Točno
19. travnja 2012. 23:01 (12 godine, 3 mjeseci)
Dokažite da za svaki trokut sa stranicama
![a](/media/m/6/d/2/6d2832265560bb67cf117009608524f6.png)
,
![b](/media/m/e/e/c/eec0d7323095a1f2101fc1a74d069df6.png)
,
![c](/media/m/e/a/3/ea344283b6fa26e4a02989dd1fb52a51.png)
i nasuprotnim kutovima
![\alpha](/media/m/f/c/3/fc35d340e96ae7906bf381cae06e4d59.png)
,
![\beta](/media/m/c/e/f/cef1e3bcf491ef3475085d09fd7d291e.png)
,
![\gamma](/media/m/2/4/a/24aca7af13a8211060a900a49ef999e9.png)
vrijedi jednakost
%V0
Dokažite da za svaki trokut sa stranicama $a$, $b$, $c$ i nasuprotnim kutovima $\alpha$, $\beta$, $\gamma$ vrijedi jednakost $$\left( \frac bc + \frac cb \right)\cos \alpha + \left( \frac ca + \frac ac \right)\cos \beta + \left( \frac ab + \frac ba \right)\cos \gamma = 3\text{.}$$
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Primjenom sinusovog poučka dobivamo
![(\dfrac{\sin{\beta}}{\sin{\gamma}} + \dfrac{\sin{\gamma}}{sin{\beta}})\cos{\alpha} + (\dfrac{\sin{\gamma}}{\sin{\alpha}} + \dfrac{\sin{\alpha}}{sin{\gamma}})\cos{\beta} + (\dfrac{\sin{\alpha}}{\sin{\beta}} + \dfrac{\sin{\beta}}{sin{\alpha}})\cos{\gamma} = 3 \newline \Leftrightarrow \sin^2{\beta}\sin{\alpha}cos{\alpha} + \sin^2{\beta}\sin{\gamma}cos{\gamma} + \sin^2{\alpha}\sin{\beta}cos{\beta} + \sin^2{\alpha}\sin{\gamma}cos{\gamma} + \sin^2{\gamma}\sin{\alpha}cos{\alpha} + \sin^2{\gamma}\sin{\beta}cos{\beta} = 3\sin{\alpha}\sin{\beta}\sin{\gamma} \newline \Leftrightarrow \sin{\alpha}\sin{\beta}(\sin{\alpha}\cos{\beta} + \sin{\beta}\cos{\alpha} ) + \sin{\alpha}\sin{\gamma}(\sin{\alpha}\cos{\gamma} + \sin{\gamma}\cos{\alpha} ) + \sin{\gamma}\sin{\beta}(\sin{\gamma}\cos{\beta} + \sin{\beta}\cos{\gamma} ) = 3\sin{\alpha}\sin{\beta}\sin{\gamma} \newline \Leftrightarrow \sin{\alpha}\sin{\beta}\sin{(\alpha + \beta)} + \sin{\gamma}\sin{\beta}\sin{(\gamma + \beta)} + \sin{\alpha}\sin{\gamma}\sin{(\alpha + \gamma)} = 3\sin{\alpha}\sin{\beta}\sin{\gamma}](/media/m/9/1/9/9196b5207a10e30b627d35fd80b905de.png)
![\sin{(\alpha + \beta)}=\sin({\pi - \gamma)} = \sin{\gamma}](/media/m/6/f/5/6f58ec5076771d825da9f0e4b0688905.png)
Uvrštavanjem dobivamo
%V0
Primjenom sinusovog poučka dobivamo
$ (\dfrac{\sin{\beta}}{\sin{\gamma}} + \dfrac{\sin{\gamma}}{sin{\beta}})\cos{\alpha} + (\dfrac{\sin{\gamma}}{\sin{\alpha}} + \dfrac{\sin{\alpha}}{sin{\gamma}})\cos{\beta} + (\dfrac{\sin{\alpha}}{\sin{\beta}} + \dfrac{\sin{\beta}}{sin{\alpha}})\cos{\gamma} = 3 \newline \Leftrightarrow \sin^2{\beta}\sin{\alpha}cos{\alpha} + \sin^2{\beta}\sin{\gamma}cos{\gamma} + \sin^2{\alpha}\sin{\beta}cos{\beta} + \sin^2{\alpha}\sin{\gamma}cos{\gamma} + \sin^2{\gamma}\sin{\alpha}cos{\alpha} + \sin^2{\gamma}\sin{\beta}cos{\beta} = 3\sin{\alpha}\sin{\beta}\sin{\gamma} \newline \Leftrightarrow \sin{\alpha}\sin{\beta}(\sin{\alpha}\cos{\beta} + \sin{\beta}\cos{\alpha} ) + \sin{\alpha}\sin{\gamma}(\sin{\alpha}\cos{\gamma} + \sin{\gamma}\cos{\alpha} ) + \sin{\gamma}\sin{\beta}(\sin{\gamma}\cos{\beta} + \sin{\beta}\cos{\gamma} ) = 3\sin{\alpha}\sin{\beta}\sin{\gamma} \newline \Leftrightarrow \sin{\alpha}\sin{\beta}\sin{(\alpha + \beta)} + \sin{\gamma}\sin{\beta}\sin{(\gamma + \beta)} + \sin{\alpha}\sin{\gamma}\sin{(\alpha + \gamma)} = 3\sin{\alpha}\sin{\beta}\sin{\gamma} $
$ \sin{(\alpha + \beta)}=\sin({\pi - \gamma)} = \sin{\gamma} $
Uvrštavanjem dobivamo
$ 3\sin{\alpha}\sin{\beta}\sin{\gamma} = 3\sin{\alpha}\sin{\beta}\sin{\gamma} $
20. travnja 2012. 17:22 | mljulj | Točno |