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Fie ABC un triunghi. Calculați:
sin A sin (B-C)+ sin B sin (C-A)+ sin C sin (A-B)​


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Răspuns:

Explicație pas cu pas:

Vezi imaginea BOIUSTEF

[tex]\it sinAsin(B-C)=sinAsinBcosC-sinAsinCcosB\ \ \ \ \ \ (1)\\ \\ sinBsin(C-A)=sinBsinCcosA-sinBsinAcosC\ \ \ \ \ \ \ (2)\\ \\ sinCsin(A-B)=sinCsinAcosB-sinCsinBcosA\ \ \ \ \ \ \ \ (3)\\ \\ \\ (1),\ (2),\ (3) \Rightarrow sinAsin(B-C)+ sinBsin(C-A)+ sinCsin(A-B)=\\ \\ = cosA(sinBsinC-sinBsinC)+cosB(sinAsinC-sinAsinC)+\\ \\ +cosC(sinAsinB-sinAsinB)=0[/tex]