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Forum » Cutia cu nisip » determinant
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Autor Mesaj
Andreea10
Grup: membru
Mesaje: 103
10 Nov 2019, 19:23

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determinant    [Editează]  [Citează] 

Delta =\begin{vmatrix}
1 & \frac{1+cos2x}{2} & ctgx\\
1 & \frac{1+cos2y}{2} &ctgy \\
1 & \frac{1+cos2z}{2} & ctgz
\end{vmatrix}=\frac{1}{2}\begin{vmatrix}
1 & \frac{1+cos2x}{1} & ctgx\\
1 & \frac{1+cos2y}{1} &ctgy \\
1 & \frac{1+cos2z}{1} & ctgz
\end{vmatrix}=\frac{1}{2}\begin{vmatrix}
1 & cos2x & ctgx\\
1 & cos2y &ctgy \\
1 & cos2z & ctgz
\end{vmatrix}=\frac{1}{2}\begin{vmatrix}
1 & cos2x & ctgx\\
0 & cos2y-cos2x &ctgy-ctgx \\
0 & cos2z-cos2x & ctgz-ctgx
\end{vmatrix}=\frac{1}{2}\begin{vmatrix}
2sin\frac{2y+2x}{2}sin\frac{2x-2y}{2} &\frac{sin\left ( x-y \right )}{sinysinx} \\
2sin\frac{2z+2x}{2}sin\frac{2z-2z}{2} &\frac{sin\left ( x-z \right )}{sinzsinx}
\end{vmatrix}=\begin{vmatrix}
sin\left ( x+y \right )sin\left ( x-y \right ) &\frac{sin\left ( x-y \right )}{sinysinx} \\
sin\left ( z+x \right )sin\left ( x-z \right )&\frac{sin\left ( x-z \right )}{sinzsinx}
\end{vmatrix}= sin\left ( x-y \right )sin\left ( x-z \right )\begin{vmatrix}
sin\left ( x+y \right )&\frac{1}{sinysinx} \\
sin\left ( z+x \right )&\frac{1}{sinzsinx}
\end{vmatrix}=\frac{sin\left ( x-y \right )sin\left ( x-z \right )}{sinx}\begin{vmatrix}
sin\left ( x+y \right ) &\frac{1}{siny} \\
sin\left ( z+x \right )& \frac{1}{sinz}
\end{vmatrix}=\frac{sin\left ( x-y \right )sin\left ( x-z \right )}{sinxsinysinz}\left [ sin\left ( x+y \right )siny-sin\left ( x+z \right )sinz \right ]
unde:
sin\left ( x+y \right )siny=\frac{1}{2}\left [ cos\left ( x+y-y \right )-cos\left ( x+y+y \right ) \right ]=\frac{1}{2}\left [ cosx-cos\left ( x+2y \right ) \right ]
si
sin\left ( x+z \right )sinz=\frac{1}{2}\left [ cos\left ( x+z-z \right )-cos\left ( x+z+z \right ) \right ]=\frac{1}{2}\left [ cosx-cos\left ( x+2z \right ) \right ]
Atunci:
\Delta =\frac{1}{2}\frac{sin\left ( x-y \right )sin\left ( x-z \right )}{sinxsinysinz}\left [ cosx-cos\left ( x+y \right )-cosx+cos\left ( x+2z \right ) \right ]=\frac{1}{2}\frac{sin\left ( x-y \right )sin\left ( x-z \right )}{sinxsinysinz}\left [ cos\left ( x+2z \right )-cos\left ( x+2y \right ) \right ]=\frac{1}{2}\frac{sin\left ( x-y \right )sin\left ( x-z \right )}{sinxsinysinz}2sin\frac{x+2z+x+2y}{2}sin\frac{x+2y-x-2z}{2}=\frac{sin\left ( x-y \right )sin\left ( x-z \right )sin\left ( y-z \right )sin\left ( x+y+z \right )}{sinxsinysinz}


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Andre


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