∀ tam giác ABC: S= √((vecto AB ² x vecto AC ²)-(vectoAB x vectoAC) ²)

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1. Giải thích các bước giải:

   Ta có:

   $A=\sqrt{(\vec{AB})^2\cdot (\vec{AC})^2-(\vec{AB}\cdot \vec{AC})^2}$

   $\to A=\sqrt{AB^2\cdot AC^2-(AB\cdot AC\cdot \cos\widehat{BAC})^2}$

   $\to A=\sqrt{AB^2\cdot AC^2-AB^2\cdot AC^2\cdot \cos^2\widehat{BAC}}$

   $\to A=\sqrt{AB^2\cdot AC^2(1- \cos^2\widehat{BAC})}$

   $\to A=\sqrt{AB^2\cdot AC^2\cdot \sin^2\widehat{BAC}}$

   $\to A=AB\cdot AC\cdot \sin\widehat{BAC}$

   Mà $S_{ABC}=\dfrac12AB\cdot AC\cdot \sin\widehat{BAC}$

   $\to S=\dfrac12A$

   $\to S=\dfrac12\sqrt{(\vec{AB})^2\cdot (\vec{AC})^2-(\vec{AB}\cdot \vec{AC})^2}$

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2. Ta có: $sin^2A+cos^2A=1$

   <=> $sinA=\sqrt{1-cos^2A}$

   <=> $\frac{1}{2}AB.AC.sinA=\frac{1}{2}AB.AC.\sqrt{1-cos^2A}$

   <=> $S=\frac{1}{2}.\sqrt{AB^2.AC^2.(1-cos^2A)}$

   <=> $S=\frac{1}{2}.\sqrt{AB^2.AC^2.-AB^2.AC^2.cos^2A)}$

   <=> $S=\frac{1}{2}.\sqrt{(\vec{AB})^2.(\vec{AC})^2-(\vec{AB}.\vec{AB})^2}$ $[DPCM]$

    

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