\alpha=\beta$.
Izvor: Općinsko natjecanje iz matematike 2006